Mixing
necessary and sufficient condition for trivial kernel of a matrix over a commutative ring
$begingroup$
In answering Do these matrix rings have non-zero elements that are neither units nor zero divisors? I was surprised how hard it was to find anything on the Web about the generalization of the following fact to commutative rings:
A square matrix over a field has trivial kernel if and only if its determinant is non-zero.
As Bill demonstrated in the above question, a related fact about fields generalizes directly to commutative rings:
A square matrix over a commutative ring is invertible if and only if its
determinant is invertible.
However, the kernel being trivial and the matrix being invertible are not equivalent for general rings, so the question arises what the proper generalization of the first fact is. Since it took me quite a lot of searching to find the answer to this rather basic question, and it's excplicitly encouraged to write a question and answer it to document something that might be useful to others, I thought I'd write this up here in an accessible form.
So my questions are: What is the relationship between the determinant of a square matrix over a commutative ring and the triviality of its kernel? Can the simple relationship that holds for fields be generalized? And (generalizing with a view to the answer) what is a necessary and sufficient condition for a (not necessarily square) matrix over a commutative ring to have trivial kernel?
linear-algebra abstract-algebra matrices commutative-algebra
edited Aug 11 '18 at 0:04
Pierre-Yves Gaillard
13.4k23184
asked Oct 11 '11 at 16:07
jorikijoriki
171k10188349
$endgroup$
add a comment |
$begingroup$
In answering Do these matrix rings have non-zero elements that are neither units nor zero divisors? I was surprised how hard it was to find anything on the Web about the generalization of the following fact to commutative rings:
A square matrix over a field has trivial kernel if and only if its determinant is non-zero.
As Bill demonstrated in the above question, a related fact about fields generalizes directly to commutative rings:
A square matrix over a commutative ring is invertible if and only if its
determinant is invertible.
However, the kernel being trivial and the matrix being invertible are not equivalent for general rings, so the question arises what the proper generalization of the first fact is. Since it took me quite a lot of searching to find the answer to this rather basic question, and it's excplicitly encouraged to write a question and answer it to document something that might be useful to others, I thought I'd write this up here in an accessible form.
So my questions are: What is the relationship between the determinant of a square matrix over a commutative ring and the triviality of its kernel? Can the simple relationship that holds for fields be generalized? And (generalizing with a view to the answer) what is a necessary and sufficient condition for a (not necessarily square) matrix over a commutative ring to have trivial kernel?
linear-algebra abstract-algebra matrices commutative-algebra
edited Aug 11 '18 at 0:04
Pierre-Yves Gaillard
13.4k23184
asked Oct 11 '11 at 16:07
jorikijoriki
171k10188349
$endgroup$
add a comment |
$begingroup$
In answering Do these matrix rings have non-zero elements that are neither units nor zero divisors? I was surprised how hard it was to find anything on the Web about the generalization of the following fact to commutative rings:
A square matrix over a field has trivial kernel if and only if its determinant is non-zero.
As Bill demonstrated in the above question, a related fact about fields generalizes directly to commutative rings:
A square matrix over a commutative ring is invertible if and only if its
determinant is invertible.
However, the kernel being trivial and the matrix being invertible are not equivalent for general rings, so the question arises what the proper generalization of the first fact is. Since it took me quite a lot of searching to find the answer to this rather basic question, and it's excplicitly encouraged to write a question and answer it to document something that might be useful to others, I thought I'd write this up here in an accessible form.
So my questions are: What is the relationship between the determinant of a square matrix over a commutative ring and the triviality of its kernel? Can the simple relationship that holds for fields be generalized? And (generalizing with a view to the answer) what is a necessary and sufficient condition for a (not necessarily square) matrix over a commutative ring to have trivial kernel?
linear-algebra abstract-algebra matrices commutative-algebra
edited Aug 11 '18 at 0:04
Pierre-Yves Gaillard
13.4k23184
asked Oct 11 '11 at 16:07
jorikijoriki
171k10188349
$endgroup$
In answering Do these matrix rings have non-zero elements that are neither units nor zero divisors? I was surprised how hard it was to find anything on the Web about the generalization of the following fact to commutative rings:
A square matrix over a field has trivial kernel if and only if its determinant is non-zero.
As Bill demonstrated in the above question, a related fact about fields generalizes directly to commutative rings:
A square matrix over a commutative ring is invertible if and only if its
determinant is invertible.
However, the kernel being trivial and the matrix being invertible are not equivalent for general rings, so the question arises what the proper generalization of the first fact is. Since it took me quite a lot of searching to find the answer to this rather basic question, and it's excplicitly encouraged to write a question and answer it to document something that might be useful to others, I thought I'd write this up here in an accessible form.
So my questions are: What is the relationship between the determinant of a square matrix over a commutative ring and the triviality of its kernel? Can the simple relationship that holds for fields be generalized? And (generalizing with a view to the answer) what is a necessary and sufficient condition for a (not necessarily square) matrix over a commutative ring to have trivial kernel?
linear-algebra abstract-algebra matrices commutative-algebra
linear-algebra abstract-algebra matrices commutative-algebra
edited Aug 11 '18 at 0:04
Pierre-Yves Gaillard
13.4k23184
asked Oct 11 '11 at 16:07
jorikijoriki
171k10188349
edited Aug 11 '18 at 0:04
Pierre-Yves Gaillard
13.4k23184
asked Oct 11 '11 at 16:07
jorikijoriki
171k10188349
edited Aug 11 '18 at 0:04
Pierre-Yves Gaillard
13.4k23184
edited Aug 11 '18 at 0:04
Pierre-Yves Gaillard
13.4k23184
edited Aug 11 '18 at 0:04
Pierre-Yves Gaillard
13.4k23184
13.4k23184
asked Oct 11 '11 at 16:07
jorikijoriki
171k10188349
asked Oct 11 '11 at 16:07
jorikijoriki
171k10188349
asked Oct 11 '11 at 16:07
jorikijoriki
171k10188349
171k10188349
add a comment |
add a comment |
2 Answers
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I found the answer in this book (in Section $6.4.14$, “Determinants, Ranks and Linear Equations”). I'd tried using a similar Laplace expansion myself but was missing the idea of using the largest dimension at which the minors are not all annihilated by the same non-zero element. I'll try to summarize the argument in somewhat less formal terms, omitting the tangential material included in the book.
Let $A$ be an $mtimes n$ matrix over a commutative ring $R$. We want to find a condition for the system of equations $Ax=0$ with $xin R^n$ to have a non-trivial solution. If $R$ is a field, various definitions of the rank of $A$ coincide, including the column rank (the dimension of the column space), the row rank (the dimension of the row space) and the determinantal rank (the dimension of the lowest non-zero minor). This is not the case for a general commutative ring. It turns out that for our present purposes a useful generalization of rank is the largest integer $k$ such that there is no non-zero element of $R$ that annihilates all minors of dimension $k$, with $k=0$ if there is no such integer.
We want to show that $Ax=0$ has a non-trivial solution if and only if $klt n$.
If $k=0$, there is a non-zero element $rin R$ which annihilates all matrix elements (the minors of dimension $1$), so there is a non-trivial solution
$$Apmatrix{r\vdots\r}=0;.$$
Now assume $0lt klt n$. If $mlt n$, we can add rows of zeros to $A$ without changing $k$ or the solution set, so we can assume $klt nle m$. There is some non-zero element $rin R$ that annihilates all minors of dimension $k+1$, and there is a minor of dimension $k$ that isn't annihilated by $r$. Without loss of generality, assume that this is the minor of the first $k$ rows and columns. Now consider the matrix formed of the first $k+1$ rows and columns of $A$, and form a solution $x$ from the $(k+1)$-th column of its adjugate by multiplying it by $r$ and padding it with zeros. By construction, the first $k$ entries of $Ax$ are determinants of a matrix with two equal rows, and thus vanish; the remaining entries are each $r$ times a minor of dimension $k+1$, and thus also vanish. But the $(k+1)$-th entry of this solution is non-zero, being $r$ times the minor of the first $k$ rows and columns, which isn't annihilated by $r$. Thus we have constructed a non-trivial solution.
In summary, if $klt n$, there is a non-trivial solution to $Ax=0$.
Now assume conversely that there is such a solution $x$. If $ngt m$, there are no minors of dimension $n$, so $klt n$. Thus we can assume $nle m$. The minors of dimension $n$ are the determinants of matrices $B$ formed by choosing any $n$ rows of $A$. Since each row of $A$ times $x$ is $0$, we have $Bx=0$, and then multiplying by the adjugate of $B$ yields $det B x=0$. Since there is at least one non-zero entry in the non-trivial solution $x$, there is at least one non-zero element of $R$ that annihilates all minors of size $n$, and thus $klt n$.
Specializing to the case $m=n$ of square matrices, we can conclude:
A system of linear equations $Ax=0$ with a square $ntimes n$ matrix
$A$ over a commutative ring $R$ has a non-trivial solution if and only
if its determinant (its only minor of dimension $n$) is annihilated by
some non-zero element of $R$, that is, if its determinant is a zero divisor or zero.
answered Oct 11 '11 at 16:08
jorikijoriki
171k10188349
$endgroup$
3
$begingroup$
The book joriki links to is A Second Semester of Linear Algebra by S. E. Payne. Here are two other links to the same book, via the author's site: PDF file. HTML page: Class Notes.
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– Pierre-Yves Gaillard
Oct 11 '11 at 17:45
add a comment |
$begingroup$
See Section III.8.7, entitled Application to Linear Equations, of Algebra, by Nicolas Bourbaki.
EDIT 1. Let $R$ be a commutative ring, let $m$ and $n$ be positive integers, let $M$ be an $R$-module, and let $A:R^nto M$ be $R$-linear.
Identify the $n$ th exterior power $Lambda^n(R^n)$ of $R^n$ to $R$ in the obvious way, so that $Lambda^n(A)$ is a map from $R$ to $Lambda^n(M)$.
Put $v_i:=Ae_i$, where $e_i$ is the $i$ th vector of the canonical basis of $R^n$. In particular we have
$$
Ax=sum_{i=1}^n x_i v_i,quadLambda^n(A) r=r v_1wedgecdotswedge v_n.
$$
(where $x_i$ is the $i$-th coordinate of $x$, and $r$ denotes any element of $Lambda^nleft(R^nright) cong R$).
If $Lambda^n(A)$ is injective, so is $A$.
In other words:
If the $v_i$ are linearly dependent, then $r v_1wedgecdotswedge v_n=0$ for some nonzero $r$ in $R$.
Indeed, for $x$ in $ker A$ we have
$$
Lambda^n(A) x_1=x_1 v_1wedge v_2wedgecdotswedge v_n=
-sum_{i=2}^n x_i v_iwedge v_2wedgecdotswedge v_n=0,
$$
and, similarly, $Lambda^n(A) x_i=0$ for all $i$.
[Edit: Old version (before Georges's comment): Assume now that $M$ embeds into $R^m$.]
Assume now that there is an $R$-linear injection $B:Mto R^m$ such that
$$
Lambda^n(B):Lambda^n(M)toLambda^n(R^m)
$$
is injective. This is always the case (for a suitable $m$) if $M$ is projective and finitely generated.
If $A$ is injective, so is $Lambda^n(A)$.
In other words:
If $r v_1wedgecdotswedge v_n=0$ for some nonzero $r$ in $R$, then the $v_i$ are linearly dependent.
The proof is given in joriki's nice answer.
This is also proved as Proposition 12 in Bourbaki's Algebra III.7.9 p. 519. Unfortunately, I don't understand Bourbaki's argument. I'd be most grateful to whoever would be kind and patient enough to explain it to me.
EDIT 2. According to the indications given by Tsit-Yuen Lam on page 150 of his book Exercises in modules and rings, the theorem is due to N. H. McCoy, and appeared first, as Theorem 1 page 288, in
- N. H. McCoy, Remarks on Divisors of Zero, The American Mathematical Monthly Vol. 49, No. 5 (May, 1942), pp. 286-295, JSTOR.
Lam also says that
- N. H. McCoy, Rings and ideals, The Carus Mathematical Monographs, no. 8, The Mathematical Association of America, 1948,
is an "excellent exposition" of the subject. See Theorem 51 page 159.
McCoy's Theorem is also stated and proved in the following texts:
Ex. 5.23.A(3) on page 149 of Lam's Exercises in modules and rings.
Theorem 2.2 page 3 in Anton Gerashenko's notes from Lam's Course: Math 274, Commutative Rings, Fall 2006: PDF file.
Theorem 1.6 in Chapter 13, entitled "Various topics", of The CRing Project. --- PDF file for Chapter 13. --- PDF file for the whole book.
Blocki, Zbigniew, An elementary proof of the McCoy theorem, J. Univ. Iagel. Acta Math.; N 30; 1993; 215-218.
Theorem 6.4.16. page 101, A Second Semester of Linear Algebra, Math 5718, by Stan Payne. PDF file.
edited Jan 20 at 12:34
darij grinberg
11.1k33167
answered Oct 11 '11 at 16:57
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
$endgroup$
$begingroup$
Thanks for this link! Note, however, that the result is proved only for square matrices there. This case was the original motivation for the question, but the question and my answer apply to the general case.
$endgroup$
– joriki
Nov 10 '11 at 9:13
$begingroup$
Dear @Pierre-Yves, I'm not quite sure that your claim (displayed in grey) "If $A$ is injective, so is $Lambda^n A$" holds if you assume only that $M$ embeds into $R^m$.There is the subtle point that exterior products don't mean the same in both spaces, in other words $Lambda ^n Mto Lambda ^n R^m$ needn't be injective. Everything is fine if $M$ is projective, though, and this is Bourbaki's assumption . But this is nitpicking: +1, needless to say.
$endgroup$
– Georges Elencwajg
Nov 12 '11 at 14:17
$begingroup$
Dear @Georges: No, this is definitely not nitpicking! Thanks a lot! I hope it's correct now. - It was my secret hope that you would read this answer. Did you see the last paragraph? I'm sure you understand Bourbaki's argument...
$endgroup$
– Pierre-Yves Gaillard
Nov 12 '11 at 15:11
$begingroup$
Dear @Pierre-Yves: No, I don't understand Bourbaki's argument either. Specifically, when he writes "it follows from no.8, Corollary 3 to Theorem 1 that $mu x_1$ is a linear combination....", I don't see how it follows. (By the way, this is Proposition 12 in my edition)
$endgroup$
– Georges Elencwajg
Nov 12 '11 at 18:19
$begingroup$
Dear @Georges: Once more you're right: it's Proposition 12 (in the link to the English edition I give, and in my French edition - it was just a typo). Thank you very much for your time and effort. I'm having the same problem as you with Bourbaki's proof. It's weird. I find joriki's formulation of the argument very nice.
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– Pierre-Yves Gaillard
Nov 12 '11 at 18:42
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2 Answers
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$begingroup$
I found the answer in this book (in Section $6.4.14$, “Determinants, Ranks and Linear Equations”). I'd tried using a similar Laplace expansion myself but was missing the idea of using the largest dimension at which the minors are not all annihilated by the same non-zero element. I'll try to summarize the argument in somewhat less formal terms, omitting the tangential material included in the book.
Let $A$ be an $mtimes n$ matrix over a commutative ring $R$. We want to find a condition for the system of equations $Ax=0$ with $xin R^n$ to have a non-trivial solution. If $R$ is a field, various definitions of the rank of $A$ coincide, including the column rank (the dimension of the column space), the row rank (the dimension of the row space) and the determinantal rank (the dimension of the lowest non-zero minor). This is not the case for a general commutative ring. It turns out that for our present purposes a useful generalization of rank is the largest integer $k$ such that there is no non-zero element of $R$ that annihilates all minors of dimension $k$, with $k=0$ if there is no such integer.
We want to show that $Ax=0$ has a non-trivial solution if and only if $klt n$.
If $k=0$, there is a non-zero element $rin R$ which annihilates all matrix elements (the minors of dimension $1$), so there is a non-trivial solution
$$Apmatrix{r\vdots\r}=0;.$$
Now assume $0lt klt n$. If $mlt n$, we can add rows of zeros to $A$ without changing $k$ or the solution set, so we can assume $klt nle m$. There is some non-zero element $rin R$ that annihilates all minors of dimension $k+1$, and there is a minor of dimension $k$ that isn't annihilated by $r$. Without loss of generality, assume that this is the minor of the first $k$ rows and columns. Now consider the matrix formed of the first $k+1$ rows and columns of $A$, and form a solution $x$ from the $(k+1)$-th column of its adjugate by multiplying it by $r$ and padding it with zeros. By construction, the first $k$ entries of $Ax$ are determinants of a matrix with two equal rows, and thus vanish; the remaining entries are each $r$ times a minor of dimension $k+1$, and thus also vanish. But the $(k+1)$-th entry of this solution is non-zero, being $r$ times the minor of the first $k$ rows and columns, which isn't annihilated by $r$. Thus we have constructed a non-trivial solution.
In summary, if $klt n$, there is a non-trivial solution to $Ax=0$.
Now assume conversely that there is such a solution $x$. If $ngt m$, there are no minors of dimension $n$, so $klt n$. Thus we can assume $nle m$. The minors of dimension $n$ are the determinants of matrices $B$ formed by choosing any $n$ rows of $A$. Since each row of $A$ times $x$ is $0$, we have $Bx=0$, and then multiplying by the adjugate of $B$ yields $det B x=0$. Since there is at least one non-zero entry in the non-trivial solution $x$, there is at least one non-zero element of $R$ that annihilates all minors of size $n$, and thus $klt n$.
Specializing to the case $m=n$ of square matrices, we can conclude:
A system of linear equations $Ax=0$ with a square $ntimes n$ matrix
$A$ over a commutative ring $R$ has a non-trivial solution if and only
if its determinant (its only minor of dimension $n$) is annihilated by
some non-zero element of $R$, that is, if its determinant is a zero divisor or zero.
answered Oct 11 '11 at 16:08
jorikijoriki
171k10188349
$endgroup$
3
$begingroup$
The book joriki links to is A Second Semester of Linear Algebra by S. E. Payne. Here are two other links to the same book, via the author's site: PDF file. HTML page: Class Notes.
$endgroup$
– Pierre-Yves Gaillard
Oct 11 '11 at 17:45
add a comment |
$begingroup$
I found the answer in this book (in Section $6.4.14$, “Determinants, Ranks and Linear Equations”). I'd tried using a similar Laplace expansion myself but was missing the idea of using the largest dimension at which the minors are not all annihilated by the same non-zero element. I'll try to summarize the argument in somewhat less formal terms, omitting the tangential material included in the book.
Let $A$ be an $mtimes n$ matrix over a commutative ring $R$. We want to find a condition for the system of equations $Ax=0$ with $xin R^n$ to have a non-trivial solution. If $R$ is a field, various definitions of the rank of $A$ coincide, including the column rank (the dimension of the column space), the row rank (the dimension of the row space) and the determinantal rank (the dimension of the lowest non-zero minor). This is not the case for a general commutative ring. It turns out that for our present purposes a useful generalization of rank is the largest integer $k$ such that there is no non-zero element of $R$ that annihilates all minors of dimension $k$, with $k=0$ if there is no such integer.
We want to show that $Ax=0$ has a non-trivial solution if and only if $klt n$.
If $k=0$, there is a non-zero element $rin R$ which annihilates all matrix elements (the minors of dimension $1$), so there is a non-trivial solution
$$Apmatrix{r\vdots\r}=0;.$$
Now assume $0lt klt n$. If $mlt n$, we can add rows of zeros to $A$ without changing $k$ or the solution set, so we can assume $klt nle m$. There is some non-zero element $rin R$ that annihilates all minors of dimension $k+1$, and there is a minor of dimension $k$ that isn't annihilated by $r$. Without loss of generality, assume that this is the minor of the first $k$ rows and columns. Now consider the matrix formed of the first $k+1$ rows and columns of $A$, and form a solution $x$ from the $(k+1)$-th column of its adjugate by multiplying it by $r$ and padding it with zeros. By construction, the first $k$ entries of $Ax$ are determinants of a matrix with two equal rows, and thus vanish; the remaining entries are each $r$ times a minor of dimension $k+1$, and thus also vanish. But the $(k+1)$-th entry of this solution is non-zero, being $r$ times the minor of the first $k$ rows and columns, which isn't annihilated by $r$. Thus we have constructed a non-trivial solution.
In summary, if $klt n$, there is a non-trivial solution to $Ax=0$.
Now assume conversely that there is such a solution $x$. If $ngt m$, there are no minors of dimension $n$, so $klt n$. Thus we can assume $nle m$. The minors of dimension $n$ are the determinants of matrices $B$ formed by choosing any $n$ rows of $A$. Since each row of $A$ times $x$ is $0$, we have $Bx=0$, and then multiplying by the adjugate of $B$ yields $det B x=0$. Since there is at least one non-zero entry in the non-trivial solution $x$, there is at least one non-zero element of $R$ that annihilates all minors of size $n$, and thus $klt n$.
Specializing to the case $m=n$ of square matrices, we can conclude:
A system of linear equations $Ax=0$ with a square $ntimes n$ matrix
$A$ over a commutative ring $R$ has a non-trivial solution if and only
if its determinant (its only minor of dimension $n$) is annihilated by
some non-zero element of $R$, that is, if its determinant is a zero divisor or zero.
answered Oct 11 '11 at 16:08
jorikijoriki
171k10188349
$endgroup$
3
$begingroup$
The book joriki links to is A Second Semester of Linear Algebra by S. E. Payne. Here are two other links to the same book, via the author's site: PDF file. HTML page: Class Notes.
$endgroup$
– Pierre-Yves Gaillard
Oct 11 '11 at 17:45
add a comment |
$begingroup$
I found the answer in this book (in Section $6.4.14$, “Determinants, Ranks and Linear Equations”). I'd tried using a similar Laplace expansion myself but was missing the idea of using the largest dimension at which the minors are not all annihilated by the same non-zero element. I'll try to summarize the argument in somewhat less formal terms, omitting the tangential material included in the book.
Let $A$ be an $mtimes n$ matrix over a commutative ring $R$. We want to find a condition for the system of equations $Ax=0$ with $xin R^n$ to have a non-trivial solution. If $R$ is a field, various definitions of the rank of $A$ coincide, including the column rank (the dimension of the column space), the row rank (the dimension of the row space) and the determinantal rank (the dimension of the lowest non-zero minor). This is not the case for a general commutative ring. It turns out that for our present purposes a useful generalization of rank is the largest integer $k$ such that there is no non-zero element of $R$ that annihilates all minors of dimension $k$, with $k=0$ if there is no such integer.
We want to show that $Ax=0$ has a non-trivial solution if and only if $klt n$.
If $k=0$, there is a non-zero element $rin R$ which annihilates all matrix elements (the minors of dimension $1$), so there is a non-trivial solution
$$Apmatrix{r\vdots\r}=0;.$$
Now assume $0lt klt n$. If $mlt n$, we can add rows of zeros to $A$ without changing $k$ or the solution set, so we can assume $klt nle m$. There is some non-zero element $rin R$ that annihilates all minors of dimension $k+1$, and there is a minor of dimension $k$ that isn't annihilated by $r$. Without loss of generality, assume that this is the minor of the first $k$ rows and columns. Now consider the matrix formed of the first $k+1$ rows and columns of $A$, and form a solution $x$ from the $(k+1)$-th column of its adjugate by multiplying it by $r$ and padding it with zeros. By construction, the first $k$ entries of $Ax$ are determinants of a matrix with two equal rows, and thus vanish; the remaining entries are each $r$ times a minor of dimension $k+1$, and thus also vanish. But the $(k+1)$-th entry of this solution is non-zero, being $r$ times the minor of the first $k$ rows and columns, which isn't annihilated by $r$. Thus we have constructed a non-trivial solution.
In summary, if $klt n$, there is a non-trivial solution to $Ax=0$.
Now assume conversely that there is such a solution $x$. If $ngt m$, there are no minors of dimension $n$, so $klt n$. Thus we can assume $nle m$. The minors of dimension $n$ are the determinants of matrices $B$ formed by choosing any $n$ rows of $A$. Since each row of $A$ times $x$ is $0$, we have $Bx=0$, and then multiplying by the adjugate of $B$ yields $det B x=0$. Since there is at least one non-zero entry in the non-trivial solution $x$, there is at least one non-zero element of $R$ that annihilates all minors of size $n$, and thus $klt n$.
Specializing to the case $m=n$ of square matrices, we can conclude:
A system of linear equations $Ax=0$ with a square $ntimes n$ matrix
$A$ over a commutative ring $R$ has a non-trivial solution if and only
if its determinant (its only minor of dimension $n$) is annihilated by
some non-zero element of $R$, that is, if its determinant is a zero divisor or zero.
answered Oct 11 '11 at 16:08
jorikijoriki
171k10188349
$endgroup$
I found the answer in this book (in Section $6.4.14$, “Determinants, Ranks and Linear Equations”). I'd tried using a similar Laplace expansion myself but was missing the idea of using the largest dimension at which the minors are not all annihilated by the same non-zero element. I'll try to summarize the argument in somewhat less formal terms, omitting the tangential material included in the book.
Let $A$ be an $mtimes n$ matrix over a commutative ring $R$. We want to find a condition for the system of equations $Ax=0$ with $xin R^n$ to have a non-trivial solution. If $R$ is a field, various definitions of the rank of $A$ coincide, including the column rank (the dimension of the column space), the row rank (the dimension of the row space) and the determinantal rank (the dimension of the lowest non-zero minor). This is not the case for a general commutative ring. It turns out that for our present purposes a useful generalization of rank is the largest integer $k$ such that there is no non-zero element of $R$ that annihilates all minors of dimension $k$, with $k=0$ if there is no such integer.
We want to show that $Ax=0$ has a non-trivial solution if and only if $klt n$.
If $k=0$, there is a non-zero element $rin R$ which annihilates all matrix elements (the minors of dimension $1$), so there is a non-trivial solution
$$Apmatrix{r\vdots\r}=0;.$$
Now assume $0lt klt n$. If $mlt n$, we can add rows of zeros to $A$ without changing $k$ or the solution set, so we can assume $klt nle m$. There is some non-zero element $rin R$ that annihilates all minors of dimension $k+1$, and there is a minor of dimension $k$ that isn't annihilated by $r$. Without loss of generality, assume that this is the minor of the first $k$ rows and columns. Now consider the matrix formed of the first $k+1$ rows and columns of $A$, and form a solution $x$ from the $(k+1)$-th column of its adjugate by multiplying it by $r$ and padding it with zeros. By construction, the first $k$ entries of $Ax$ are determinants of a matrix with two equal rows, and thus vanish; the remaining entries are each $r$ times a minor of dimension $k+1$, and thus also vanish. But the $(k+1)$-th entry of this solution is non-zero, being $r$ times the minor of the first $k$ rows and columns, which isn't annihilated by $r$. Thus we have constructed a non-trivial solution.
In summary, if $klt n$, there is a non-trivial solution to $Ax=0$.
Now assume conversely that there is such a solution $x$. If $ngt m$, there are no minors of dimension $n$, so $klt n$. Thus we can assume $nle m$. The minors of dimension $n$ are the determinants of matrices $B$ formed by choosing any $n$ rows of $A$. Since each row of $A$ times $x$ is $0$, we have $Bx=0$, and then multiplying by the adjugate of $B$ yields $det B x=0$. Since there is at least one non-zero entry in the non-trivial solution $x$, there is at least one non-zero element of $R$ that annihilates all minors of size $n$, and thus $klt n$.
Specializing to the case $m=n$ of square matrices, we can conclude:
A system of linear equations $Ax=0$ with a square $ntimes n$ matrix
$A$ over a commutative ring $R$ has a non-trivial solution if and only
if its determinant (its only minor of dimension $n$) is annihilated by
some non-zero element of $R$, that is, if its determinant is a zero divisor or zero.
answered Oct 11 '11 at 16:08
jorikijoriki
171k10188349
edited Aug 11 '18 at 15:57
answered Oct 11 '11 at 16:08
jorikijoriki
171k10188349
answered Oct 11 '11 at 16:08
jorikijoriki
171k10188349
answered Oct 11 '11 at 16:08
jorikijoriki
171k10188349
171k10188349
3
$begingroup$
The book joriki links to is A Second Semester of Linear Algebra by S. E. Payne. Here are two other links to the same book, via the author's site: PDF file. HTML page: Class Notes.
$endgroup$
– Pierre-Yves Gaillard
Oct 11 '11 at 17:45
add a comment |
3
$begingroup$
The book joriki links to is A Second Semester of Linear Algebra by S. E. Payne. Here are two other links to the same book, via the author's site: PDF file. HTML page: Class Notes.
$endgroup$
– Pierre-Yves Gaillard
Oct 11 '11 at 17:45
3
3
$begingroup$
The book joriki links to is A Second Semester of Linear Algebra by S. E. Payne. Here are two other links to the same book, via the author's site: PDF file. HTML page: Class Notes.
$endgroup$
– Pierre-Yves Gaillard
Oct 11 '11 at 17:45
$begingroup$
The book joriki links to is A Second Semester of Linear Algebra by S. E. Payne. Here are two other links to the same book, via the author's site: PDF file. HTML page: Class Notes.
$endgroup$
– Pierre-Yves Gaillard
Oct 11 '11 at 17:45
add a comment |
$begingroup$
See Section III.8.7, entitled Application to Linear Equations, of Algebra, by Nicolas Bourbaki.
EDIT 1. Let $R$ be a commutative ring, let $m$ and $n$ be positive integers, let $M$ be an $R$-module, and let $A:R^nto M$ be $R$-linear.
Identify the $n$ th exterior power $Lambda^n(R^n)$ of $R^n$ to $R$ in the obvious way, so that $Lambda^n(A)$ is a map from $R$ to $Lambda^n(M)$.
Put $v_i:=Ae_i$, where $e_i$ is the $i$ th vector of the canonical basis of $R^n$. In particular we have
$$
Ax=sum_{i=1}^n x_i v_i,quadLambda^n(A) r=r v_1wedgecdotswedge v_n.
$$
(where $x_i$ is the $i$-th coordinate of $x$, and $r$ denotes any element of $Lambda^nleft(R^nright) cong R$).
If $Lambda^n(A)$ is injective, so is $A$.
In other words:
If the $v_i$ are linearly dependent, then $r v_1wedgecdotswedge v_n=0$ for some nonzero $r$ in $R$.
Indeed, for $x$ in $ker A$ we have
$$
Lambda^n(A) x_1=x_1 v_1wedge v_2wedgecdotswedge v_n=
-sum_{i=2}^n x_i v_iwedge v_2wedgecdotswedge v_n=0,
$$
and, similarly, $Lambda^n(A) x_i=0$ for all $i$.
[Edit: Old version (before Georges's comment): Assume now that $M$ embeds into $R^m$.]
Assume now that there is an $R$-linear injection $B:Mto R^m$ such that
$$
Lambda^n(B):Lambda^n(M)toLambda^n(R^m)
$$
is injective. This is always the case (for a suitable $m$) if $M$ is projective and finitely generated.
If $A$ is injective, so is $Lambda^n(A)$.
In other words:
If $r v_1wedgecdotswedge v_n=0$ for some nonzero $r$ in $R$, then the $v_i$ are linearly dependent.
The proof is given in joriki's nice answer.
This is also proved as Proposition 12 in Bourbaki's Algebra III.7.9 p. 519. Unfortunately, I don't understand Bourbaki's argument. I'd be most grateful to whoever would be kind and patient enough to explain it to me.
EDIT 2. According to the indications given by Tsit-Yuen Lam on page 150 of his book Exercises in modules and rings, the theorem is due to N. H. McCoy, and appeared first, as Theorem 1 page 288, in
- N. H. McCoy, Remarks on Divisors of Zero, The American Mathematical Monthly Vol. 49, No. 5 (May, 1942), pp. 286-295, JSTOR.
Lam also says that
- N. H. McCoy, Rings and ideals, The Carus Mathematical Monographs, no. 8, The Mathematical Association of America, 1948,
is an "excellent exposition" of the subject. See Theorem 51 page 159.
McCoy's Theorem is also stated and proved in the following texts:
Ex. 5.23.A(3) on page 149 of Lam's Exercises in modules and rings.
Theorem 2.2 page 3 in Anton Gerashenko's notes from Lam's Course: Math 274, Commutative Rings, Fall 2006: PDF file.
Theorem 1.6 in Chapter 13, entitled "Various topics", of The CRing Project. --- PDF file for Chapter 13. --- PDF file for the whole book.
Blocki, Zbigniew, An elementary proof of the McCoy theorem, J. Univ. Iagel. Acta Math.; N 30; 1993; 215-218.
Theorem 6.4.16. page 101, A Second Semester of Linear Algebra, Math 5718, by Stan Payne. PDF file.
edited Jan 20 at 12:34
darij grinberg
11.1k33167
answered Oct 11 '11 at 16:57
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
$endgroup$
$begingroup$
Thanks for this link! Note, however, that the result is proved only for square matrices there. This case was the original motivation for the question, but the question and my answer apply to the general case.
$endgroup$
– joriki
Nov 10 '11 at 9:13
$begingroup$
Dear @Pierre-Yves, I'm not quite sure that your claim (displayed in grey) "If $A$ is injective, so is $Lambda^n A$" holds if you assume only that $M$ embeds into $R^m$.There is the subtle point that exterior products don't mean the same in both spaces, in other words $Lambda ^n Mto Lambda ^n R^m$ needn't be injective. Everything is fine if $M$ is projective, though, and this is Bourbaki's assumption . But this is nitpicking: +1, needless to say.
$endgroup$
– Georges Elencwajg
Nov 12 '11 at 14:17
$begingroup$
Dear @Georges: No, this is definitely not nitpicking! Thanks a lot! I hope it's correct now. - It was my secret hope that you would read this answer. Did you see the last paragraph? I'm sure you understand Bourbaki's argument...
$endgroup$
– Pierre-Yves Gaillard
Nov 12 '11 at 15:11
$begingroup$
Dear @Pierre-Yves: No, I don't understand Bourbaki's argument either. Specifically, when he writes "it follows from no.8, Corollary 3 to Theorem 1 that $mu x_1$ is a linear combination....", I don't see how it follows. (By the way, this is Proposition 12 in my edition)
$endgroup$
– Georges Elencwajg
Nov 12 '11 at 18:19
$begingroup$
Dear @Georges: Once more you're right: it's Proposition 12 (in the link to the English edition I give, and in my French edition - it was just a typo). Thank you very much for your time and effort. I'm having the same problem as you with Bourbaki's proof. It's weird. I find joriki's formulation of the argument very nice.
$endgroup$
– Pierre-Yves Gaillard
Nov 12 '11 at 18:42
add a comment |
$begingroup$
See Section III.8.7, entitled Application to Linear Equations, of Algebra, by Nicolas Bourbaki.
EDIT 1. Let $R$ be a commutative ring, let $m$ and $n$ be positive integers, let $M$ be an $R$-module, and let $A:R^nto M$ be $R$-linear.
Identify the $n$ th exterior power $Lambda^n(R^n)$ of $R^n$ to $R$ in the obvious way, so that $Lambda^n(A)$ is a map from $R$ to $Lambda^n(M)$.
Put $v_i:=Ae_i$, where $e_i$ is the $i$ th vector of the canonical basis of $R^n$. In particular we have
$$
Ax=sum_{i=1}^n x_i v_i,quadLambda^n(A) r=r v_1wedgecdotswedge v_n.
$$
(where $x_i$ is the $i$-th coordinate of $x$, and $r$ denotes any element of $Lambda^nleft(R^nright) cong R$).
If $Lambda^n(A)$ is injective, so is $A$.
In other words:
If the $v_i$ are linearly dependent, then $r v_1wedgecdotswedge v_n=0$ for some nonzero $r$ in $R$.
Indeed, for $x$ in $ker A$ we have
$$
Lambda^n(A) x_1=x_1 v_1wedge v_2wedgecdotswedge v_n=
-sum_{i=2}^n x_i v_iwedge v_2wedgecdotswedge v_n=0,
$$
and, similarly, $Lambda^n(A) x_i=0$ for all $i$.
[Edit: Old version (before Georges's comment): Assume now that $M$ embeds into $R^m$.]
Assume now that there is an $R$-linear injection $B:Mto R^m$ such that
$$
Lambda^n(B):Lambda^n(M)toLambda^n(R^m)
$$
is injective. This is always the case (for a suitable $m$) if $M$ is projective and finitely generated.
If $A$ is injective, so is $Lambda^n(A)$.
In other words:
If $r v_1wedgecdotswedge v_n=0$ for some nonzero $r$ in $R$, then the $v_i$ are linearly dependent.
The proof is given in joriki's nice answer.
This is also proved as Proposition 12 in Bourbaki's Algebra III.7.9 p. 519. Unfortunately, I don't understand Bourbaki's argument. I'd be most grateful to whoever would be kind and patient enough to explain it to me.
EDIT 2. According to the indications given by Tsit-Yuen Lam on page 150 of his book Exercises in modules and rings, the theorem is due to N. H. McCoy, and appeared first, as Theorem 1 page 288, in
- N. H. McCoy, Remarks on Divisors of Zero, The American Mathematical Monthly Vol. 49, No. 5 (May, 1942), pp. 286-295, JSTOR.
Lam also says that
- N. H. McCoy, Rings and ideals, The Carus Mathematical Monographs, no. 8, The Mathematical Association of America, 1948,
is an "excellent exposition" of the subject. See Theorem 51 page 159.
McCoy's Theorem is also stated and proved in the following texts:
Ex. 5.23.A(3) on page 149 of Lam's Exercises in modules and rings.
Theorem 2.2 page 3 in Anton Gerashenko's notes from Lam's Course: Math 274, Commutative Rings, Fall 2006: PDF file.
Theorem 1.6 in Chapter 13, entitled "Various topics", of The CRing Project. --- PDF file for Chapter 13. --- PDF file for the whole book.
Blocki, Zbigniew, An elementary proof of the McCoy theorem, J. Univ. Iagel. Acta Math.; N 30; 1993; 215-218.
Theorem 6.4.16. page 101, A Second Semester of Linear Algebra, Math 5718, by Stan Payne. PDF file.
edited Jan 20 at 12:34
darij grinberg
11.1k33167
answered Oct 11 '11 at 16:57
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
$endgroup$
$begingroup$
Thanks for this link! Note, however, that the result is proved only for square matrices there. This case was the original motivation for the question, but the question and my answer apply to the general case.
$endgroup$
– joriki
Nov 10 '11 at 9:13
$begingroup$
Dear @Pierre-Yves, I'm not quite sure that your claim (displayed in grey) "If $A$ is injective, so is $Lambda^n A$" holds if you assume only that $M$ embeds into $R^m$.There is the subtle point that exterior products don't mean the same in both spaces, in other words $Lambda ^n Mto Lambda ^n R^m$ needn't be injective. Everything is fine if $M$ is projective, though, and this is Bourbaki's assumption . But this is nitpicking: +1, needless to say.
$endgroup$
– Georges Elencwajg
Nov 12 '11 at 14:17
$begingroup$
Dear @Georges: No, this is definitely not nitpicking! Thanks a lot! I hope it's correct now. - It was my secret hope that you would read this answer. Did you see the last paragraph? I'm sure you understand Bourbaki's argument...
$endgroup$
– Pierre-Yves Gaillard
Nov 12 '11 at 15:11
$begingroup$
Dear @Pierre-Yves: No, I don't understand Bourbaki's argument either. Specifically, when he writes "it follows from no.8, Corollary 3 to Theorem 1 that $mu x_1$ is a linear combination....", I don't see how it follows. (By the way, this is Proposition 12 in my edition)
$endgroup$
– Georges Elencwajg
Nov 12 '11 at 18:19
$begingroup$
Dear @Georges: Once more you're right: it's Proposition 12 (in the link to the English edition I give, and in my French edition - it was just a typo). Thank you very much for your time and effort. I'm having the same problem as you with Bourbaki's proof. It's weird. I find joriki's formulation of the argument very nice.
$endgroup$
– Pierre-Yves Gaillard
Nov 12 '11 at 18:42
add a comment |
$begingroup$
See Section III.8.7, entitled Application to Linear Equations, of Algebra, by Nicolas Bourbaki.
EDIT 1. Let $R$ be a commutative ring, let $m$ and $n$ be positive integers, let $M$ be an $R$-module, and let $A:R^nto M$ be $R$-linear.
Identify the $n$ th exterior power $Lambda^n(R^n)$ of $R^n$ to $R$ in the obvious way, so that $Lambda^n(A)$ is a map from $R$ to $Lambda^n(M)$.
Put $v_i:=Ae_i$, where $e_i$ is the $i$ th vector of the canonical basis of $R^n$. In particular we have
$$
Ax=sum_{i=1}^n x_i v_i,quadLambda^n(A) r=r v_1wedgecdotswedge v_n.
$$
(where $x_i$ is the $i$-th coordinate of $x$, and $r$ denotes any element of $Lambda^nleft(R^nright) cong R$).
If $Lambda^n(A)$ is injective, so is $A$.
In other words:
If the $v_i$ are linearly dependent, then $r v_1wedgecdotswedge v_n=0$ for some nonzero $r$ in $R$.
Indeed, for $x$ in $ker A$ we have
$$
Lambda^n(A) x_1=x_1 v_1wedge v_2wedgecdotswedge v_n=
-sum_{i=2}^n x_i v_iwedge v_2wedgecdotswedge v_n=0,
$$
and, similarly, $Lambda^n(A) x_i=0$ for all $i$.
[Edit: Old version (before Georges's comment): Assume now that $M$ embeds into $R^m$.]
Assume now that there is an $R$-linear injection $B:Mto R^m$ such that
$$
Lambda^n(B):Lambda^n(M)toLambda^n(R^m)
$$
is injective. This is always the case (for a suitable $m$) if $M$ is projective and finitely generated.
If $A$ is injective, so is $Lambda^n(A)$.
In other words:
If $r v_1wedgecdotswedge v_n=0$ for some nonzero $r$ in $R$, then the $v_i$ are linearly dependent.
The proof is given in joriki's nice answer.
This is also proved as Proposition 12 in Bourbaki's Algebra III.7.9 p. 519. Unfortunately, I don't understand Bourbaki's argument. I'd be most grateful to whoever would be kind and patient enough to explain it to me.
EDIT 2. According to the indications given by Tsit-Yuen Lam on page 150 of his book Exercises in modules and rings, the theorem is due to N. H. McCoy, and appeared first, as Theorem 1 page 288, in
- N. H. McCoy, Remarks on Divisors of Zero, The American Mathematical Monthly Vol. 49, No. 5 (May, 1942), pp. 286-295, JSTOR.
Lam also says that
- N. H. McCoy, Rings and ideals, The Carus Mathematical Monographs, no. 8, The Mathematical Association of America, 1948,
is an "excellent exposition" of the subject. See Theorem 51 page 159.
McCoy's Theorem is also stated and proved in the following texts:
Ex. 5.23.A(3) on page 149 of Lam's Exercises in modules and rings.
Theorem 2.2 page 3 in Anton Gerashenko's notes from Lam's Course: Math 274, Commutative Rings, Fall 2006: PDF file.
Theorem 1.6 in Chapter 13, entitled "Various topics", of The CRing Project. --- PDF file for Chapter 13. --- PDF file for the whole book.
Blocki, Zbigniew, An elementary proof of the McCoy theorem, J. Univ. Iagel. Acta Math.; N 30; 1993; 215-218.
Theorem 6.4.16. page 101, A Second Semester of Linear Algebra, Math 5718, by Stan Payne. PDF file.
edited Jan 20 at 12:34
darij grinberg
11.1k33167
answered Oct 11 '11 at 16:57
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
$endgroup$
See Section III.8.7, entitled Application to Linear Equations, of Algebra, by Nicolas Bourbaki.
EDIT 1. Let $R$ be a commutative ring, let $m$ and $n$ be positive integers, let $M$ be an $R$-module, and let $A:R^nto M$ be $R$-linear.
Identify the $n$ th exterior power $Lambda^n(R^n)$ of $R^n$ to $R$ in the obvious way, so that $Lambda^n(A)$ is a map from $R$ to $Lambda^n(M)$.
Put $v_i:=Ae_i$, where $e_i$ is the $i$ th vector of the canonical basis of $R^n$. In particular we have
$$
Ax=sum_{i=1}^n x_i v_i,quadLambda^n(A) r=r v_1wedgecdotswedge v_n.
$$
(where $x_i$ is the $i$-th coordinate of $x$, and $r$ denotes any element of $Lambda^nleft(R^nright) cong R$).
If $Lambda^n(A)$ is injective, so is $A$.
In other words:
If the $v_i$ are linearly dependent, then $r v_1wedgecdotswedge v_n=0$ for some nonzero $r$ in $R$.
Indeed, for $x$ in $ker A$ we have
$$
Lambda^n(A) x_1=x_1 v_1wedge v_2wedgecdotswedge v_n=
-sum_{i=2}^n x_i v_iwedge v_2wedgecdotswedge v_n=0,
$$
and, similarly, $Lambda^n(A) x_i=0$ for all $i$.
[Edit: Old version (before Georges's comment): Assume now that $M$ embeds into $R^m$.]
Assume now that there is an $R$-linear injection $B:Mto R^m$ such that
$$
Lambda^n(B):Lambda^n(M)toLambda^n(R^m)
$$
is injective. This is always the case (for a suitable $m$) if $M$ is projective and finitely generated.
If $A$ is injective, so is $Lambda^n(A)$.
In other words:
If $r v_1wedgecdotswedge v_n=0$ for some nonzero $r$ in $R$, then the $v_i$ are linearly dependent.
The proof is given in joriki's nice answer.
This is also proved as Proposition 12 in Bourbaki's Algebra III.7.9 p. 519. Unfortunately, I don't understand Bourbaki's argument. I'd be most grateful to whoever would be kind and patient enough to explain it to me.
EDIT 2. According to the indications given by Tsit-Yuen Lam on page 150 of his book Exercises in modules and rings, the theorem is due to N. H. McCoy, and appeared first, as Theorem 1 page 288, in
- N. H. McCoy, Remarks on Divisors of Zero, The American Mathematical Monthly Vol. 49, No. 5 (May, 1942), pp. 286-295, JSTOR.
Lam also says that
- N. H. McCoy, Rings and ideals, The Carus Mathematical Monographs, no. 8, The Mathematical Association of America, 1948,
is an "excellent exposition" of the subject. See Theorem 51 page 159.
McCoy's Theorem is also stated and proved in the following texts:
Ex. 5.23.A(3) on page 149 of Lam's Exercises in modules and rings.
Theorem 2.2 page 3 in Anton Gerashenko's notes from Lam's Course: Math 274, Commutative Rings, Fall 2006: PDF file.
Theorem 1.6 in Chapter 13, entitled "Various topics", of The CRing Project. --- PDF file for Chapter 13. --- PDF file for the whole book.
Blocki, Zbigniew, An elementary proof of the McCoy theorem, J. Univ. Iagel. Acta Math.; N 30; 1993; 215-218.
Theorem 6.4.16. page 101, A Second Semester of Linear Algebra, Math 5718, by Stan Payne. PDF file.
edited Jan 20 at 12:34
darij grinberg
11.1k33167
answered Oct 11 '11 at 16:57
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
edited Jan 20 at 12:34
darij grinberg
11.1k33167
edited Jan 20 at 12:34
darij grinberg
11.1k33167
edited Jan 20 at 12:34
darij grinberg
11.1k33167
11.1k33167
answered Oct 11 '11 at 16:57
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
answered Oct 11 '11 at 16:57
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
answered Oct 11 '11 at 16:57
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
13.4k23184
$begingroup$
Thanks for this link! Note, however, that the result is proved only for square matrices there. This case was the original motivation for the question, but the question and my answer apply to the general case.
$endgroup$
– joriki
Nov 10 '11 at 9:13
$begingroup$
Dear @Pierre-Yves, I'm not quite sure that your claim (displayed in grey) "If $A$ is injective, so is $Lambda^n A$" holds if you assume only that $M$ embeds into $R^m$.There is the subtle point that exterior products don't mean the same in both spaces, in other words $Lambda ^n Mto Lambda ^n R^m$ needn't be injective. Everything is fine if $M$ is projective, though, and this is Bourbaki's assumption . But this is nitpicking: +1, needless to say.
$endgroup$
– Georges Elencwajg
Nov 12 '11 at 14:17
$begingroup$
Dear @Georges: No, this is definitely not nitpicking! Thanks a lot! I hope it's correct now. - It was my secret hope that you would read this answer. Did you see the last paragraph? I'm sure you understand Bourbaki's argument...
$endgroup$
– Pierre-Yves Gaillard
Nov 12 '11 at 15:11
$begingroup$
Dear @Pierre-Yves: No, I don't understand Bourbaki's argument either. Specifically, when he writes "it follows from no.8, Corollary 3 to Theorem 1 that $mu x_1$ is a linear combination....", I don't see how it follows. (By the way, this is Proposition 12 in my edition)
$endgroup$
– Georges Elencwajg
Nov 12 '11 at 18:19
$begingroup$
Dear @Georges: Once more you're right: it's Proposition 12 (in the link to the English edition I give, and in my French edition - it was just a typo). Thank you very much for your time and effort. I'm having the same problem as you with Bourbaki's proof. It's weird. I find joriki's formulation of the argument very nice.
$endgroup$
– Pierre-Yves Gaillard
Nov 12 '11 at 18:42
add a comment |
$begingroup$
Thanks for this link! Note, however, that the result is proved only for square matrices there. This case was the original motivation for the question, but the question and my answer apply to the general case.
$endgroup$
– joriki
Nov 10 '11 at 9:13
$begingroup$
Dear @Pierre-Yves, I'm not quite sure that your claim (displayed in grey) "If $A$ is injective, so is $Lambda^n A$" holds if you assume only that $M$ embeds into $R^m$.There is the subtle point that exterior products don't mean the same in both spaces, in other words $Lambda ^n Mto Lambda ^n R^m$ needn't be injective. Everything is fine if $M$ is projective, though, and this is Bourbaki's assumption . But this is nitpicking: +1, needless to say.
$endgroup$
– Georges Elencwajg
Nov 12 '11 at 14:17
$begingroup$
Dear @Georges: No, this is definitely not nitpicking! Thanks a lot! I hope it's correct now. - It was my secret hope that you would read this answer. Did you see the last paragraph? I'm sure you understand Bourbaki's argument...
$endgroup$
– Pierre-Yves Gaillard
Nov 12 '11 at 15:11
$begingroup$
Dear @Pierre-Yves: No, I don't understand Bourbaki's argument either. Specifically, when he writes "it follows from no.8, Corollary 3 to Theorem 1 that $mu x_1$ is a linear combination....", I don't see how it follows. (By the way, this is Proposition 12 in my edition)
$endgroup$
– Georges Elencwajg
Nov 12 '11 at 18:19
$begingroup$
Dear @Georges: Once more you're right: it's Proposition 12 (in the link to the English edition I give, and in my French edition - it was just a typo). Thank you very much for your time and effort. I'm having the same problem as you with Bourbaki's proof. It's weird. I find joriki's formulation of the argument very nice.
$endgroup$
– Pierre-Yves Gaillard
Nov 12 '11 at 18:42
$begingroup$
Thanks for this link! Note, however, that the result is proved only for square matrices there. This case was the original motivation for the question, but the question and my answer apply to the general case.
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– joriki
Nov 10 '11 at 9:13
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Thanks for this link! Note, however, that the result is proved only for square matrices there. This case was the original motivation for the question, but the question and my answer apply to the general case.
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– joriki
Nov 10 '11 at 9:13
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Dear @Pierre-Yves, I'm not quite sure that your claim (displayed in grey) "If $A$ is injective, so is $Lambda^n A$" holds if you assume only that $M$ embeds into $R^m$.There is the subtle point that exterior products don't mean the same in both spaces, in other words $Lambda ^n Mto Lambda ^n R^m$ needn't be injective. Everything is fine if $M$ is projective, though, and this is Bourbaki's assumption . But this is nitpicking: +1, needless to say.
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– Georges Elencwajg
Nov 12 '11 at 14:17
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Dear @Pierre-Yves, I'm not quite sure that your claim (displayed in grey) "If $A$ is injective, so is $Lambda^n A$" holds if you assume only that $M$ embeds into $R^m$.There is the subtle point that exterior products don't mean the same in both spaces, in other words $Lambda ^n Mto Lambda ^n R^m$ needn't be injective. Everything is fine if $M$ is projective, though, and this is Bourbaki's assumption . But this is nitpicking: +1, needless to say.
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– Georges Elencwajg
Nov 12 '11 at 14:17
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Dear @Georges: No, this is definitely not nitpicking! Thanks a lot! I hope it's correct now. - It was my secret hope that you would read this answer. Did you see the last paragraph? I'm sure you understand Bourbaki's argument...
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– Pierre-Yves Gaillard
Nov 12 '11 at 15:11
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Dear @Georges: No, this is definitely not nitpicking! Thanks a lot! I hope it's correct now. - It was my secret hope that you would read this answer. Did you see the last paragraph? I'm sure you understand Bourbaki's argument...
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– Pierre-Yves Gaillard
Nov 12 '11 at 15:11
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Dear @Pierre-Yves: No, I don't understand Bourbaki's argument either. Specifically, when he writes "it follows from no.8, Corollary 3 to Theorem 1 that $mu x_1$ is a linear combination....", I don't see how it follows. (By the way, this is Proposition 12 in my edition)
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– Georges Elencwajg
Nov 12 '11 at 18:19
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Dear @Pierre-Yves: No, I don't understand Bourbaki's argument either. Specifically, when he writes "it follows from no.8, Corollary 3 to Theorem 1 that $mu x_1$ is a linear combination....", I don't see how it follows. (By the way, this is Proposition 12 in my edition)
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– Georges Elencwajg
Nov 12 '11 at 18:19
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Dear @Georges: Once more you're right: it's Proposition 12 (in the link to the English edition I give, and in my French edition - it was just a typo). Thank you very much for your time and effort. I'm having the same problem as you with Bourbaki's proof. It's weird. I find joriki's formulation of the argument very nice.
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– Pierre-Yves Gaillard
Nov 12 '11 at 18:42
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Dear @Georges: Once more you're right: it's Proposition 12 (in the link to the English edition I give, and in my French edition - it was just a typo). Thank you very much for your time and effort. I'm having the same problem as you with Bourbaki's proof. It's weird. I find joriki's formulation of the argument very nice.
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– Pierre-Yves Gaillard
Nov 12 '11 at 18:42
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