Mixing
Prove that $A$ is invertible iff $det(A)neq 0$ with Cauchy-Binet theorem.
$begingroup$
Let $A$ a matrix $ntimes n$ over $mathbb{R}$.
I'm trying to prove that A is invertible if and only if $det(A)neq0$ using the Cauchy-Binet theorem.
I know that the Cauchy -Binet theorem is $$det(A B)=det(A)cdot det(B)$$
But for now, I couldn't think of any solutions to solve the proof.
linear-algebra proof-explanation determinant
edited Jan 9 at 20:26
user376343
3,5483827
asked Jan 9 at 19:33
KevinKevin
13311
$endgroup$
add a comment |
$begingroup$
Let $A$ a matrix $ntimes n$ over $mathbb{R}$.
I'm trying to prove that A is invertible if and only if $det(A)neq0$ using the Cauchy-Binet theorem.
I know that the Cauchy -Binet theorem is $$det(A B)=det(A)cdot det(B)$$
But for now, I couldn't think of any solutions to solve the proof.
linear-algebra proof-explanation determinant
edited Jan 9 at 20:26
user376343
3,5483827
asked Jan 9 at 19:33
KevinKevin
13311
$endgroup$
add a comment |
$begingroup$
Let $A$ a matrix $ntimes n$ over $mathbb{R}$.
I'm trying to prove that A is invertible if and only if $det(A)neq0$ using the Cauchy-Binet theorem.
I know that the Cauchy -Binet theorem is $$det(A B)=det(A)cdot det(B)$$
But for now, I couldn't think of any solutions to solve the proof.
linear-algebra proof-explanation determinant
edited Jan 9 at 20:26
user376343
3,5483827
asked Jan 9 at 19:33
KevinKevin
13311
$endgroup$
Let $A$ a matrix $ntimes n$ over $mathbb{R}$.
I'm trying to prove that A is invertible if and only if $det(A)neq0$ using the Cauchy-Binet theorem.
I know that the Cauchy -Binet theorem is $$det(A B)=det(A)cdot det(B)$$
But for now, I couldn't think of any solutions to solve the proof.
linear-algebra proof-explanation determinant
linear-algebra proof-explanation determinant
edited Jan 9 at 20:26
user376343
3,5483827
asked Jan 9 at 19:33
KevinKevin
13311
edited Jan 9 at 20:26
user376343
3,5483827
asked Jan 9 at 19:33
KevinKevin
13311
edited Jan 9 at 20:26
user376343
3,5483827
edited Jan 9 at 20:26
user376343
3,5483827
edited Jan 9 at 20:26
user376343
3,5483827
3,5483827
asked Jan 9 at 19:33
KevinKevin
13311
asked Jan 9 at 19:33
KevinKevin
13311
asked Jan 9 at 19:33
KevinKevin
13311
13311
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose that $A$ is invertible. Then there is a matrix $B$ such that $AB=I$. The Cauchy-Binet theorem then implies that
$$1=det(I)=det(AB)=det(A)det(B)$$
so that $det(A)neq 0neq det(B)$.
Conversely, suppose $det(A)neq 0$. Then $B=frac{1}{detA}adj(A)$ satisfies $AB=I$, so that $A$ is invertible. Here $adj(A)$ is the adjugate of the matrix $A$.
Edit: Given the comments on my answer, I'm including a link to a thread about left/right inverses:
If $AB = I$ then $BA = I$
answered Jan 9 at 19:42
pwerthpwerth
3,113417
$endgroup$
1
$begingroup$
In my experience, some professors are incredibly pedantic with the definition of matrix inverse; you need not only specify that there exists $B$ such that $AB = I$, but also that $BA = I$ as well. That is, $AB = I = BA$.
$endgroup$
– Decaf-Math
Jan 9 at 19:58
$begingroup$
Either that or append a proof that $AB=Ito BA=I$ for square matrices on finite-dimensional spaces.
$endgroup$
– J.G.
Jan 9 at 21:12
add a comment |
$begingroup$
If $A$ is invertible, then $AA^{-1}=I$ (identity matrix), so
$$
1=det I=det(AA^{-1})=det Adet(A^{-1})
$$
and therefore $det Ane0$.
The converse doesn't follow from Binet's theorem, but rather from the fact that the determinant is multilinear and alternating on the columns of a matrix.
Fact 1. If $A$ has a zero column, then $det A=det A+det A$, so $det A=0$.
Fact 2. If $A$ has two identical columns, then $det A=-det A$, by swapping them, so $det A=0$.
Fact 3. If $A$ is not invertible, then $det A=0$.
Since we want to show that $det A=0$, possibly with a column swap we can assume that the last column is a linear combination of the other $n-1$ columns. Say $A=[a_1 dots a_{n-1} a_n]$, with
$$
a_n=c_1a_1+dots+c_{n-1}a_{n-1}
$$
Then, by multilinearity and facts 1 and 2, we have
$$
0=det[a_1 dots a_{n-1} 0]=
det A
-c_1det[a_1 dots a_{n-1} a_1]
-dots
-c_{n-1}det[a_1 dots a_{n-1} a_{n-1}]
$$
answered Jan 9 at 22:58
egregegreg
181k1485203
$endgroup$
$begingroup$
A somewhat similar approach for the converse uses the fact that row and column operations can be achieved by multiplying (on the left or right) by elementary matrices.
$endgroup$
– Cheerful Parsnip
Jan 9 at 23:02
$begingroup$
@CheerfulParsnip Indeed, in my course I don't mention multilinearity, but rather define the determinant to be “invariant” by elementary column operations (or, equivalently, row operations): multiplying a column by $c$ multiplies the determinant by $c$; adding to a column another column multiplied by $d$ doesn't change the determinant; swapping two columns multiplies the determinant by $-1$. I find this better because the course puts great emphasis on elimination.
$endgroup$
– egreg
Jan 9 at 23:11
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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votes
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votes
$begingroup$
Suppose that $A$ is invertible. Then there is a matrix $B$ such that $AB=I$. The Cauchy-Binet theorem then implies that
$$1=det(I)=det(AB)=det(A)det(B)$$
so that $det(A)neq 0neq det(B)$.
Conversely, suppose $det(A)neq 0$. Then $B=frac{1}{detA}adj(A)$ satisfies $AB=I$, so that $A$ is invertible. Here $adj(A)$ is the adjugate of the matrix $A$.
Edit: Given the comments on my answer, I'm including a link to a thread about left/right inverses:
If $AB = I$ then $BA = I$
answered Jan 9 at 19:42
pwerthpwerth
3,113417
$endgroup$
1
$begingroup$
In my experience, some professors are incredibly pedantic with the definition of matrix inverse; you need not only specify that there exists $B$ such that $AB = I$, but also that $BA = I$ as well. That is, $AB = I = BA$.
$endgroup$
– Decaf-Math
Jan 9 at 19:58
$begingroup$
Either that or append a proof that $AB=Ito BA=I$ for square matrices on finite-dimensional spaces.
$endgroup$
– J.G.
Jan 9 at 21:12
add a comment |
$begingroup$
Suppose that $A$ is invertible. Then there is a matrix $B$ such that $AB=I$. The Cauchy-Binet theorem then implies that
$$1=det(I)=det(AB)=det(A)det(B)$$
so that $det(A)neq 0neq det(B)$.
Conversely, suppose $det(A)neq 0$. Then $B=frac{1}{detA}adj(A)$ satisfies $AB=I$, so that $A$ is invertible. Here $adj(A)$ is the adjugate of the matrix $A$.
Edit: Given the comments on my answer, I'm including a link to a thread about left/right inverses:
If $AB = I$ then $BA = I$
answered Jan 9 at 19:42
pwerthpwerth
3,113417
$endgroup$
1
$begingroup$
In my experience, some professors are incredibly pedantic with the definition of matrix inverse; you need not only specify that there exists $B$ such that $AB = I$, but also that $BA = I$ as well. That is, $AB = I = BA$.
$endgroup$
– Decaf-Math
Jan 9 at 19:58
$begingroup$
Either that or append a proof that $AB=Ito BA=I$ for square matrices on finite-dimensional spaces.
$endgroup$
– J.G.
Jan 9 at 21:12
add a comment |
$begingroup$
Suppose that $A$ is invertible. Then there is a matrix $B$ such that $AB=I$. The Cauchy-Binet theorem then implies that
$$1=det(I)=det(AB)=det(A)det(B)$$
so that $det(A)neq 0neq det(B)$.
Conversely, suppose $det(A)neq 0$. Then $B=frac{1}{detA}adj(A)$ satisfies $AB=I$, so that $A$ is invertible. Here $adj(A)$ is the adjugate of the matrix $A$.
Edit: Given the comments on my answer, I'm including a link to a thread about left/right inverses:
If $AB = I$ then $BA = I$
answered Jan 9 at 19:42
pwerthpwerth
3,113417
$endgroup$
Suppose that $A$ is invertible. Then there is a matrix $B$ such that $AB=I$. The Cauchy-Binet theorem then implies that
$$1=det(I)=det(AB)=det(A)det(B)$$
so that $det(A)neq 0neq det(B)$.
Conversely, suppose $det(A)neq 0$. Then $B=frac{1}{detA}adj(A)$ satisfies $AB=I$, so that $A$ is invertible. Here $adj(A)$ is the adjugate of the matrix $A$.
Edit: Given the comments on my answer, I'm including a link to a thread about left/right inverses:
If $AB = I$ then $BA = I$
answered Jan 9 at 19:42
pwerthpwerth
3,113417
edited Jan 9 at 21:34
answered Jan 9 at 19:42
pwerthpwerth
3,113417
answered Jan 9 at 19:42
pwerthpwerth
3,113417
answered Jan 9 at 19:42
pwerthpwerth
3,113417
3,113417
1
$begingroup$
In my experience, some professors are incredibly pedantic with the definition of matrix inverse; you need not only specify that there exists $B$ such that $AB = I$, but also that $BA = I$ as well. That is, $AB = I = BA$.
$endgroup$
– Decaf-Math
Jan 9 at 19:58
$begingroup$
Either that or append a proof that $AB=Ito BA=I$ for square matrices on finite-dimensional spaces.
$endgroup$
– J.G.
Jan 9 at 21:12
add a comment |
1
$begingroup$
In my experience, some professors are incredibly pedantic with the definition of matrix inverse; you need not only specify that there exists $B$ such that $AB = I$, but also that $BA = I$ as well. That is, $AB = I = BA$.
$endgroup$
– Decaf-Math
Jan 9 at 19:58
$begingroup$
Either that or append a proof that $AB=Ito BA=I$ for square matrices on finite-dimensional spaces.
$endgroup$
– J.G.
Jan 9 at 21:12
1
1
$begingroup$
In my experience, some professors are incredibly pedantic with the definition of matrix inverse; you need not only specify that there exists $B$ such that $AB = I$, but also that $BA = I$ as well. That is, $AB = I = BA$.
$endgroup$
– Decaf-Math
Jan 9 at 19:58
$begingroup$
In my experience, some professors are incredibly pedantic with the definition of matrix inverse; you need not only specify that there exists $B$ such that $AB = I$, but also that $BA = I$ as well. That is, $AB = I = BA$.
$endgroup$
– Decaf-Math
Jan 9 at 19:58
$begingroup$
Either that or append a proof that $AB=Ito BA=I$ for square matrices on finite-dimensional spaces.
$endgroup$
– J.G.
Jan 9 at 21:12
$begingroup$
Either that or append a proof that $AB=Ito BA=I$ for square matrices on finite-dimensional spaces.
$endgroup$
– J.G.
Jan 9 at 21:12
add a comment |
$begingroup$
If $A$ is invertible, then $AA^{-1}=I$ (identity matrix), so
$$
1=det I=det(AA^{-1})=det Adet(A^{-1})
$$
and therefore $det Ane0$.
The converse doesn't follow from Binet's theorem, but rather from the fact that the determinant is multilinear and alternating on the columns of a matrix.
Fact 1. If $A$ has a zero column, then $det A=det A+det A$, so $det A=0$.
Fact 2. If $A$ has two identical columns, then $det A=-det A$, by swapping them, so $det A=0$.
Fact 3. If $A$ is not invertible, then $det A=0$.
Since we want to show that $det A=0$, possibly with a column swap we can assume that the last column is a linear combination of the other $n-1$ columns. Say $A=[a_1 dots a_{n-1} a_n]$, with
$$
a_n=c_1a_1+dots+c_{n-1}a_{n-1}
$$
Then, by multilinearity and facts 1 and 2, we have
$$
0=det[a_1 dots a_{n-1} 0]=
det A
-c_1det[a_1 dots a_{n-1} a_1]
-dots
-c_{n-1}det[a_1 dots a_{n-1} a_{n-1}]
$$
answered Jan 9 at 22:58
egregegreg
181k1485203
$endgroup$
$begingroup$
A somewhat similar approach for the converse uses the fact that row and column operations can be achieved by multiplying (on the left or right) by elementary matrices.
$endgroup$
– Cheerful Parsnip
Jan 9 at 23:02
$begingroup$
@CheerfulParsnip Indeed, in my course I don't mention multilinearity, but rather define the determinant to be “invariant” by elementary column operations (or, equivalently, row operations): multiplying a column by $c$ multiplies the determinant by $c$; adding to a column another column multiplied by $d$ doesn't change the determinant; swapping two columns multiplies the determinant by $-1$. I find this better because the course puts great emphasis on elimination.
$endgroup$
– egreg
Jan 9 at 23:11
add a comment |
$begingroup$
If $A$ is invertible, then $AA^{-1}=I$ (identity matrix), so
$$
1=det I=det(AA^{-1})=det Adet(A^{-1})
$$
and therefore $det Ane0$.
The converse doesn't follow from Binet's theorem, but rather from the fact that the determinant is multilinear and alternating on the columns of a matrix.
Fact 1. If $A$ has a zero column, then $det A=det A+det A$, so $det A=0$.
Fact 2. If $A$ has two identical columns, then $det A=-det A$, by swapping them, so $det A=0$.
Fact 3. If $A$ is not invertible, then $det A=0$.
Since we want to show that $det A=0$, possibly with a column swap we can assume that the last column is a linear combination of the other $n-1$ columns. Say $A=[a_1 dots a_{n-1} a_n]$, with
$$
a_n=c_1a_1+dots+c_{n-1}a_{n-1}
$$
Then, by multilinearity and facts 1 and 2, we have
$$
0=det[a_1 dots a_{n-1} 0]=
det A
-c_1det[a_1 dots a_{n-1} a_1]
-dots
-c_{n-1}det[a_1 dots a_{n-1} a_{n-1}]
$$
answered Jan 9 at 22:58
egregegreg
181k1485203
$endgroup$
$begingroup$
A somewhat similar approach for the converse uses the fact that row and column operations can be achieved by multiplying (on the left or right) by elementary matrices.
$endgroup$
– Cheerful Parsnip
Jan 9 at 23:02
$begingroup$
@CheerfulParsnip Indeed, in my course I don't mention multilinearity, but rather define the determinant to be “invariant” by elementary column operations (or, equivalently, row operations): multiplying a column by $c$ multiplies the determinant by $c$; adding to a column another column multiplied by $d$ doesn't change the determinant; swapping two columns multiplies the determinant by $-1$. I find this better because the course puts great emphasis on elimination.
$endgroup$
– egreg
Jan 9 at 23:11
add a comment |
$begingroup$
If $A$ is invertible, then $AA^{-1}=I$ (identity matrix), so
$$
1=det I=det(AA^{-1})=det Adet(A^{-1})
$$
and therefore $det Ane0$.
The converse doesn't follow from Binet's theorem, but rather from the fact that the determinant is multilinear and alternating on the columns of a matrix.
Fact 1. If $A$ has a zero column, then $det A=det A+det A$, so $det A=0$.
Fact 2. If $A$ has two identical columns, then $det A=-det A$, by swapping them, so $det A=0$.
Fact 3. If $A$ is not invertible, then $det A=0$.
Since we want to show that $det A=0$, possibly with a column swap we can assume that the last column is a linear combination of the other $n-1$ columns. Say $A=[a_1 dots a_{n-1} a_n]$, with
$$
a_n=c_1a_1+dots+c_{n-1}a_{n-1}
$$
Then, by multilinearity and facts 1 and 2, we have
$$
0=det[a_1 dots a_{n-1} 0]=
det A
-c_1det[a_1 dots a_{n-1} a_1]
-dots
-c_{n-1}det[a_1 dots a_{n-1} a_{n-1}]
$$
answered Jan 9 at 22:58
egregegreg
181k1485203
$endgroup$
If $A$ is invertible, then $AA^{-1}=I$ (identity matrix), so
$$
1=det I=det(AA^{-1})=det Adet(A^{-1})
$$
and therefore $det Ane0$.
The converse doesn't follow from Binet's theorem, but rather from the fact that the determinant is multilinear and alternating on the columns of a matrix.
Fact 1. If $A$ has a zero column, then $det A=det A+det A$, so $det A=0$.
Fact 2. If $A$ has two identical columns, then $det A=-det A$, by swapping them, so $det A=0$.
Fact 3. If $A$ is not invertible, then $det A=0$.
Since we want to show that $det A=0$, possibly with a column swap we can assume that the last column is a linear combination of the other $n-1$ columns. Say $A=[a_1 dots a_{n-1} a_n]$, with
$$
a_n=c_1a_1+dots+c_{n-1}a_{n-1}
$$
Then, by multilinearity and facts 1 and 2, we have
$$
0=det[a_1 dots a_{n-1} 0]=
det A
-c_1det[a_1 dots a_{n-1} a_1]
-dots
-c_{n-1}det[a_1 dots a_{n-1} a_{n-1}]
$$
answered Jan 9 at 22:58
egregegreg
181k1485203
answered Jan 9 at 22:58
egregegreg
181k1485203
answered Jan 9 at 22:58
egregegreg
181k1485203
answered Jan 9 at 22:58
egregegreg
181k1485203
181k1485203
$begingroup$
A somewhat similar approach for the converse uses the fact that row and column operations can be achieved by multiplying (on the left or right) by elementary matrices.
$endgroup$
– Cheerful Parsnip
Jan 9 at 23:02
$begingroup$
@CheerfulParsnip Indeed, in my course I don't mention multilinearity, but rather define the determinant to be “invariant” by elementary column operations (or, equivalently, row operations): multiplying a column by $c$ multiplies the determinant by $c$; adding to a column another column multiplied by $d$ doesn't change the determinant; swapping two columns multiplies the determinant by $-1$. I find this better because the course puts great emphasis on elimination.
$endgroup$
– egreg
Jan 9 at 23:11
add a comment |
$begingroup$
A somewhat similar approach for the converse uses the fact that row and column operations can be achieved by multiplying (on the left or right) by elementary matrices.
$endgroup$
– Cheerful Parsnip
Jan 9 at 23:02
$begingroup$
@CheerfulParsnip Indeed, in my course I don't mention multilinearity, but rather define the determinant to be “invariant” by elementary column operations (or, equivalently, row operations): multiplying a column by $c$ multiplies the determinant by $c$; adding to a column another column multiplied by $d$ doesn't change the determinant; swapping two columns multiplies the determinant by $-1$. I find this better because the course puts great emphasis on elimination.
$endgroup$
– egreg
Jan 9 at 23:11
$begingroup$
A somewhat similar approach for the converse uses the fact that row and column operations can be achieved by multiplying (on the left or right) by elementary matrices.
$endgroup$
– Cheerful Parsnip
Jan 9 at 23:02
$begingroup$
A somewhat similar approach for the converse uses the fact that row and column operations can be achieved by multiplying (on the left or right) by elementary matrices.
$endgroup$
– Cheerful Parsnip
Jan 9 at 23:02
$begingroup$
@CheerfulParsnip Indeed, in my course I don't mention multilinearity, but rather define the determinant to be “invariant” by elementary column operations (or, equivalently, row operations): multiplying a column by $c$ multiplies the determinant by $c$; adding to a column another column multiplied by $d$ doesn't change the determinant; swapping two columns multiplies the determinant by $-1$. I find this better because the course puts great emphasis on elimination.
$endgroup$
– egreg
Jan 9 at 23:11
$begingroup$
@CheerfulParsnip Indeed, in my course I don't mention multilinearity, but rather define the determinant to be “invariant” by elementary column operations (or, equivalently, row operations): multiplying a column by $c$ multiplies the determinant by $c$; adding to a column another column multiplied by $d$ doesn't change the determinant; swapping two columns multiplies the determinant by $-1$. I find this better because the course puts great emphasis on elimination.
$endgroup$
– egreg
Jan 9 at 23:11
add a comment |
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