Mixing
Prove that the rank of $(1-I)$ is $n$
$begingroup$
The rank of $(1-I_n)$, where $1$ is the $n times n$ all-1 matrix and $I_n$ the $n times n$ identity matrix, seems to be $n$.
How to prove this concisely?
linear-algebra matrices
edited Jun 20 '13 at 18:57
Davide Giraudo
128k17156268
asked Jun 20 '13 at 18:02
rankyranky
61
$endgroup$
add a comment |
$begingroup$
The rank of $(1-I_n)$, where $1$ is the $n times n$ all-1 matrix and $I_n$ the $n times n$ identity matrix, seems to be $n$.
How to prove this concisely?
linear-algebra matrices
edited Jun 20 '13 at 18:57
Davide Giraudo
128k17156268
asked Jun 20 '13 at 18:02
rankyranky
61
$endgroup$
2
$begingroup$
Well if you look at a example. When you subtract a matrix filled with ones from a matrix with 1`s going down in a diagonal this creates all columns that are linearly independent due to the zeroes appearing one position lower from column 1 to column N.
$endgroup$
– user60887
Jun 20 '13 at 18:11
$begingroup$
Maybe easiest is to assume given that $r(A + B) leq r(A) + r(B)$ (subadditivity of rank)? By then choosing $A = 1$ and $B = -(1-I)$, this gives $r(I) leq r(1) + r(-(1-I))$ or $n leq 1 + r(-(1-I))$ and hence $r(1-I) = r(-(1-I)) geq n-1$.
$endgroup$
– ranky
Jun 20 '13 at 18:24
add a comment |
$begingroup$
The rank of $(1-I_n)$, where $1$ is the $n times n$ all-1 matrix and $I_n$ the $n times n$ identity matrix, seems to be $n$.
How to prove this concisely?
linear-algebra matrices
edited Jun 20 '13 at 18:57
Davide Giraudo
128k17156268
asked Jun 20 '13 at 18:02
rankyranky
61
$endgroup$
The rank of $(1-I_n)$, where $1$ is the $n times n$ all-1 matrix and $I_n$ the $n times n$ identity matrix, seems to be $n$.
How to prove this concisely?
linear-algebra matrices
linear-algebra matrices
edited Jun 20 '13 at 18:57
Davide Giraudo
128k17156268
asked Jun 20 '13 at 18:02
rankyranky
61
edited Jun 20 '13 at 18:57
Davide Giraudo
128k17156268
asked Jun 20 '13 at 18:02
rankyranky
61
edited Jun 20 '13 at 18:57
Davide Giraudo
128k17156268
edited Jun 20 '13 at 18:57
Davide Giraudo
128k17156268
edited Jun 20 '13 at 18:57
Davide Giraudo
128k17156268
128k17156268
asked Jun 20 '13 at 18:02
rankyranky
61
asked Jun 20 '13 at 18:02
rankyranky
61
asked Jun 20 '13 at 18:02
rankyranky
61
61
2
$begingroup$
Well if you look at a example. When you subtract a matrix filled with ones from a matrix with 1`s going down in a diagonal this creates all columns that are linearly independent due to the zeroes appearing one position lower from column 1 to column N.
$endgroup$
– user60887
Jun 20 '13 at 18:11
$begingroup$
Maybe easiest is to assume given that $r(A + B) leq r(A) + r(B)$ (subadditivity of rank)? By then choosing $A = 1$ and $B = -(1-I)$, this gives $r(I) leq r(1) + r(-(1-I))$ or $n leq 1 + r(-(1-I))$ and hence $r(1-I) = r(-(1-I)) geq n-1$.
$endgroup$
– ranky
Jun 20 '13 at 18:24
add a comment |
2
$begingroup$
Well if you look at a example. When you subtract a matrix filled with ones from a matrix with 1`s going down in a diagonal this creates all columns that are linearly independent due to the zeroes appearing one position lower from column 1 to column N.
$endgroup$
– user60887
Jun 20 '13 at 18:11
$begingroup$
Maybe easiest is to assume given that $r(A + B) leq r(A) + r(B)$ (subadditivity of rank)? By then choosing $A = 1$ and $B = -(1-I)$, this gives $r(I) leq r(1) + r(-(1-I))$ or $n leq 1 + r(-(1-I))$ and hence $r(1-I) = r(-(1-I)) geq n-1$.
$endgroup$
– ranky
Jun 20 '13 at 18:24
2
2
$begingroup$
Well if you look at a example. When you subtract a matrix filled with ones from a matrix with 1`s going down in a diagonal this creates all columns that are linearly independent due to the zeroes appearing one position lower from column 1 to column N.
$endgroup$
– user60887
Jun 20 '13 at 18:11
$begingroup$
Well if you look at a example. When you subtract a matrix filled with ones from a matrix with 1`s going down in a diagonal this creates all columns that are linearly independent due to the zeroes appearing one position lower from column 1 to column N.
$endgroup$
– user60887
Jun 20 '13 at 18:11
$begingroup$
Maybe easiest is to assume given that $r(A + B) leq r(A) + r(B)$ (subadditivity of rank)? By then choosing $A = 1$ and $B = -(1-I)$, this gives $r(I) leq r(1) + r(-(1-I))$ or $n leq 1 + r(-(1-I))$ and hence $r(1-I) = r(-(1-I)) geq n-1$.
$endgroup$
– ranky
Jun 20 '13 at 18:24
$begingroup$
Maybe easiest is to assume given that $r(A + B) leq r(A) + r(B)$ (subadditivity of rank)? By then choosing $A = 1$ and $B = -(1-I)$, this gives $r(I) leq r(1) + r(-(1-I))$ or $n leq 1 + r(-(1-I))$ and hence $r(1-I) = r(-(1-I)) geq n-1$.
$endgroup$
– ranky
Jun 20 '13 at 18:24
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Let's call $J$ the matrix with all coefficients equal to $1$. Its eigenvalues are $n$ and $0$: in fact $J$ has rank $1$ and so $0$ is an eigenvalue of multiplicity $n-1$; clearly $Jv=nv$ where $v$ is the "all $1$" vector.
So $1$ is not a root of the characteristic polynomial of $J$, that is,
$$
det(J-XI_n)
$$
which means that $det(J-I_n)ne0$.
Of course we assume $n>1$, otherwise the assertion is false.
answered Jun 20 '13 at 18:07
egregegreg
185k1486207
$endgroup$
add a comment |
$begingroup$
If the rank were not to be $n$, there exists a non-zero vector $x$ such that
$$(ee^T - I)x = 0$$
i.e.,
$$sum_{overset{i=1}{i neq j}}^n x_i = 0$$ for all $j in {1,2,ldots,n}$. Can this be true for $x neq 0$ ?
Also, as an aside, the inverse of $(ee^T - I)$ is given by the Sherman-Morrison formula or the more general Woodbury formula
$$- left(I + dfrac{ee^T}{1-e^Te}right) = - left(I - dfrac{ee^T}{n-1}right)$$
In general, if $A$ is invertible, then $A+uv^T$ is invertible when $1+v^TA^{-1}u neq 0$. In your case, this corresponds to the condition that $n-1 neq 0$, i.e., $n neq 1$.
$endgroup$
add a comment |
$begingroup$
If $A$ is any $ntimes n$ matrix of rank$~1$, then it has an eigenspace for$~0$ of dimension$~n-1$, and the eigenvalue$~0$ has at least that (algebraic) multiplicity; the final eigenvalue is $deftr{operatorname{tr}}tr A$ (as that is the sum of the eigenvalues). Therefore $A-lambda I$ with $lambdaneq0$ is invertible unless $lambda=tr A$, in which case $defrk{operatorname{rk}}rk(A-lambda I)=n-1$. In your case if $A$ is the matrix you confusingly call$~1$, you have $tr A=n$, so for $lambda=1$ you get that $A-I$ is invertible unless $n=1$, in which case it has rank$~n-1=0$.
answered Jun 20 '13 at 19:26
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
$endgroup$
add a comment |
$begingroup$
$newcommand{rank}{operatorname{rank}}$Hint: The matrix $A=[a_{ij}]_{ntimes n}$is full column rank ($rank(A)=n$)if and only if it's invertible.
$$det(1-I_n)=(n-1)(-1)^{n-1}neq0$$
use row echelon operations to Compute $det(B)$$$forall i, ~~ 2le ile n :~~~~~~~C _1-C _i,~~~~~~ R_1+R_i$$
wherein $R$ stands for row and $C$ stands for column. Now;$$det(B)=detbegin{pmatrix}
r & a & a & cdots &a \
a & r& a & cdots & a \
vdots & vdots& vdots & ddots & vdots \
a & a & a & cdots & r
end{pmatrix}=(r+(n-1)a)(r-a)^{n-1}$$
edited Jan 30 at 19:36
Martin Sleziak
44.9k10122277
answered Jun 20 '13 at 18:08
M.HM.H
7,32211654
$endgroup$
add a comment |
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4 Answers
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4 Answers
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$begingroup$
Let's call $J$ the matrix with all coefficients equal to $1$. Its eigenvalues are $n$ and $0$: in fact $J$ has rank $1$ and so $0$ is an eigenvalue of multiplicity $n-1$; clearly $Jv=nv$ where $v$ is the "all $1$" vector.
So $1$ is not a root of the characteristic polynomial of $J$, that is,
$$
det(J-XI_n)
$$
which means that $det(J-I_n)ne0$.
Of course we assume $n>1$, otherwise the assertion is false.
answered Jun 20 '13 at 18:07
egregegreg
185k1486207
$endgroup$
add a comment |
$begingroup$
Let's call $J$ the matrix with all coefficients equal to $1$. Its eigenvalues are $n$ and $0$: in fact $J$ has rank $1$ and so $0$ is an eigenvalue of multiplicity $n-1$; clearly $Jv=nv$ where $v$ is the "all $1$" vector.
So $1$ is not a root of the characteristic polynomial of $J$, that is,
$$
det(J-XI_n)
$$
which means that $det(J-I_n)ne0$.
Of course we assume $n>1$, otherwise the assertion is false.
answered Jun 20 '13 at 18:07
egregegreg
185k1486207
$endgroup$
add a comment |
$begingroup$
Let's call $J$ the matrix with all coefficients equal to $1$. Its eigenvalues are $n$ and $0$: in fact $J$ has rank $1$ and so $0$ is an eigenvalue of multiplicity $n-1$; clearly $Jv=nv$ where $v$ is the "all $1$" vector.
So $1$ is not a root of the characteristic polynomial of $J$, that is,
$$
det(J-XI_n)
$$
which means that $det(J-I_n)ne0$.
Of course we assume $n>1$, otherwise the assertion is false.
answered Jun 20 '13 at 18:07
egregegreg
185k1486207
$endgroup$
Let's call $J$ the matrix with all coefficients equal to $1$. Its eigenvalues are $n$ and $0$: in fact $J$ has rank $1$ and so $0$ is an eigenvalue of multiplicity $n-1$; clearly $Jv=nv$ where $v$ is the "all $1$" vector.
So $1$ is not a root of the characteristic polynomial of $J$, that is,
$$
det(J-XI_n)
$$
which means that $det(J-I_n)ne0$.
Of course we assume $n>1$, otherwise the assertion is false.
answered Jun 20 '13 at 18:07
egregegreg
185k1486207
answered Jun 20 '13 at 18:07
egregegreg
185k1486207
answered Jun 20 '13 at 18:07
egregegreg
185k1486207
answered Jun 20 '13 at 18:07
egregegreg
185k1486207
185k1486207
add a comment |
add a comment |
$begingroup$
If the rank were not to be $n$, there exists a non-zero vector $x$ such that
$$(ee^T - I)x = 0$$
i.e.,
$$sum_{overset{i=1}{i neq j}}^n x_i = 0$$ for all $j in {1,2,ldots,n}$. Can this be true for $x neq 0$ ?
Also, as an aside, the inverse of $(ee^T - I)$ is given by the Sherman-Morrison formula or the more general Woodbury formula
$$- left(I + dfrac{ee^T}{1-e^Te}right) = - left(I - dfrac{ee^T}{n-1}right)$$
In general, if $A$ is invertible, then $A+uv^T$ is invertible when $1+v^TA^{-1}u neq 0$. In your case, this corresponds to the condition that $n-1 neq 0$, i.e., $n neq 1$.
$endgroup$
add a comment |
$begingroup$
If the rank were not to be $n$, there exists a non-zero vector $x$ such that
$$(ee^T - I)x = 0$$
i.e.,
$$sum_{overset{i=1}{i neq j}}^n x_i = 0$$ for all $j in {1,2,ldots,n}$. Can this be true for $x neq 0$ ?
Also, as an aside, the inverse of $(ee^T - I)$ is given by the Sherman-Morrison formula or the more general Woodbury formula
$$- left(I + dfrac{ee^T}{1-e^Te}right) = - left(I - dfrac{ee^T}{n-1}right)$$
In general, if $A$ is invertible, then $A+uv^T$ is invertible when $1+v^TA^{-1}u neq 0$. In your case, this corresponds to the condition that $n-1 neq 0$, i.e., $n neq 1$.
$endgroup$
add a comment |
$begingroup$
If the rank were not to be $n$, there exists a non-zero vector $x$ such that
$$(ee^T - I)x = 0$$
i.e.,
$$sum_{overset{i=1}{i neq j}}^n x_i = 0$$ for all $j in {1,2,ldots,n}$. Can this be true for $x neq 0$ ?
Also, as an aside, the inverse of $(ee^T - I)$ is given by the Sherman-Morrison formula or the more general Woodbury formula
$$- left(I + dfrac{ee^T}{1-e^Te}right) = - left(I - dfrac{ee^T}{n-1}right)$$
In general, if $A$ is invertible, then $A+uv^T$ is invertible when $1+v^TA^{-1}u neq 0$. In your case, this corresponds to the condition that $n-1 neq 0$, i.e., $n neq 1$.
$endgroup$
If the rank were not to be $n$, there exists a non-zero vector $x$ such that
$$(ee^T - I)x = 0$$
i.e.,
$$sum_{overset{i=1}{i neq j}}^n x_i = 0$$ for all $j in {1,2,ldots,n}$. Can this be true for $x neq 0$ ?
Also, as an aside, the inverse of $(ee^T - I)$ is given by the Sherman-Morrison formula or the more general Woodbury formula
$$- left(I + dfrac{ee^T}{1-e^Te}right) = - left(I - dfrac{ee^T}{n-1}right)$$
In general, if $A$ is invertible, then $A+uv^T$ is invertible when $1+v^TA^{-1}u neq 0$. In your case, this corresponds to the condition that $n-1 neq 0$, i.e., $n neq 1$.
answered Jun 20 '13 at 18:08
user17762
add a comment |
add a comment |
$begingroup$
If $A$ is any $ntimes n$ matrix of rank$~1$, then it has an eigenspace for$~0$ of dimension$~n-1$, and the eigenvalue$~0$ has at least that (algebraic) multiplicity; the final eigenvalue is $deftr{operatorname{tr}}tr A$ (as that is the sum of the eigenvalues). Therefore $A-lambda I$ with $lambdaneq0$ is invertible unless $lambda=tr A$, in which case $defrk{operatorname{rk}}rk(A-lambda I)=n-1$. In your case if $A$ is the matrix you confusingly call$~1$, you have $tr A=n$, so for $lambda=1$ you get that $A-I$ is invertible unless $n=1$, in which case it has rank$~n-1=0$.
answered Jun 20 '13 at 19:26
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
$endgroup$
add a comment |
$begingroup$
If $A$ is any $ntimes n$ matrix of rank$~1$, then it has an eigenspace for$~0$ of dimension$~n-1$, and the eigenvalue$~0$ has at least that (algebraic) multiplicity; the final eigenvalue is $deftr{operatorname{tr}}tr A$ (as that is the sum of the eigenvalues). Therefore $A-lambda I$ with $lambdaneq0$ is invertible unless $lambda=tr A$, in which case $defrk{operatorname{rk}}rk(A-lambda I)=n-1$. In your case if $A$ is the matrix you confusingly call$~1$, you have $tr A=n$, so for $lambda=1$ you get that $A-I$ is invertible unless $n=1$, in which case it has rank$~n-1=0$.
answered Jun 20 '13 at 19:26
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
$endgroup$
add a comment |
$begingroup$
If $A$ is any $ntimes n$ matrix of rank$~1$, then it has an eigenspace for$~0$ of dimension$~n-1$, and the eigenvalue$~0$ has at least that (algebraic) multiplicity; the final eigenvalue is $deftr{operatorname{tr}}tr A$ (as that is the sum of the eigenvalues). Therefore $A-lambda I$ with $lambdaneq0$ is invertible unless $lambda=tr A$, in which case $defrk{operatorname{rk}}rk(A-lambda I)=n-1$. In your case if $A$ is the matrix you confusingly call$~1$, you have $tr A=n$, so for $lambda=1$ you get that $A-I$ is invertible unless $n=1$, in which case it has rank$~n-1=0$.
answered Jun 20 '13 at 19:26
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
$endgroup$
If $A$ is any $ntimes n$ matrix of rank$~1$, then it has an eigenspace for$~0$ of dimension$~n-1$, and the eigenvalue$~0$ has at least that (algebraic) multiplicity; the final eigenvalue is $deftr{operatorname{tr}}tr A$ (as that is the sum of the eigenvalues). Therefore $A-lambda I$ with $lambdaneq0$ is invertible unless $lambda=tr A$, in which case $defrk{operatorname{rk}}rk(A-lambda I)=n-1$. In your case if $A$ is the matrix you confusingly call$~1$, you have $tr A=n$, so for $lambda=1$ you get that $A-I$ is invertible unless $n=1$, in which case it has rank$~n-1=0$.
answered Jun 20 '13 at 19:26
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
edited Jun 20 '13 at 19:34
answered Jun 20 '13 at 19:26
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
answered Jun 20 '13 at 19:26
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
answered Jun 20 '13 at 19:26
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
88.7k5111230
add a comment |
add a comment |
$begingroup$
$newcommand{rank}{operatorname{rank}}$Hint: The matrix $A=[a_{ij}]_{ntimes n}$is full column rank ($rank(A)=n$)if and only if it's invertible.
$$det(1-I_n)=(n-1)(-1)^{n-1}neq0$$
use row echelon operations to Compute $det(B)$$$forall i, ~~ 2le ile n :~~~~~~~C _1-C _i,~~~~~~ R_1+R_i$$
wherein $R$ stands for row and $C$ stands for column. Now;$$det(B)=detbegin{pmatrix}
r & a & a & cdots &a \
a & r& a & cdots & a \
vdots & vdots& vdots & ddots & vdots \
a & a & a & cdots & r
end{pmatrix}=(r+(n-1)a)(r-a)^{n-1}$$
edited Jan 30 at 19:36
Martin Sleziak
44.9k10122277
answered Jun 20 '13 at 18:08
M.HM.H
7,32211654
$endgroup$
add a comment |
$begingroup$
$newcommand{rank}{operatorname{rank}}$Hint: The matrix $A=[a_{ij}]_{ntimes n}$is full column rank ($rank(A)=n$)if and only if it's invertible.
$$det(1-I_n)=(n-1)(-1)^{n-1}neq0$$
use row echelon operations to Compute $det(B)$$$forall i, ~~ 2le ile n :~~~~~~~C _1-C _i,~~~~~~ R_1+R_i$$
wherein $R$ stands for row and $C$ stands for column. Now;$$det(B)=detbegin{pmatrix}
r & a & a & cdots &a \
a & r& a & cdots & a \
vdots & vdots& vdots & ddots & vdots \
a & a & a & cdots & r
end{pmatrix}=(r+(n-1)a)(r-a)^{n-1}$$
edited Jan 30 at 19:36
Martin Sleziak
44.9k10122277
answered Jun 20 '13 at 18:08
M.HM.H
7,32211654
$endgroup$
add a comment |
$begingroup$
$newcommand{rank}{operatorname{rank}}$Hint: The matrix $A=[a_{ij}]_{ntimes n}$is full column rank ($rank(A)=n$)if and only if it's invertible.
$$det(1-I_n)=(n-1)(-1)^{n-1}neq0$$
use row echelon operations to Compute $det(B)$$$forall i, ~~ 2le ile n :~~~~~~~C _1-C _i,~~~~~~ R_1+R_i$$
wherein $R$ stands for row and $C$ stands for column. Now;$$det(B)=detbegin{pmatrix}
r & a & a & cdots &a \
a & r& a & cdots & a \
vdots & vdots& vdots & ddots & vdots \
a & a & a & cdots & r
end{pmatrix}=(r+(n-1)a)(r-a)^{n-1}$$
edited Jan 30 at 19:36
Martin Sleziak
44.9k10122277
answered Jun 20 '13 at 18:08
M.HM.H
7,32211654
$endgroup$
$newcommand{rank}{operatorname{rank}}$Hint: The matrix $A=[a_{ij}]_{ntimes n}$is full column rank ($rank(A)=n$)if and only if it's invertible.
$$det(1-I_n)=(n-1)(-1)^{n-1}neq0$$
use row echelon operations to Compute $det(B)$$$forall i, ~~ 2le ile n :~~~~~~~C _1-C _i,~~~~~~ R_1+R_i$$
wherein $R$ stands for row and $C$ stands for column. Now;$$det(B)=detbegin{pmatrix}
r & a & a & cdots &a \
a & r& a & cdots & a \
vdots & vdots& vdots & ddots & vdots \
a & a & a & cdots & r
end{pmatrix}=(r+(n-1)a)(r-a)^{n-1}$$
edited Jan 30 at 19:36
Martin Sleziak
44.9k10122277
answered Jun 20 '13 at 18:08
M.HM.H
7,32211654
edited Jan 30 at 19:36
Martin Sleziak
44.9k10122277
edited Jan 30 at 19:36
Martin Sleziak
44.9k10122277
edited Jan 30 at 19:36
Martin Sleziak
44.9k10122277
44.9k10122277
answered Jun 20 '13 at 18:08
M.HM.H
7,32211654
answered Jun 20 '13 at 18:08
M.HM.H
7,32211654
answered Jun 20 '13 at 18:08
M.HM.H
7,32211654
7,32211654
add a comment |
add a comment |
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2
$begingroup$
Well if you look at a example. When you subtract a matrix filled with ones from a matrix with 1`s going down in a diagonal this creates all columns that are linearly independent due to the zeroes appearing one position lower from column 1 to column N.
$endgroup$
– user60887
Jun 20 '13 at 18:11
$begingroup$
Maybe easiest is to assume given that $r(A + B) leq r(A) + r(B)$ (subadditivity of rank)? By then choosing $A = 1$ and $B = -(1-I)$, this gives $r(I) leq r(1) + r(-(1-I))$ or $n leq 1 + r(-(1-I))$ and hence $r(1-I) = r(-(1-I)) geq n-1$.
$endgroup$
– ranky
Jun 20 '13 at 18:24