matrices and determinant
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Let $Ain mathcal{M}_{n,p}(mathbb{R})$ and $G={}^{t!}AA$
We assume that $operatorname{rk}(A)=p$
To show that $det(G)>0$ the argument provided is $det G=(det A)^2>0$, unfortunately I don't understand, because for me this argument holds only if $Ain mathcal{M}_{p}(mathbb{R})$ (a square matrix)
linear-algebra determinant
edited Nov 16 at 22:20
egreg
174k1383198
asked Nov 16 at 22:03
Stu
1,1011313
|
show 2 more comments
up vote
2
down vote
favorite
Let $Ain mathcal{M}_{n,p}(mathbb{R})$ and $G={}^{t!}AA$
We assume that $operatorname{rk}(A)=p$
To show that $det(G)>0$ the argument provided is $det G=(det A)^2>0$, unfortunately I don't understand, because for me this argument holds only if $Ain mathcal{M}_{p}(mathbb{R})$ (a square matrix)
linear-algebra determinant
edited Nov 16 at 22:20
egreg
174k1383198
asked Nov 16 at 22:03
Stu
1,1011313
$G$ doesn't make sense... I'm not familiar with putting the transpose on the left of the matrix. Do you mean $G=A^tA$? Also how do you know that $det(A)neq 0$? Finally, what is your question?
– Squirtle
Nov 16 at 22:06
What is $rg(A)$?
– Robert Lewis
Nov 16 at 22:13
1
sorry but $$rg(A)=p$$
– Stu
Nov 16 at 22:14
Google "Cauchy-Binet identity".
– darij grinberg
Nov 16 at 22:14
2
@Stu The statement is false, unless you assume $nle p$. For $n>p$, the matrix $G$ is not invertible, so its determinant is $0$. If the rank of $A$ is $p$ and $nle p$, then $G$ is indeed invertible.
– egreg
Nov 16 at 22:24
|
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $Ain mathcal{M}_{n,p}(mathbb{R})$ and $G={}^{t!}AA$
We assume that $operatorname{rk}(A)=p$
To show that $det(G)>0$ the argument provided is $det G=(det A)^2>0$, unfortunately I don't understand, because for me this argument holds only if $Ain mathcal{M}_{p}(mathbb{R})$ (a square matrix)
linear-algebra determinant
edited Nov 16 at 22:20
egreg
174k1383198
asked Nov 16 at 22:03
Stu
1,1011313
Let $Ain mathcal{M}_{n,p}(mathbb{R})$ and $G={}^{t!}AA$
We assume that $operatorname{rk}(A)=p$
To show that $det(G)>0$ the argument provided is $det G=(det A)^2>0$, unfortunately I don't understand, because for me this argument holds only if $Ain mathcal{M}_{p}(mathbb{R})$ (a square matrix)
linear-algebra determinant
linear-algebra determinant
edited Nov 16 at 22:20
egreg
174k1383198
asked Nov 16 at 22:03
Stu
1,1011313
edited Nov 16 at 22:20
egreg
174k1383198
asked Nov 16 at 22:03
Stu
1,1011313
edited Nov 16 at 22:20
egreg
174k1383198
edited Nov 16 at 22:20
egreg
174k1383198
edited Nov 16 at 22:20
egreg
174k1383198
174k1383198
asked Nov 16 at 22:03
Stu
1,1011313
asked Nov 16 at 22:03
Stu
1,1011313
asked Nov 16 at 22:03
Stu
1,1011313
1,1011313
$G$ doesn't make sense... I'm not familiar with putting the transpose on the left of the matrix. Do you mean $G=A^tA$? Also how do you know that $det(A)neq 0$? Finally, what is your question?
– Squirtle
Nov 16 at 22:06
What is $rg(A)$?
– Robert Lewis
Nov 16 at 22:13
1
sorry but $$rg(A)=p$$
– Stu
Nov 16 at 22:14
Google "Cauchy-Binet identity".
– darij grinberg
Nov 16 at 22:14
2
@Stu The statement is false, unless you assume $nle p$. For $n>p$, the matrix $G$ is not invertible, so its determinant is $0$. If the rank of $A$ is $p$ and $nle p$, then $G$ is indeed invertible.
– egreg
Nov 16 at 22:24
|
show 2 more comments
$G$ doesn't make sense... I'm not familiar with putting the transpose on the left of the matrix. Do you mean $G=A^tA$? Also how do you know that $det(A)neq 0$? Finally, what is your question?
– Squirtle
Nov 16 at 22:06
What is $rg(A)$?
– Robert Lewis
Nov 16 at 22:13
1
sorry but $$rg(A)=p$$
– Stu
Nov 16 at 22:14
Google "Cauchy-Binet identity".
– darij grinberg
Nov 16 at 22:14
2
@Stu The statement is false, unless you assume $nle p$. For $n>p$, the matrix $G$ is not invertible, so its determinant is $0$. If the rank of $A$ is $p$ and $nle p$, then $G$ is indeed invertible.
– egreg
Nov 16 at 22:24
$G$ doesn't make sense... I'm not familiar with putting the transpose on the left of the matrix. Do you mean $G=A^tA$? Also how do you know that $det(A)neq 0$? Finally, what is your question?
– Squirtle
Nov 16 at 22:06
$G$ doesn't make sense... I'm not familiar with putting the transpose on the left of the matrix. Do you mean $G=A^tA$? Also how do you know that $det(A)neq 0$? Finally, what is your question?
– Squirtle
Nov 16 at 22:06
What is $rg(A)$?
– Robert Lewis
Nov 16 at 22:13
What is $rg(A)$?
– Robert Lewis
Nov 16 at 22:13
1
1
sorry but $$rg(A)=p$$
– Stu
Nov 16 at 22:14
sorry but $$rg(A)=p$$
– Stu
Nov 16 at 22:14
Google "Cauchy-Binet identity".
– darij grinberg
Nov 16 at 22:14
Google "Cauchy-Binet identity".
– darij grinberg
Nov 16 at 22:14
2
2
@Stu The statement is false, unless you assume $nle p$. For $n>p$, the matrix $G$ is not invertible, so its determinant is $0$. If the rank of $A$ is $p$ and $nle p$, then $G$ is indeed invertible.
– egreg
Nov 16 at 22:24
@Stu The statement is false, unless you assume $nle p$. For $n>p$, the matrix $G$ is not invertible, so its determinant is $0$. If the rank of $A$ is $p$ and $nle p$, then $G$ is indeed invertible.
– egreg
Nov 16 at 22:24
|
show 2 more comments
1 Answer
1
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oldest
votes
up vote
3
down vote
accepted
The expression $(det A)^2$ indeed doesn't make sense because $A$ is not necessarily square. As pointed out in a comment, you may use Cauchy-Binet formula to prove the statement. Alternatively, as $A$ has full column rank, if $xinmathbb R^p$ is nonzero, then $Axne0$ and $x^TGx=x^TA^TAx=|Ax|^2>0$. Therefore $G$ is symmetric positive definite and it has a positive determinant.
answered Nov 17 at 1:45
user1551
70.5k566125
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The expression $(det A)^2$ indeed doesn't make sense because $A$ is not necessarily square. As pointed out in a comment, you may use Cauchy-Binet formula to prove the statement. Alternatively, as $A$ has full column rank, if $xinmathbb R^p$ is nonzero, then $Axne0$ and $x^TGx=x^TA^TAx=|Ax|^2>0$. Therefore $G$ is symmetric positive definite and it has a positive determinant.
answered Nov 17 at 1:45
user1551
70.5k566125
add a comment |
up vote
3
down vote
accepted
The expression $(det A)^2$ indeed doesn't make sense because $A$ is not necessarily square. As pointed out in a comment, you may use Cauchy-Binet formula to prove the statement. Alternatively, as $A$ has full column rank, if $xinmathbb R^p$ is nonzero, then $Axne0$ and $x^TGx=x^TA^TAx=|Ax|^2>0$. Therefore $G$ is symmetric positive definite and it has a positive determinant.
answered Nov 17 at 1:45
user1551
70.5k566125
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The expression $(det A)^2$ indeed doesn't make sense because $A$ is not necessarily square. As pointed out in a comment, you may use Cauchy-Binet formula to prove the statement. Alternatively, as $A$ has full column rank, if $xinmathbb R^p$ is nonzero, then $Axne0$ and $x^TGx=x^TA^TAx=|Ax|^2>0$. Therefore $G$ is symmetric positive definite and it has a positive determinant.
answered Nov 17 at 1:45
user1551
70.5k566125
The expression $(det A)^2$ indeed doesn't make sense because $A$ is not necessarily square. As pointed out in a comment, you may use Cauchy-Binet formula to prove the statement. Alternatively, as $A$ has full column rank, if $xinmathbb R^p$ is nonzero, then $Axne0$ and $x^TGx=x^TA^TAx=|Ax|^2>0$. Therefore $G$ is symmetric positive definite and it has a positive determinant.
answered Nov 17 at 1:45
user1551
70.5k566125
answered Nov 17 at 1:45
user1551
70.5k566125
answered Nov 17 at 1:45
user1551
70.5k566125
answered Nov 17 at 1:45
user1551
70.5k566125
70.5k566125
add a comment |
add a comment |
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$G$ doesn't make sense... I'm not familiar with putting the transpose on the left of the matrix. Do you mean $G=A^tA$? Also how do you know that $det(A)neq 0$? Finally, what is your question?
– Squirtle
Nov 16 at 22:06
What is $rg(A)$?
– Robert Lewis
Nov 16 at 22:13
1
sorry but $$rg(A)=p$$
– Stu
Nov 16 at 22:14
Google "Cauchy-Binet identity".
– darij grinberg
Nov 16 at 22:14
2
@Stu The statement is false, unless you assume $nle p$. For $n>p$, the matrix $G$ is not invertible, so its determinant is $0$. If the rank of $A$ is $p$ and $nle p$, then $G$ is indeed invertible.
– egreg
Nov 16 at 22:24