For matrix $A_{ntimes n},X_{ntimes p}$, $rank(X)=p$. Prove that if $M(X)subset M(A)$, $X^TAX>0$.
0
$begingroup$
Since $M(X)subset M(A)$, I know that there exist a matrix B, s.t. $X=AB$.
Since $rank(X)=p$，I know that X is full rank, which means $Xy=0$ only has zero solution.
But I don't know how to complete the proof.
linear-algebra
asked Dec 22 '18 at 11:47
cholechole
333
$endgroup$
add a comment |
0
$begingroup$
Since $M(X)subset M(A)$, I know that there exist a matrix B, s.t. $X=AB$.
Since $rank(X)=p$，I know that X is full rank, which means $Xy=0$ only has zero solution.
But I don't know how to complete the proof.
linear-algebra
asked Dec 22 '18 at 11:47
cholechole
333
$endgroup$
$begingroup$
Hi what is $M$?
$endgroup$
– user122049
Dec 22 '18 at 11:53
$begingroup$
What's $M(X)$? Also by $X^{mathrm T}AX >0$, do you mean that $X^{mathrm T}AX$ is positive definite?
$endgroup$
– xbh
Dec 22 '18 at 11:54
$begingroup$
M(X) is the linear space formed by X.
$endgroup$
– chole
Dec 22 '18 at 13:31
$begingroup$
I think it means that $X^TAX$ is positive definite.
$endgroup$
– chole
Dec 22 '18 at 13:32
add a comment |
0
0
0
$begingroup$
Since $M(X)subset M(A)$, I know that there exist a matrix B, s.t. $X=AB$.
Since $rank(X)=p$，I know that X is full rank, which means $Xy=0$ only has zero solution.
But I don't know how to complete the proof.
linear-algebra
asked Dec 22 '18 at 11:47
cholechole
333
$endgroup$
Since $M(X)subset M(A)$, I know that there exist a matrix B, s.t. $X=AB$.
Since $rank(X)=p$，I know that X is full rank, which means $Xy=0$ only has zero solution.
But I don't know how to complete the proof.
linear-algebra
linear-algebra
asked Dec 22 '18 at 11:47
cholechole
333
asked Dec 22 '18 at 11:47
cholechole
333
asked Dec 22 '18 at 11:47
cholechole
333
asked Dec 22 '18 at 11:47
cholechole
333
asked Dec 22 '18 at 11:47
cholechole
333
333
$begingroup$
Hi what is $M$?
$endgroup$
– user122049
Dec 22 '18 at 11:53
$begingroup$
What's $M(X)$? Also by $X^{mathrm T}AX >0$, do you mean that $X^{mathrm T}AX$ is positive definite?
$endgroup$
– xbh
Dec 22 '18 at 11:54
$begingroup$
M(X) is the linear space formed by X.
$endgroup$
– chole
Dec 22 '18 at 13:31
$begingroup$
I think it means that $X^TAX$ is positive definite.
$endgroup$
– chole
Dec 22 '18 at 13:32
add a comment |
$begingroup$
Hi what is $M$?
$endgroup$
– user122049
Dec 22 '18 at 11:53
$begingroup$
What's $M(X)$? Also by $X^{mathrm T}AX >0$, do you mean that $X^{mathrm T}AX$ is positive definite?
$endgroup$
– xbh
Dec 22 '18 at 11:54
$begingroup$
M(X) is the linear space formed by X.
$endgroup$
– chole
Dec 22 '18 at 13:31
$begingroup$
I think it means that $X^TAX$ is positive definite.
$endgroup$
– chole
Dec 22 '18 at 13:32
$begingroup$
Hi what is $M$?
$endgroup$
– user122049
Dec 22 '18 at 11:53
$begingroup$
Hi what is $M$?
$endgroup$
– user122049
Dec 22 '18 at 11:53
$begingroup$
What's $M(X)$? Also by $X^{mathrm T}AX >0$, do you mean that $X^{mathrm T}AX$ is positive definite?
$endgroup$
– xbh
Dec 22 '18 at 11:54
$begingroup$
What's $M(X)$? Also by $X^{mathrm T}AX >0$, do you mean that $X^{mathrm T}AX$ is positive definite?
$endgroup$
– xbh
Dec 22 '18 at 11:54
$begingroup$
M(X) is the linear space formed by X.
$endgroup$
– chole
Dec 22 '18 at 13:31
$begingroup$
M(X) is the linear space formed by X.
$endgroup$
– chole
Dec 22 '18 at 13:31
$begingroup$
I think it means that $X^TAX$ is positive definite.
$endgroup$
– chole
Dec 22 '18 at 13:32
$begingroup$
I think it means that $X^TAX$ is positive definite.
$endgroup$
– chole
Dec 22 '18 at 13:32
add a comment |
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$begingroup$
Hi what is $M$?
$endgroup$
– user122049
Dec 22 '18 at 11:53
$begingroup$
What's $M(X)$? Also by $X^{mathrm T}AX >0$, do you mean that $X^{mathrm T}AX$ is positive definite?
$endgroup$
– xbh
Dec 22 '18 at 11:54
$begingroup$
M(X) is the linear space formed by X.
$endgroup$
– chole
Dec 22 '18 at 13:31
$begingroup$
I think it means that $X^TAX$ is positive definite.
$endgroup$
– chole
Dec 22 '18 at 13:32