Positive Definiteness problem
$begingroup$
Consider the positive definite matrix $B succ 0$ and the matrix ( not necessarily square) $A$. what can we say abut the positive definiteness of:
$$ A^prime B A$$
My hunch is that this is positive semi definite due to the lack of constraints on $A$ meaning there is a vector $x neq 0$ such that $Ax=0$.
begin{align}
x^prime A^prime B A x &=(Ax)^prime B(Ax) \
&rightarrow trace(B||Ax||^2) ge 0
end{align}
the eigenvalues of this are greater than or equal to zero.
linear-algebra positive-definite
asked Dec 3 '18 at 11:44
p32fr4p32fr4
3717
$endgroup$
add a comment |
$begingroup$
Consider the positive definite matrix $B succ 0$ and the matrix ( not necessarily square) $A$. what can we say abut the positive definiteness of:
$$ A^prime B A$$
My hunch is that this is positive semi definite due to the lack of constraints on $A$ meaning there is a vector $x neq 0$ such that $Ax=0$.
begin{align}
x^prime A^prime B A x &=(Ax)^prime B(Ax) \
&rightarrow trace(B||Ax||^2) ge 0
end{align}
the eigenvalues of this are greater than or equal to zero.
linear-algebra positive-definite
asked Dec 3 '18 at 11:44
p32fr4p32fr4
3717
$endgroup$
add a comment |
$begingroup$
Consider the positive definite matrix $B succ 0$ and the matrix ( not necessarily square) $A$. what can we say abut the positive definiteness of:
$$ A^prime B A$$
My hunch is that this is positive semi definite due to the lack of constraints on $A$ meaning there is a vector $x neq 0$ such that $Ax=0$.
begin{align}
x^prime A^prime B A x &=(Ax)^prime B(Ax) \
&rightarrow trace(B||Ax||^2) ge 0
end{align}
the eigenvalues of this are greater than or equal to zero.
linear-algebra positive-definite
asked Dec 3 '18 at 11:44
p32fr4p32fr4
3717
$endgroup$
Consider the positive definite matrix $B succ 0$ and the matrix ( not necessarily square) $A$. what can we say abut the positive definiteness of:
$$ A^prime B A$$
My hunch is that this is positive semi definite due to the lack of constraints on $A$ meaning there is a vector $x neq 0$ such that $Ax=0$.
begin{align}
x^prime A^prime B A x &=(Ax)^prime B(Ax) \
&rightarrow trace(B||Ax||^2) ge 0
end{align}
the eigenvalues of this are greater than or equal to zero.
linear-algebra positive-definite
linear-algebra positive-definite
asked Dec 3 '18 at 11:44
p32fr4p32fr4
3717
asked Dec 3 '18 at 11:44
p32fr4p32fr4
3717
asked Dec 3 '18 at 11:44
p32fr4p32fr4
3717
asked Dec 3 '18 at 11:44
p32fr4p32fr4
3717
asked Dec 3 '18 at 11:44
p32fr4p32fr4
3717
3717
add a comment |
add a comment |
1 Answer
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$begingroup$
You are right, but there is an easier way to see it. Why not denote $Ax$ as $y$ so you don't know $y$ equals 0 or not, but you can always have $y'Aygeq 0$.
answered Dec 3 '18 at 12:00
U2647U2647
447
$endgroup$
$begingroup$
That's great thank you. I was pretty sure but just needed the second opinion!
$endgroup$
– p32fr4
Dec 3 '18 at 12:08
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are right, but there is an easier way to see it. Why not denote $Ax$ as $y$ so you don't know $y$ equals 0 or not, but you can always have $y'Aygeq 0$.
answered Dec 3 '18 at 12:00
U2647U2647
447
$endgroup$
$begingroup$
That's great thank you. I was pretty sure but just needed the second opinion!
$endgroup$
– p32fr4
Dec 3 '18 at 12:08
add a comment |
$begingroup$
You are right, but there is an easier way to see it. Why not denote $Ax$ as $y$ so you don't know $y$ equals 0 or not, but you can always have $y'Aygeq 0$.
answered Dec 3 '18 at 12:00
U2647U2647
447
$endgroup$
$begingroup$
That's great thank you. I was pretty sure but just needed the second opinion!
$endgroup$
– p32fr4
Dec 3 '18 at 12:08
add a comment |
$begingroup$
You are right, but there is an easier way to see it. Why not denote $Ax$ as $y$ so you don't know $y$ equals 0 or not, but you can always have $y'Aygeq 0$.
answered Dec 3 '18 at 12:00
U2647U2647
447
$endgroup$
You are right, but there is an easier way to see it. Why not denote $Ax$ as $y$ so you don't know $y$ equals 0 or not, but you can always have $y'Aygeq 0$.
answered Dec 3 '18 at 12:00
U2647U2647
447
answered Dec 3 '18 at 12:00
U2647U2647
447
answered Dec 3 '18 at 12:00
U2647U2647
447
answered Dec 3 '18 at 12:00
U2647U2647
447
447
$begingroup$
That's great thank you. I was pretty sure but just needed the second opinion!
$endgroup$
– p32fr4
Dec 3 '18 at 12:08
add a comment |
$begingroup$
That's great thank you. I was pretty sure but just needed the second opinion!
$endgroup$
– p32fr4
Dec 3 '18 at 12:08
$begingroup$
That's great thank you. I was pretty sure but just needed the second opinion!
$endgroup$
– p32fr4
Dec 3 '18 at 12:08
$begingroup$
That's great thank you. I was pretty sure but just needed the second opinion!
$endgroup$
– p32fr4
Dec 3 '18 at 12:08
add a comment |
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