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Negating off-diagonal blocks retains positive-semidefiniteness?
Is always possible to find a generalized eigenvector for the Jordan basis M?Calculating matrix normLTI system state transition matrixPositive Semidefiniteness on off diagonal pertibationEigenvalues and eigenvectors example on Strang's bookCompute preimage of linear transformation$f(x) = x^TMx$, using Lagrange multipliers to prove SVD decompositionA is a non-invertible matrix, for which values $lambda in mathbbR$ does the matrix equation $AX=lambda X$ have non-trivial solutions?Entering and leaving variables in this linear programming problemHow to prove a matrix is positive semidefinite?
$begingroup$
I am trying to follow some notes that state
$$
M=
beginbmatrix
A&B^T\B&C
endbmatrix
succeq 0
Longleftrightarrow
M'=
beginbmatrix
A&-B^T\-B&C
endbmatrix
succeq 0$$
and I want to prove this to myself (just the $Rightarrow$ direction because it's trivial to go the other way once one direction is proven).
Clearly, $
beginbmatrix
A&B^T\B&C
endbmatrix
succeq 0Longrightarrow A,Csucceq0
$
by computing $x^TMx$ (for $x=beginbmatrixv\0endbmatrix$ and $beginbmatrix0\vendbmatrix$) which is $geq0 forall v$ by definition.
Then for a general $x=beginbmatrixx_1\x_2endbmatrix$,
$$
beginbmatrix
x_1^T&x_2^T
endbmatrix
M
beginbmatrix
x_1\x_2
endbmatrix
=
underbracex_1^TAx_1_geq0
+
underbracex_2^TCx_2_geq0
+2x_2^TBx_1geq2x_2^TBx_1
$$
and
$$
beginbmatrix
x_1^T&x_2^T
endbmatrix
M'
beginbmatrix
x_1\x_2
endbmatrix
=x_1^TAx_1+x_2^TCx_2-2x_2^TBx_1geq-2x_2^TBx_1.
$$
Then I get stuck.
linear-algebra positive-semidefinite block-matrices
asked Mar 17 at 21:57
DanDan
927
$endgroup$
add a comment |
$begingroup$
I am trying to follow some notes that state
$$
M=
beginbmatrix
A&B^T\B&C
endbmatrix
succeq 0
Longleftrightarrow
M'=
beginbmatrix
A&-B^T\-B&C
endbmatrix
succeq 0$$
and I want to prove this to myself (just the $Rightarrow$ direction because it's trivial to go the other way once one direction is proven).
Clearly, $
beginbmatrix
A&B^T\B&C
endbmatrix
succeq 0Longrightarrow A,Csucceq0
$
by computing $x^TMx$ (for $x=beginbmatrixv\0endbmatrix$ and $beginbmatrix0\vendbmatrix$) which is $geq0 forall v$ by definition.
Then for a general $x=beginbmatrixx_1\x_2endbmatrix$,
$$
beginbmatrix
x_1^T&x_2^T
endbmatrix
M
beginbmatrix
x_1\x_2
endbmatrix
=
underbracex_1^TAx_1_geq0
+
underbracex_2^TCx_2_geq0
+2x_2^TBx_1geq2x_2^TBx_1
$$
and
$$
beginbmatrix
x_1^T&x_2^T
endbmatrix
M'
beginbmatrix
x_1\x_2
endbmatrix
=x_1^TAx_1+x_2^TCx_2-2x_2^TBx_1geq-2x_2^TBx_1.
$$
Then I get stuck.
linear-algebra positive-semidefinite block-matrices
asked Mar 17 at 21:57
DanDan
927
$endgroup$
add a comment |
$begingroup$
I am trying to follow some notes that state
$$
M=
beginbmatrix
A&B^T\B&C
endbmatrix
succeq 0
Longleftrightarrow
M'=
beginbmatrix
A&-B^T\-B&C
endbmatrix
succeq 0$$
and I want to prove this to myself (just the $Rightarrow$ direction because it's trivial to go the other way once one direction is proven).
Clearly, $
beginbmatrix
A&B^T\B&C
endbmatrix
succeq 0Longrightarrow A,Csucceq0
$
by computing $x^TMx$ (for $x=beginbmatrixv\0endbmatrix$ and $beginbmatrix0\vendbmatrix$) which is $geq0 forall v$ by definition.
Then for a general $x=beginbmatrixx_1\x_2endbmatrix$,
$$
beginbmatrix
x_1^T&x_2^T
endbmatrix
M
beginbmatrix
x_1\x_2
endbmatrix
=
underbracex_1^TAx_1_geq0
+
underbracex_2^TCx_2_geq0
+2x_2^TBx_1geq2x_2^TBx_1
$$
and
$$
beginbmatrix
x_1^T&x_2^T
endbmatrix
M'
beginbmatrix
x_1\x_2
endbmatrix
=x_1^TAx_1+x_2^TCx_2-2x_2^TBx_1geq-2x_2^TBx_1.
$$
Then I get stuck.
linear-algebra positive-semidefinite block-matrices
asked Mar 17 at 21:57
DanDan
927
$endgroup$
I am trying to follow some notes that state
$$
M=
beginbmatrix
A&B^T\B&C
endbmatrix
succeq 0
Longleftrightarrow
M'=
beginbmatrix
A&-B^T\-B&C
endbmatrix
succeq 0$$
and I want to prove this to myself (just the $Rightarrow$ direction because it's trivial to go the other way once one direction is proven).
Clearly, $
beginbmatrix
A&B^T\B&C
endbmatrix
succeq 0Longrightarrow A,Csucceq0
$
by computing $x^TMx$ (for $x=beginbmatrixv\0endbmatrix$ and $beginbmatrix0\vendbmatrix$) which is $geq0 forall v$ by definition.
Then for a general $x=beginbmatrixx_1\x_2endbmatrix$,
$$
beginbmatrix
x_1^T&x_2^T
endbmatrix
M
beginbmatrix
x_1\x_2
endbmatrix
=
underbracex_1^TAx_1_geq0
+
underbracex_2^TCx_2_geq0
+2x_2^TBx_1geq2x_2^TBx_1
$$
and
$$
beginbmatrix
x_1^T&x_2^T
endbmatrix
M'
beginbmatrix
x_1\x_2
endbmatrix
=x_1^TAx_1+x_2^TCx_2-2x_2^TBx_1geq-2x_2^TBx_1.
$$
Then I get stuck.
linear-algebra positive-semidefinite block-matrices
linear-algebra positive-semidefinite block-matrices
asked Mar 17 at 21:57
DanDan
927
asked Mar 17 at 21:57
DanDan
927
asked Mar 17 at 21:57
DanDan
927
asked Mar 17 at 21:57
DanDan
927
asked Mar 17 at 21:57
DanDan
927
927
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just apply your assumption on $M$ to the vector $[-x_1^T,x_2^T]$.
answered Mar 17 at 22:11
GReyesGReyes
2,31315
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
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oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Just apply your assumption on $M$ to the vector $[-x_1^T,x_2^T]$.
answered Mar 17 at 22:11
GReyesGReyes
2,31315
$endgroup$
add a comment |
$begingroup$
Just apply your assumption on $M$ to the vector $[-x_1^T,x_2^T]$.
answered Mar 17 at 22:11
GReyesGReyes
2,31315
$endgroup$
add a comment |
$begingroup$
Just apply your assumption on $M$ to the vector $[-x_1^T,x_2^T]$.
answered Mar 17 at 22:11
GReyesGReyes
2,31315
$endgroup$
Just apply your assumption on $M$ to the vector $[-x_1^T,x_2^T]$.
answered Mar 17 at 22:11
GReyesGReyes
2,31315
answered Mar 17 at 22:11
GReyesGReyes
2,31315
answered Mar 17 at 22:11
GReyesGReyes
2,31315
answered Mar 17 at 22:11
GReyesGReyes
2,31315
2,31315
add a comment |
add a comment |
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