Positive semidefinite block matrix is PSD with negated off-diagonal blocks?
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I am trying to follow some notes that imply
$$
M=
begin{bmatrix}
A&B^T\B&C
end{bmatrix}
succeq 0
Longleftrightarrow
M'=
begin{bmatrix}
A&-B^T\-B&C
end{bmatrix}
succeq 0$$
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).
So far I have that $A,Csucceq0$ by computing $x^TMx$ (for certain $x$) which is $geq0 forall x$ by definition. Therefore,
$$
begin{bmatrix}
x1&x2
end{bmatrix}
M
begin{bmatrix}
x1\x2
end{bmatrix}
=
underbrace{x_1^TAx_1}_{geq0}
+
underbrace{x_2^TCx_2}_{geq0}
+2x_2^TBx_1geq2x_2^TBx_1
$$
and
$$
begin{bmatrix}
x1&x2
end{bmatrix}
M'
begin{bmatrix}
x1\x2
end{bmatrix}
=x_1^TAx_1+x_2^TCx_2-2x_2^TBx_1geq-2x_2^TBx_1.
$$
Then I get stuck, but I feel like I'm very close to the conclusion
positive-semidefinite block-matrices
asked Nov 23 at 15:45
Dan
154
add a comment |
up vote
0
down vote
favorite
I am trying to follow some notes that imply
$$
M=
begin{bmatrix}
A&B^T\B&C
end{bmatrix}
succeq 0
Longleftrightarrow
M'=
begin{bmatrix}
A&-B^T\-B&C
end{bmatrix}
succeq 0$$
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).
So far I have that $A,Csucceq0$ by computing $x^TMx$ (for certain $x$) which is $geq0 forall x$ by definition. Therefore,
$$
begin{bmatrix}
x1&x2
end{bmatrix}
M
begin{bmatrix}
x1\x2
end{bmatrix}
=
underbrace{x_1^TAx_1}_{geq0}
+
underbrace{x_2^TCx_2}_{geq0}
+2x_2^TBx_1geq2x_2^TBx_1
$$
and
$$
begin{bmatrix}
x1&x2
end{bmatrix}
M'
begin{bmatrix}
x1\x2
end{bmatrix}
=x_1^TAx_1+x_2^TCx_2-2x_2^TBx_1geq-2x_2^TBx_1.
$$
Then I get stuck, but I feel like I'm very close to the conclusion
positive-semidefinite block-matrices
asked Nov 23 at 15:45
Dan
154
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to follow some notes that imply
$$
M=
begin{bmatrix}
A&B^T\B&C
end{bmatrix}
succeq 0
Longleftrightarrow
M'=
begin{bmatrix}
A&-B^T\-B&C
end{bmatrix}
succeq 0$$
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).
So far I have that $A,Csucceq0$ by computing $x^TMx$ (for certain $x$) which is $geq0 forall x$ by definition. Therefore,
$$
begin{bmatrix}
x1&x2
end{bmatrix}
M
begin{bmatrix}
x1\x2
end{bmatrix}
=
underbrace{x_1^TAx_1}_{geq0}
+
underbrace{x_2^TCx_2}_{geq0}
+2x_2^TBx_1geq2x_2^TBx_1
$$
and
$$
begin{bmatrix}
x1&x2
end{bmatrix}
M'
begin{bmatrix}
x1\x2
end{bmatrix}
=x_1^TAx_1+x_2^TCx_2-2x_2^TBx_1geq-2x_2^TBx_1.
$$
Then I get stuck, but I feel like I'm very close to the conclusion
positive-semidefinite block-matrices
asked Nov 23 at 15:45
Dan
154
I am trying to follow some notes that imply
$$
M=
begin{bmatrix}
A&B^T\B&C
end{bmatrix}
succeq 0
Longleftrightarrow
M'=
begin{bmatrix}
A&-B^T\-B&C
end{bmatrix}
succeq 0$$
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).
So far I have that $A,Csucceq0$ by computing $x^TMx$ (for certain $x$) which is $geq0 forall x$ by definition. Therefore,
$$
begin{bmatrix}
x1&x2
end{bmatrix}
M
begin{bmatrix}
x1\x2
end{bmatrix}
=
underbrace{x_1^TAx_1}_{geq0}
+
underbrace{x_2^TCx_2}_{geq0}
+2x_2^TBx_1geq2x_2^TBx_1
$$
and
$$
begin{bmatrix}
x1&x2
end{bmatrix}
M'
begin{bmatrix}
x1\x2
end{bmatrix}
=x_1^TAx_1+x_2^TCx_2-2x_2^TBx_1geq-2x_2^TBx_1.
$$
Then I get stuck, but I feel like I'm very close to the conclusion
positive-semidefinite block-matrices
positive-semidefinite block-matrices
asked Nov 23 at 15:45
Dan
154
asked Nov 23 at 15:45
Dan
154
edited Nov 23 at 17:59
asked Nov 23 at 15:45
Dan
154
asked Nov 23 at 15:45
Dan
154
asked Nov 23 at 15:45
Dan
154
154
add a comment |
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