# Block Matrices Positive Definiteness

• May 29th 2010, 06:48 AM
thetouristbr
Block Matrices Positive Definiteness
Hello everyone,

I got stuck in the following problem, any help would surely be appreciated:

I have to prove that a block matrix of the form:

$\mathbf{R}=\left[\begin{array}{cc}\mathbf{B} & -\mathbf{C}\\\mathbf{C} & \mathbf{B}\end{array}\right]$

is positive definite, where $\mathbf{B}$ is symmetric and positive definite, and $\mathbf{C}$ is anti-symmetric, that is, $\mathbf{C}^{T}=-\mathbf{C}$.

When I take the following nonzero vector:

$\mathbf{y}=\left[\begin{array}{c}
\mathbf{y}_{1}\\
\mathbf{y}_{2}\end{array}\right]$

and try to check the usual condition $\mathbf{y}^{H}\mathbf{R}\mathbf{y}>0$, I get the following:

$\mathbf{y}^{H}\mathbf{R}\mathbf{y}=\left[\begin{array}{cc}
\mathbf{y}_{1}^{H} & \mathbf{y}_{2}^{H}\end{array}\right]\left[\begin{array}{cc}
\mathbf{B} & \mathbf{C}^{T}\\
\mathbf{C} & \mathbf{B}\end{array}\right]\left[\begin{array}{c}
\mathbf{y}_{1}\\
\mathbf{y}_{2}\end{array}\right]=$

$=\mathbf{y}_{1}^{H}\mathbf{B}\mathbf{y}_{1}+\mathb f{y}_{2}^{H}\mathbf{B}\mathbf{y}_{2}+\mathbf{y}_{1 }^{H}\mathbf{C}^{T}\mathbf{y}_{2}+\mathbf{y}_{2}^{ H}\mathbf{C}\mathbf{y}_{1}$

But:
$\mathbf{y}_{1}^{H}\mathbf{C}^{T}\mathbf{y}_{2}=(\m athbf{y}_{2}^{H}\mathbf{C}\mathbf{y}_{1})^{H}$

And $\mathbf{y}_{2}^{H}\mathbf{C}\mathbf{y}_{1}$ is a scalar, so:

$\mathbf{y}_{1}^{H}\mathbf{C}^{T}\mathbf{y}_{2}=(\m athbf{y}_{2}^{H}\mathbf{C}\mathbf{y}_{1})^{*}$

Substituting above, we have:

$\mathbf{y}_{1}^{H}\mathbf{B}\mathbf{y}_{1}+\mathbf {y}_{2}^{H}\mathbf{B}\mathbf{y}_{2}+(\mathbf{y}_{2 }^{H}\mathbf{C}\mathbf{y}_{1})^{*}+\mathbf{y}_{2}^ {H}\mathbf{C}\mathbf{y}_{1}=$

$=\mathbf{y}_{1}^{H}\mathbf{B}\mathbf{y}_{1}+\mathb f{y}_{2}^{H}\mathbf{B}\mathbf{y}_{2}+2\textrm{Re}\ {\mathbf{y}_{2}^{H}\mathbf{C}\mathbf{y}_{1}\}$

Now, how can I prove that the real number $\mathbf{y}_{1}^{H}\mathbf{B}\mathbf{y}_{1}+\mathbf {y}_{2}^{H}\mathbf{B}\mathbf{y}_{2}+2\textrm{Re}\{ \mathbf{y}_{2}^{H}\mathbf{C}\mathbf{y}_{1}\}$ is always positive?

For sure, $\mathbf{y}_{1}^{H}\mathbf{B}\mathbf{y}_{1}$ and $\mathbf{y}_{2}^{H}\mathbf{B}\mathbf{y}_{2}$ are both positive, or at least one of them is positive and the other is zero, because $\mathbf{B}$ is positive definite.

Thanks,

Henrique
• May 30th 2010, 08:58 AM
Opalg
Quote:

Originally Posted by thetouristbr
Hello everyone,

I got stuck in the following problem, any help would surely be appreciated:

I have to prove that a block matrix of the form:

$\mathbf{R}=\left[\begin{array}{cc}\mathbf{B} & -\mathbf{C}\\\mathbf{C} & \mathbf{B}\end{array}\right]$

is positive definite, where $\mathbf{B}$ is symmetric and positive definite, and $\mathbf{C}$ is anti-symmetric, that is, $\mathbf{C}^{T}=-\mathbf{C}$.

When I take the following nonzero vector:

$\mathbf{y}=\left[\begin{array}{c}
\mathbf{y}_{1}\\
\mathbf{y}_{2}\end{array}\right]$

and try to check the usual condition $\mathbf{y}^{H}\mathbf{R}\mathbf{y}>0$, I get the following:

$\mathbf{y}^{H}\mathbf{R}\mathbf{y}=\left[\begin{array}{cc}
\mathbf{y}_{1}^{H} & \mathbf{y}_{2}^{H}\end{array}\right]\left[\begin{array}{cc}
\mathbf{B} & \mathbf{C}^{T}\\
\mathbf{C} & \mathbf{B}\end{array}\right]\left[\begin{array}{c}
\mathbf{y}_{1}\\
\mathbf{y}_{2}\end{array}\right]=\ldots.$

That result is false. For example, take 2×2 matrices $\mathbf{B} = \mathbf{0}$ and $\mathbf{C} = \begin{bmatrix}0&1\\-1&0\end{bmatrix}$. Let $\mathbf{y}_{1} = \begin{bmatrix}1\\0\end{bmatrix}$ and $\mathbf{y}_{2} = \begin{bmatrix}0\\1\end{bmatrix}$. Then $\mathbf{y}^{\textsc{h}}\mathbf{R}\mathbf{y} = -2$.

If you don't like taking $\mathbf{B} = \mathbf{0}$ (because it's not strictly speaking positive definite) then you can replace it with a small positive multiple of the 2×2 identity matrix, and you will still find that $\mathbf{y}^{\textsc{h}}\mathbf{R}\mathbf{y}$ is negative.
• May 30th 2010, 09:26 AM
Opalg
Quote:

Originally Posted by Opalg
That result is false. For example, take 2×2 matrices $\mathbf{B} = \mathbf{0}$ and $\mathbf{C} = \begin{bmatrix}0&1\\-1&0\end{bmatrix}$. Let $\mathbf{y}_{1} = \begin{bmatrix}1\\0\end{bmatrix}$ and $\mathbf{y}_{2} = \begin{bmatrix}0\\1\end{bmatrix}$. Then $\mathbf{y}^{\textsc{h}}\mathbf{R}\mathbf{y} = -2$.

If you don't like taking $\mathbf{B} = \mathbf{0}$ (because it's not strictly speaking positive definite) then you can replace it with a small positive multiple of the 2×2 identity matrix, and you will still find that $\mathbf{y}^{\textsc{h}}\mathbf{R}\mathbf{y}$ is negative.

On the other hand, if the scalars are supposed to be real, and if $\mathbf{R} = \begin{bmatrix}\mathbf{B}&\mathbf{C}\\\mathbf{C}&\ mathbf{B}\end{bmatrix}$ (without the minus sign in the top right corner), then you can check that the result holds. That is probably what was intended.