# Thread: Block Matrices Positive Definiteness

1. ## 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:

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

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

When I take the following nonzero vector:

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

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

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

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

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

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

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

Substituting above, we have:

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

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

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

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

Thanks,

Henrique

2. 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:

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

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

When I take the following nonzero vector:

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

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

$\displaystyle \mathbf{y}^{H}\mathbf{R}\mathbf{y}=\left[\begin{array}{cc} \mathbf{y}_{1}^{H} & \mathbf{y}_{2}^{H}\end{array}\right]\left[\begin{array}{cc} \mathbf{B} & \mathbf{C}^{T}\\ \mathbf{C} & \mathbf{B}\end{array}\right]\left[\begin{array}{c} \mathbf{y}_{1}\\ \mathbf{y}_{2}\end{array}\right]=\ldots.$
That result is false. For example, take 2×2 matrices $\displaystyle \mathbf{B} = \mathbf{0}$ and $\displaystyle \mathbf{C} = \begin{bmatrix}0&1\\-1&0\end{bmatrix}$. Let $\displaystyle \mathbf{y}_{1} = \begin{bmatrix}1\\0\end{bmatrix}$ and $\displaystyle \mathbf{y}_{2} = \begin{bmatrix}0\\1\end{bmatrix}$. Then $\displaystyle \mathbf{y}^{\textsc{h}}\mathbf{R}\mathbf{y} = -2$.

If you don't like taking $\displaystyle \mathbf{B} = \mathbf{0}$ (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 $\displaystyle \mathbf{y}^{\textsc{h}}\mathbf{R}\mathbf{y}$ is negative.

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

If you don't like taking $\displaystyle \mathbf{B} = \mathbf{0}$ (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 $\displaystyle \mathbf{y}^{\textsc{h}}\mathbf{R}\mathbf{y}$ is negative.
On the other hand, if the scalars are supposed to be real, and if $\displaystyle \mathbf{R} = \begin{bmatrix}\mathbf{B}&\mathbf{C}\\\mathbf{C}&\ mathbf{B}\end{bmatrix}$ (without the minus sign in the top right corner), then you can check that the result holds. That is probably what was intended.