# If D is a diagonal matrix, when is B D B^T diagonal?

1. Dec 11, 2013

### Constantinos

Hey! So here's the question:

1. The problem statement, all variables and given/known data
Let

$\mathbf{B} \in \mathbb{R}^{n \times n}$ be some square matrix we can choose and

$\mathbf{D} \in \mathbb{R}^{n \times n}$ be some given diagonal matrix with positive diagonal elements.

For what matrices $\mathbf{B}$ is the product

$\mathbf{BDB}^{T}$

a diagonal matrix?

2. Relevant equations
Anything goes here I guess. It could be something very easy or requiring theorems (SVD ?)

3. The attempt at a solution
Well I tried to do this for $\mathbf{B}, \mathbf{D} \in \mathbb{R}^{2 \times 2}$ and the matrix that satisfies the above could be diagonal, anti-diagonal, or even have a row zeroed (but not column). So it can't be something simple, like just saying that it should be diagonal (which is rather obvious)

If anyone has anything to suggest feel free, thanks!

Last edited: Dec 11, 2013
2. Dec 11, 2013

### Simon Bridge

It wants you to discover what kind of matrix B has to be in general ... i.e. would it be "singular", "self adjoint", that sort of thing. What are some types of matrixes you have been learning about lately? Ones that have a special relationship with their transpose?

Note: if D is diagonal, then D=dt I: d=tr(D)

3. Dec 12, 2013

### Constantinos

Well I haven't been learning about any matrices lately! This question arose on a stochastic processes exercise, were a certain Z = BX + d where X is a random vector of uncorrelated Gaussian random variables. The exercise asks what must Β be for the Ζ variables to still remain uncorrelated. This amounts to the problem above, i.e find Β such that:

$\mathbf{B} Cov(X) \mathbf{B}^T$ is diagonal. $Cov(X)$ is diagonal since the Gaussians are uncorrelated.

I guess I could just answer it *can* be diagonal, or anti-diagonal, but there seem to be more matrices with this property. I am just wondering if there is a closed formula to describe them.

Now that I think of it,

$\mathbf{D} = Cov(X) = \mathbf{SS}^T$

where

$\mathbf{S} = \begin{bmatrix} \sigma_1 & 0 & \cdots & 0 \\ 0 & \sigma_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \sigma_n \end{bmatrix}$

So we have:

$\mathbf{BDB}^T = \mathbf{BS}\mathbf{S}^T\mathbf{B}^T = (\mathbf{BS})(\mathbf{BS})^T = \mathbf{A}\mathbf{A}^T, ~~ \mathbf{A} \equiv \mathbf{BS}$

So now the question becomes, for what matrices $\mathbf{B}$ is the above $\mathbf{A}\mathbf{A}^T$ diagonal? I'm not aware of any theorems to continue from here on.

Last edited: Dec 12, 2013