# How to Prove a Matrix is Diagonal?

1. Nov 9, 2011

### weetabixharry

I have a complex matrix, $\textbf{A}$, whose columns are linearly independent. In other words, $\textbf{A}$ is either tall or square and $\left( \textbf{A}^H\textbf{A}\right)^{-1}$ exists (where $\left(\right)^H$ denotes conjugate transpose). I am trying to prove that the matrix:

$\textbf{B} \triangleq \left( \textbf{A}^H\textbf{A}\right)$

must be diagonal, based on the following:

$\textbf{A}= diag(\underline{\lambda})\textbf{A}\textbf{A}^H \textbf{A} \textbf{A}^H \textbf{A}$

for some real diagonal matrix $diag(\underline{\lambda})$. It may or may not also be useful to note that $\textbf{A}$ is also subject to the constraint:

$\underline{diag}(\textbf{A}\textbf{A}^H) = \underline{1}$

by which I mean that all the diagonal entries of $(\textbf{A}\textbf{A}^H)$ are equal to 1 (i.e. the Euclidean norms of the rows of $\textbf{A}$ are all 1).

I have deduced all sorts of properties of $\textbf{A}$, but strongly believe that it should be possible to show that $\textbf{B}$ is diagonal... but a proof escapes me. Any help is greatly appreciated!

Last edited: Nov 9, 2011
2. Nov 9, 2011

### micromass

I might misunderstand your problem but

$$\left(\begin{array}{cc} 1 & 2\\ 3 & 4\end{array}\right)^H\left(\begin{array}{cc} 1 & 2\\ 3 & 4\end{array}\right) = \left(\begin{array}{cc} 10 & 14\\ 14 & 20\end{array}\right)$$

This is not diagonal. It IS hermition though (as can easily be proven).

3. Nov 9, 2011

### weetabixharry

I'm not sure what the consequences of that are in this context. The matrix you suggested cannot satisfy either of the equations:

$\textbf{A}= diag(\underline{\lambda})\textbf{A}\textbf{A}^H \textbf{A} \textbf{A}^H \textbf{A}$

$\underline{diag}(\textbf{A}\textbf{A}^H) = \underline{1}$

4. Nov 9, 2011

### I like Serena

Does this mean that $(\textbf{A}\textbf{A}^H)$ is the identity matrix?

5. Nov 9, 2011

### weetabixharry

It could be any matrix with ones on the diagonal. For example:

$(\textbf{A}\textbf{A}^H) = % \left[ \begin{array}{lll} 1 & 3 & 2 \\ 3 & 1 & 7 \\ 2 & 7 & 1% \end{array}% \right]$

would be suitable in this sense.

6. Nov 9, 2011

### I like Serena

Oh, okay, so are the Euclidean norms of the rows of A not 1?

7. Nov 9, 2011

### weetabixharry

Yes, the Euclidean norms of the rows are 1.

Consider, for example:

$\textbf{A}=\left[ \begin{array}{ll} 1 & 0 \\ 1 & 0 \\ 0 & 1% \end{array}% \right] ,\quad \underline{\lambda }=\left[ \begin{array}{l} 0.25 \\ 0.25 \\ 1% \end{array}% \right]$

The norms of all rows of $\textbf{A}$ are equal to one, but $\textbf{A}\textbf{A}^H$ is not the identity matrix:

$\textbf{A}\textbf{A}^H=\left[ \begin{array}{lll} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1% \end{array}% \right]$

(but $\textbf{A}^H\textbf{A}$ is diagonal, and I want to show that this must always be true).

Last edited: Nov 9, 2011