Product of a matrix and its Hermitian transpose?

Question

Suppose I need to find a matrix B such that

$B^H B = A$

and $A = \begin{bmatrix}4 &0& 0\\ 0 &1 &i\\ 0 &-i& 1\\\end{bmatrix}$

How do I proceed with a Product of a matrix and its Hermitian transpose ?

score 1 · Answer 1 · answered Dec 03 '13 at 00:31

1

Cholesky factorization is the easiest way to go. http://en.wikipedia.org/wiki/Cholesky_decomposition

answered Dec 03 '13 at 00:31

Stephen Montgomery-Smith

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Would you mind clarifying, I am a bit behind.. – user1502776 Dec 03 '13 at 04:01
It is a bit involved to explain here. But google "cholesky decomposition example" or something like that, and you should find some youtube videos or other documents that give examples. You example is a little trickier because of the complex entries, so do an example with real entries first. – Stephen Montgomery-Smith Dec 03 '13 at 04:06

score 0 · Accepted Answer · answered Dec 03 '13 at 09:55

This will illustrate how to handle your specific, small matrix, but the other post points to a general and practical method.

Diagonalize A. It must have non-negative eigenvalues or $B^{H}B=A$ is not possible. The eigenvalues of $A$ are $4,2,0$. Check that $A$ has a basis of eigenvectors: $$ A\left[\begin{array}{c}1 \\ 0 \\ 0\end{array}\right] = 4\left[\begin{array}{c}1 \\ 0 \\ 0\end{array}\right],\;\; A\left[\begin{array}{c}0 \\ i \\ 1\end{array}\right] = 2\left[\begin{array}{c}0 \\ i \\ 1\end{array}\right],\;\; A\left[\begin{array}{c}0 \\ i \\ -1\end{array}\right] = 0\left[\begin{array}{c}0 \\ i \\ -1\end{array}\right] $$ The eigenvectors are automatically orthogonal for distinct eigenvalues. Therefore, the normalized eigenvectors give a unitary transition matrix $$ U = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \frac{i}{\sqrt{2}} & \frac{i}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{array}\right]. $$ That is $U^{H}U=UU^{H}=I$, and $$ U^{H}AU = \left[\begin{array}{ccc}4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{array}\right],\;\;\; A = U\left[\begin{array}{ccc}4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{array}\right]U^{H} = U\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & 0\end{array}\right]U^{H} $$ So one matrix that will work for $B$ is $$ B=\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & 0\end{array}\right]U^{H} $$ However, the better choice for $B$ may be the unique Hermitian matrix ($B=B^{H}$) with non-negative eigenvales for which $B^{2}=A$: $$ B=U\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & 0\end{array}\right]U^{H} $$

Product of a matrix and its Hermitian transpose?

2 Answers2