Find a matrix $B$ such that $B^* B=A$ for a given Hermitian $A$

Question

Let

$$A=\left[ \begin{array}{ccc} 4 & 0 & 0 \\ 0 & 1 & i \\ 0 & -i & 1 \end{array} \right].$$

Find a matrix $ B $ such that $B^*B$ =$A$ (star means conjugate transpose of $B$).

I think that $A$ is hermitian, and so $A^* =A$ Also, We can edit this as $BB^*=A$. But I couldn't solve it completely.

Answer 1

First of all, you can't take the Hermitian of a product of matrices like that.

$$(B^*B)^*=B^*B\neq BB^* $$

You can examine the $4$ and $\left( \begin{array}{cc} 1 & i \\ -i & 1 \\ \end{array} \right)$ separately since the off diagonal elements are zero. I was able to simply guess which matrix would give $\left( \begin{array}{cc} 1 & i \\ -i & 1 \\ \end{array} \right)$ when multiplied with its conjugate transpose.

$$B=\left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & -i & 1 \\ \end{array} \right)$$

Answer 2

Let $A$ be an $n\times n$ Hermitian matrix. There must therefore be an orthonormal basis of eigenvectors corresponding to real eigenvalues. Write this as $$A = \sum_k \lambda_k P_{v_k},$$ where $\lambda_k\in\mathbb R$ and $P_{v_k}\equiv v_k v_k^*$ is the orthoprojection onto the vector $v_k$.

Let $B$ be some $m\times n$ matrix whose SVD reads $$B = \sum_k s_k (u_k w_k^*),$$ for some $s_k\ge0$ and $u_k\in \mathbb C^m, w_k\in\mathbb C^m$ and $\{u_k\}_k,\{w_k\}_k$ sets of orthonormal vectors.

Then, $B^* B$ reads $$B^* B = \sum_k s_k^2 (w_k w_k^*) = \sum_k s_k^2 P_{w_k}.$$ If we want $B^* B=A$, we thus need $\sum_k\lambda_k P_{v_k}=\sum_k s_k^2 P_{w_k}$. Being $\{v_k\}_k,\{w_k\}_k$ both sets of orthonormal vectors, this is only possible if, up to some relabeling, $\lambda_k=s_k^2$.

If $A$ (and thus $B^* B$) is non-degenerate, then we can also immediately conclude that $P_{v_k}=P_{w_k}$. More generally, there might be degenerate eigenvalues. In this case, the uniqueness of the spectral decomposition of a matrix tells us that the eigenspaces of $A$ and $B^* B$ corresponding to the same eigenvalue must be the same.

In conclusion, if $B^*B=A$ for $A$ Hermitian, then $B$ must have the form $$B = \sum_k \sqrt{\lambda_k} (w_k v_k^*),$$ for some choice of orthonormal vectors $\{w_k\}_k\subset\mathbb C^m$.