# “Generalized Unitary Matrix”

A unitary matrix is an $N\times N$ (square) complex-valued matrix $\mathbf{A}$ satisfying:

$\mathbf{AA^\dagger} = \mathbf{I}_N$

where $\mathbf{I}_N$ is the identity matrix and $^\dagger$ stands for conjugate transpose operator.

Suppose an $N\times N$ (square) complex-valued matrix $\mathbf{B}$ satisfies:

$\mathbf{BB^\dagger} = \mathbf{D}_N$

where $\mathbf{D}_N$ is some diagonal matrix.

What can be said about $\mathbf{B}$ ?

I am particularly interested in the condition number or similar properties reflecting a degree of orthogonality or isometry property.

• Is $\mathbf{AA^\dagger}$ $A$ multiplied by its pseudo-inverse, or is it $A$ multiplied by its conjugate transpose? – Ben Grossmann Aug 16 '13 at 4:48
• conjugate transpose. Sorry for failing to mention that. – O. Souihli Aug 16 '13 at 5:02
• For what it's worth, if $B$ is invertible then there exists a positive definite diagonal matrix $C$ such that $CB$ is unitary (namely $C=\sqrt{D_N^{-1}}$). – Jonas Meyer Aug 16 '13 at 5:06
• On top of Jonas' comment, if you are using conjugate transpose, then $D_N$ has to be nonnegative because the diagonal entries of $BB^\dagger$ is taking the norm-squared of the rows of $B$. If there are $0$ entries in $D_N$, then the full row has to be zero. And of course, the rows are orthogonal to each other. – Evan Aug 16 '13 at 5:13
• On top of Evan's comment : your property just states that the column vectors of B are orthogonal, with (squared) norms given on the diagonal. – Bertrand R Aug 16 '13 at 6:22

We have $$AA^\dagger=D_N$$ iff all rows of $A$ are pairwise orthogonal. It is not necessarily the case that the columns of $A$ are pairwise orthogonal. As a counterexample, consider $$A = \begin{bmatrix} 1 & -1\\ 2 & 2 \end{bmatrix}$$ Which satisfies the about property, but does not have orthogonal columns.
Similarly $$A^\dagger A=D_N$$ iff $A$ has orthogonal columns, but under these circumstances, $A$ may not have orthogonal rows.
It is interesting to note that any matrix for which $$AA^\dagger = D_N$$ Can be decomposed into the form $$A = DU$$ for some diagonal matrix $D$ and some unitary matrix $U$. Similarly, any matrix for which $$A^\dagger A = D_N$$ Can be decomposed into the form $$A = UD$$ for some diagonal matrix $D$ and some unitary matrix $U$.