Does a unitary matrix have such property?

[Haorong Wu]

TL;DR Summary

I need prove a property of unitary matrices, but without success.

Hi. I'm learning Quantum Calculation. There is a section about controlled operations on multiple qubits. The textbook doesn't express explicitly but I can infer the following statement:

If $U$ is a unitary matrix, and $V^2=U$, then $V^ \dagger V=V V ^ \dagger=I$.

I had hard time proving it. I only can prove that if $V$ is reversible, then $\left (V^ \dagger V \right ) \left ( V V ^ \dagger \right )=I$.

I hope the statement is true, otherwise my inference would be wrong.

 

[StoneTemplePython]

Summary: I need prove a property of unitary matrices, but without success...

If $U$ is a unitary matrix, and $V^2=U$, then $V^ \dagger V=V V ^ \dagger=I$...

It isn't true. Any involutive $V$ that isn't unitarily diagonalizable serves as a counterexample. Using blocked matrices, you can build up more complicated counterexamples from here.

 

[Haorong Wu]

It isn't true. Any involutive $V$ that isn't unitarily diagonalizable serves as a counterexample. Using blocked matrices, you can build up more complicated counterexamples from here.

Thanks, StoneTemplePython. I have to review my inference. I must make some mistakes.

 

[StoneTemplePython]

Thanks, StoneTemplePython. I have to review my inference. I must make some mistakes.

for an explicit example of this, consider
$\mathbf S = \left[\begin{matrix}1 & 1 & 1\\0 & 1 & 1\\0 & 0 & 1\end{matrix}\right]$
$\mathbf D = \left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & -1\end{matrix}\right]$

and
$\mathbf V = \mathbf{SDS}^{-1} = \left[\begin{matrix}1 & 0 & -2\\0 & 1 & -2\\0 & 0 & -1\end{matrix}\right]$

you have

$\mathbf V^2 = \mathbf I = \left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right]$

but

$\mathbf I \neq \mathbf V^* \mathbf V = \left[\begin{matrix}1 & 0 & -2\\0 & 1 & -2\\-2 & -2 & 9\end{matrix}\right] \neq \left[\begin{matrix}5 & 4 & 2\\4 & 5 & 2\\2 & 2 & 1\end{matrix}\right] = \mathbf {VV}^*$
- - - -
an even easier counter example would be to just eyeball
$\left[\begin{matrix}1 & c\\ 0 & -1\end{matrix}\right]$
for some $c \neq 0$
it is diagonalizable and all eigenvalues are -1 or +1, so when you square the matrix you get the identity matrix (which is unitary). But a Triangular matrix commutes with its conjugate transpose iff it is diagonal -- it would require $c =0$ here. This is a standard result for normal matrices and something worth proving to yourself.

 

[Haorong Wu]

for an explicit example of this, consider
$\mathbf S = \left[\begin{matrix}1 & 1 & 1\\0 & 1 & 1\\0 & 0 & 1\end{matrix}\right]$
$\mathbf D = \left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & -1\end{matrix}\right]$

and
$\mathbf V = \mathbf{SDS}^{-1} = \left[\begin{matrix}1 & 0 & -2\\0 & 1 & -2\\0 & 0 & -1\end{matrix}\right]$

you have

$\mathbf V^2 = \mathbf I = \left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right]$

but

$\mathbf I \neq \mathbf V^* \mathbf V = \left[\begin{matrix}1 & 0 & -2\\0 & 1 & -2\\-2 & -2 & 9\end{matrix}\right] \neq \left[\begin{matrix}5 & 4 & 2\\4 & 5 & 2\\2 & 2 & 1\end{matrix}\right] = \mathbf {VV}^*$
- - - -
an even easier counter example would be to just eyeball
$\left[\begin{matrix}1 & c\\ 0 & -1\end{matrix}\right]$
for some $c \neq 0$
it is diagonalizable and all eigenvalues are -1 or +1, so when you square the matrix you get the identity matrix (which is unitary). But a Triangular matrix commutes with its conjugate transpose iff it is diagonal -- it would require $c =0$ here. This is a standard result for normal matrices and something worth proving to yourself.

Thanks, StoneTemplePython, with those detailed examples.