# Does a unitary matrix have such property?

• I
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
Gold Member
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.

Last edited:
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
Gold Member
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.

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.