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I've been learning about unitary diagonalization and this question has been on my mind for a while, unanswered by the text I'm using.

Would it be true to say that if a square matrix represents an involution, i.e. $A^2=I$ (or $A=A^{-1}$), then it is Hermitian ($A=A^*$) and Unitary ($A^{-1}=A^*$). The reverse implication is definitely true, but is it an equivalence?

If so, then why? And if not, could it be true for real matrices? (Involution $\Rightarrow$ Symmetric & Orthogonal)

Thanks for help!

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Assuming $A$ is square, yes. –  Tapu Mar 27 '13 at 8:01
    
Oh, yes, sorry, I added that to the question. Thanks for clearing that up, still, why would that be true? –  Dahn Jahn Mar 27 '13 at 8:04
    
In the real case, I'd use the term orthogonal rather than unitary. –  Sammy Black Mar 27 '13 at 8:12
    
Thanks, the sudden introduction of all the new names for matrices still confuses me a little. –  Dahn Jahn Mar 27 '13 at 8:14
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If $A$ satisfies the condition, then any matrix similar to it has this property. Hence, I think it is not true. –  Wayson Kong Mar 27 '13 at 8:26
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2 Answers

up vote 5 down vote accepted

First, let $A \in M_n(\mathbb{C})$, and suppose that $A^2 = I$. I claim that $A$ is a self-adjoint unitary if and only if $A$ is normal. On the one hand, if $A$ is self-adjoint and unitary, then it is trivially normal. On the other hand, suppose that $A$ is normal. Then by the finite-dimensional spectral theorem, $A = U \Delta U^\ast$ for $\Delta$ a diagonal matrix (i.e., with diagonal entries the eigenvalues of $A$ counted with multiplicity), and $U$ a unitary. Since $A^2 = I$, it therefore follows that $\Delta^2 = I$, so that the diagonal entries of $\Delta$ (viz, eigenvalues of $A$) are $\pm 1$. Thus, $\Delta$ is a self-adjoint unitary, so that since $A = U \Delta U^\ast$ for $U$ unitary, $A$ too is a self-adjoint unitary.

Let us now see that the normality condition is not redundant. Consider $$ A = \begin{pmatrix} 1 & 1\\ 0 & -1\end{pmatrix}, $$ and observe that $$ A^2 = \begin{pmatrix} 1 & 1\\ 0 & -1\end{pmatrix}\begin{pmatrix} 1 & 1\\ 0 & -1\end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I_2. $$ However, $A$ is neither self-adjoint nor unitary, and indeed $$ A A^\ast = \begin{pmatrix} 1 & 1\\ 0 & -1\end{pmatrix}\begin{pmatrix} 1 & 0\\ 1 & -1\end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix},\\ A^\ast A = \begin{pmatrix} 1 & 0\\ 1 & -1\end{pmatrix}\begin{pmatrix} 1 & 1\\ 0 & -1\end{pmatrix} = \begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix} \neq A A^\ast, $$ as expected.

Let us now consider $A \in M_n(\mathbb{R})$, and suppose that $A^2 = I$. I claim that $A$ is symmetric and orthogonal if and only if $A^T A = A A^T$. By viewing $A$ as a complex matrix, we see that $A$ is self-adjoint and unitary if and only if $A$ is normal; since $A$ is real, this means that $A$ is symmetric and orthogonal if and only if $A^T A = A A^T$. Once more, the explicit example above shows that the "normality" condition is not redundant.

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Thanks a lot, an amazing answer! –  Dahn Jahn Mar 27 '13 at 8:46
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If $A^2=I$, you know that $A$ is diagonalizable and its eigenvalues are $\pm 1$. To find a counterexample, an idea is to take $A= P \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) P^{-1}$, where $P$ is not orthogonal. For example, $P= \left( \begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array} \right)$.

And indeed, $A= \left( \begin{array}{cc} 1 & -4 \\ 0 & -1 \end{array} \right)$ is a counterexample.

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