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Recall that a matrix $A\in \mathbb{C}^{n\times n}$ is normal if $AA^{*}=A^{*}A$ where $A^*:=\bar{A}^T.$ Let $A\in \mathbb{R}^{n\times n}.$

  1. Show that not all unitary matrices are orthogonal.
  2. Use 1. to conclude that not every normal matrix in $\mathbb{R}^{n\times n}$ is orthogonally similar to a diagonal matrix.

My idea for 1.:

We want to show that $AA^T\neq A^TA \, \forall A=UBU^*$ where $U$ is a unitary matrix and $B$ is a diagonal matrix. $U^*=\bar{U}^T.$ Then we have \begin{align} AA^T\\ &=UBU^*(UBU^*)^T\\ &=UBU^*\bar{U}BU^T \end{align} So, intuitively since $U^*\bar{U}\neq \bar{U}U^*$, we have that $AA^T\neq A^TA .$ How do I come up with a clever counterexample?

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    $\begingroup$ Why not look at a simple example, like $$\begin{pmatrix}0&i\\-i&0\end{pmatrix}$$ $\endgroup$
    – J. M. ain't a mathematician
    Jul 10, 2012 at 17:59
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    $\begingroup$ @POTUS You asked for an non-orthogonal counterexample, not a non-normal. J.M.'s example works. $\endgroup$
    – Cocopuffs
    Jul 10, 2012 at 18:26
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    $\begingroup$ Then why not a more general unitary matrix, like $$\frac1{\sqrt 2}\begin{pmatrix}e^{-it}&-e^{it}\\e^{-it}&e^{it}\end{pmatrix}$$ $\endgroup$
    – J. M. ain't a mathematician
    Jul 10, 2012 at 18:29
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    $\begingroup$ How about $A = \begin{bmatrix} i \end{bmatrix}$. Then $A^* A = I$, but $A^T A = -I$. $\endgroup$
    – copper.hat
    Jul 10, 2012 at 18:51
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    $\begingroup$ You can if all entries are real... $\endgroup$
    – copper.hat
    Jul 10, 2012 at 20:11

2 Answers 2

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Let $a, b \in \mathbb{C}\setminus\mathbb{R}$ such that $$ a^2+b^2 \in \mathbb{C}\setminus\mathbb{R},\ |a|^2+|b|^2=1, \ a\bar{b} \in \mathbb{R}. $$ Setting $$ A=\left[\begin{array}{cc}a&b\cr -b&a\end{array}\right], $$ we have $$ AA^*=\left[\begin{array}{cc}a&b\cr -b&a\end{array}\right]\cdot\left[\begin{array}{cc}\bar{a}&-\bar{b}\cr \bar{b}&\bar{a}\end{array}\right] =\left[\begin{array}{cc}1&0\cr 0&1\end{array}\right] , $$ but $$ AA^T=\left[\begin{array}{cc}a&b\cr -b&a\end{array}\right]\cdot\left[\begin{array}{cc}a&-b\cr b&a\end{array}\right] =\left[\begin{array}{cc}a^2+b^2&0\cr 0&a^2+b^2\end{array}\right] \ne \left[\begin{array}{cc}1&0\cr 0&1\end{array}\right]. $$

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Observe that a matrix $A$ is unitary iff $AA^*=A^*A=I$. Thus, every unitary matrix is normal. $A$ is an orthogonal matrix iff $AA^T=A^TA=I$ (every orthogonal matrix is also normal).

This way, if $A$ is an unitary and orthogonal matrix, then $A^*A=I=A^TA$, which means that $A^*=A^T$. But $A^*=\overline{A}^T$. Then, for an unitary orthogonal matrix $\overline{A^T}=A^T$. This means that all entries in $A^T$, and therefore in $A$, must be real. Just simply take an unitary matrix with a non real entry.

For example $i.B$ with $B$ orthogonal.

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