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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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Why not look at a simple example, like $$\begin{pmatrix}0&i\\-i&0\end{pmatrix}$$ – J. M. Jul 10 '12 at 17:59
@J.M. That doesn't work: $AA^T=-I_2=A^TA.$ – Lyapunov Jul 10 '12 at 18:21
@POTUS You asked for an non-orthogonal counterexample, not a non-normal. J.M.'s example works. – Cocopuffs Jul 10 '12 at 18:26
Then why not a more general unitary matrix, like $$\frac1{\sqrt 2}\begin{pmatrix}e^{-it}&-e^{it}\\e^{-it}&e^{it}\end{pmatrix}$$ – J. M. Jul 10 '12 at 18:29
You can if all entries are real... – copper.hat Jul 10 '12 at 20:11

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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