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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 $$\begin{pmatrix}0&i\\-i&0\end{pmatrix}$$2012-07-10
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    @J.M. That doesn't work: $AA^T=-I_2=A^TA.$2012-07-10
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    @POTUS You asked for an non-orthogonal counterexample, not a non-normal. J.M.'s example works.2012-07-10
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    @POTUS It works since $-I_2 \ne I_2$!2012-07-10
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    Then why not a more general unitary matrix, like $$\frac1{\sqrt 2}\begin{pmatrix}e^{-it}&-e^{it}\\e^{-it}&e^{it}\end{pmatrix}$$2012-07-10
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    How about $A = \begin{bmatrix} i \end{bmatrix}$. Then $A^* A = I$, but $A^T A = -I$.2012-07-10
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    If $A\in \mathbb{R}^{n\times n}$ and $A$ is unitary, you cannot say $A$ is orthogonal?2012-07-10
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    You can if all entries are real...2012-07-10

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