# Not all unitary matrices are orthogonal.

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?

• Why not look at a simple example, like $$\begin{pmatrix}0&i\\-i&0\end{pmatrix}$$ Jul 10, 2012 at 17:59
• @POTUS You asked for an non-orthogonal counterexample, not a non-normal. J.M.'s example works. Jul 10, 2012 at 18:26
• Then why not a more general unitary matrix, like $$\frac1{\sqrt 2}\begin{pmatrix}e^{-it}&-e^{it}\\e^{-it}&e^{it}\end{pmatrix}$$ Jul 10, 2012 at 18:29
• How about $A = \begin{bmatrix} i \end{bmatrix}$. Then $A^* A = I$, but $A^T A = -I$. Jul 10, 2012 at 18:51
• You can if all entries are real... Jul 10, 2012 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 $$AA^T=\left[\begin{array}{cc}a&b\cr -b&a\end{array}\right]\cdot\left[\begin{array}{cc}a&-b\cr b&a\end{array}\right] =\left[\begin{array}{cc}a^2+b^2&0\cr 0&a^2+b^2\end{array}\right] \ne \left[\begin{array}{cc}1&0\cr 0&1\end{array}\right].$$
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.