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I have proved that a $2\times2$ real normal matrix $A$ is either symmetric or of the form $$ \begin{pmatrix}a&b\\-b&a\end{pmatrix} $$ Now, I want to prove that there exists $r>0,\ \theta\in(0,\pi)\cup(\pi,2\pi)$ such that $a = r\cos\theta,\ b = r\sin\theta$. From the proof that $A$ is of the above form, I have concluded that $b\neq0$. For the point $(a,b)\in\mathbb{R}^2$, we know its polar form is of $a = r\cos\theta,\ b=r\sin\theta$ for some $\theta$. Since $b\neq0$, the point is not on thef $x$-axis and hence $\theta\neq0,\pi,2\pi$. Also, since $b\neq0$ we know for sure that $r\neq0$.

Does this proof hold? Also, is this if and only if? I mean, if $A$ is of the above form with $b\neq0$ does this imply that $A$ is rotation and scaling?

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  • $\begingroup$ The identity is a normal matrix with $b=0$. $\endgroup$ – hjhjhj57 Jun 12 '15 at 7:12
  • $\begingroup$ $a$ could be zero. $\endgroup$ – JonTrav Jun 12 '15 at 7:13
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I don't think your characterization is complete. A $2\times 2$ real matrix is normal if, and only if, it's of the form: $$\begin{pmatrix} a&b\\b&d \end{pmatrix},$$ or $$\begin{pmatrix} a&b\\-b&a \end{pmatrix}.$$ Now, for the second type of normal matrices you can do what you want. Consider the vector $v:=(a,b)\in\mathbb{R}^2$, then there's always an unique angle $\theta\in[0,2\pi)$ such that $\frac{v}{||v||}=(\cos\theta,\sin\theta)$. By removing one point from the possible angles you're removing a viable possibility, so you shouldn't do it.

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