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I found a slick way to prove that every real normal matrix is orthogonally equivalent to a block diagonal form consisting of diagonal parts and $2 \times 2$ parts of the form $$\begin{pmatrix}a & b \\ -b & a\end{pmatrix}.$$

This seems awfully close to proving FTA - if every real polynomial is the characteristic polynomial of a normal matrix, then every real polynomial consequently splits over $\mathbb C$. So my question is:

Is there a slick way to find a real normal matrix given a characteristic polynomial?

Note that the companion matrix is not normal in general.

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  • $\begingroup$ Added topology because I used compactness of $S^n$ in my proof of this. So I'm not averse to a topological procedure for this question. $\endgroup$ – Dustan Levenstein Oct 20 at 0:39
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    $\begingroup$ Just as a sanity check, if you've already used compactness of $S^n$, then you've used enough analysis to prove FTA, so it's reasonable to expect you can finish with just easy algebra. $\endgroup$ – Eric Wofsey Oct 20 at 1:18
  • $\begingroup$ @EricWofsey I hesitantly agree. $\endgroup$ – Dustan Levenstein Oct 20 at 1:21
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    $\begingroup$ So, for what it's worth, just doing brute force calculations in the $2\times 2$ case, the result is not true over $\mathbb{Q}$ and to prove it over $\mathbb{R}$ you very much need to use the fact that every positive real has a square root. This suggests that the general case might not really be any easier than proving FTA. $\endgroup$ – Eric Wofsey Oct 20 at 1:55
  • $\begingroup$ @EricWofsey Good catch on it not being true over $\mathbb Q$. Feel free to post that counterexample and I'll go ahead and accept it. $\endgroup$ – Dustan Levenstein Oct 20 at 1:58
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As evidence that there is no easy solution (and in particular, no purely algebraic solution), it is not true over $\mathbb{Q}$ that every monic polynomial is the characteristic polynomial of a normal matrix. Indeed, direct calculations show that a $2\times 2$ matrix $$\begin{pmatrix} a & b \\ c & d\end{pmatrix}$$ is normal iff either $b=c$ or $b=-c$ and $a=d$. In the first case, the characteristic polynomial is $$x^2-(a+d)x+ad-b^2$$ and in the second case the characteristic polynomial is $$x^2-2ax+a^2+b^2.$$ It is easy to see that over $\mathbb{Q}$ these do not produce all possible monic quadratics; for instance, it is impossible to get $x^2+2$.

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