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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.

Dustan Levenstein
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  • 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. – Dustan Levenstein Oct 20 '19 at 00:39
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    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. – Eric Wofsey Oct 20 '19 at 01:18
  • @EricWofsey I hesitantly agree. – Dustan Levenstein Oct 20 '19 at 01:21
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    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. – Eric Wofsey Oct 20 '19 at 01:55
  • @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. – Dustan Levenstein Oct 20 '19 at 01:58

2 Answers2

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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 $x^2+2$.

Eric Wofsey
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  • Thank you for your answer Eric! After having sunk an unreasonable amount of time into this problem over the last year, I have conclusively determined that an algebraic solution can't exist. See my answer. :) – Dustan Levenstein Nov 12 '20 at 17:57
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An algebraic solution cannot exist; you can adjoin the eigenvalues of rational symmetric matrices to $\mathbb Q$ to obtain a field which satisfies the spectral theorem (symmetric matrices diagonalize orthogonally) but is not a real closed field.

Definition. Let $F \subseteq K$ be an extension of ordered fields. Then $\alpha \in K$ is symmetric over $F$ if $\alpha$ is an eigenvalue of a symmetric matrix with entries in $F$. $K$ is a symmetric extension of $F$ if every element of $K$ is symmetric over $F$.

Lemma. Let $A$ be an $n \times n$ symmetric matrix over an ordered field $F$, and let $p, q \in F[x]$ be coprime polynomials. Then $\ker p(A)$ and $\ker q(A)$ are orthogonal subspaces of $F^n$, and therefore intersect trivially.

Proof. Let $ap+bq=1$ and use the substitution $a(A)p(A)$ to assume WLOG that $p(x) = x$ and $q(x) = 1-x$. Then use the fact that the kernel of a symmetric matrix is orthogonal to its image.

Lemma. A symmetric matrix over an ordered field is semisimple.

Proof. Apply the spectral theorem to a real closure of the field of coefficients.

Theorem. Let $A$ be a symmetric matrix over an ordered field $F$. Then $A$ is orthogonally equivalent to a block diagonal matrix in which each block has irreducible characteristic polynomial.

Lemma Let $A, B$ be symmetric matrices over an ordered field $F$, of size $n \times n$ and $m \times m$ respectively. Then polynomials in $A \otimes I_m$ and $I_n \otimes B$ witness the fact that the symmetric elements of $K$ in an extension $F \subseteq K$ form a subfield of $K$.

Lemma Let $F \subseteq K \subseteq L$ be a chain of extensions of ordered fields. If $K/F$ is symmetric and $L/K$ is symmetric, then $L/F$ is symmetric.

Thus we can speak of a "symmetric closure" of $\mathbb Q$ in $\mathbb R$, consisting of roots of irreducible polynomials over $\mathbb Q$ which are the characteristic polynomials of rational symmetric matrices. In particular, every root of such a polynomial must be real, so $x^4-2$ does not have a root in the symmetric closure of $\mathbb Q$. Since $\sqrt[4]{2}$ must exist in a real closed field, this proves that there are fields satisfying the spectral theorem which are not real closed.

This answer suggests that it is an unsolved problem to determine which rational polynomials are the characteristic polynomial of a symmetric matrix.

Dustan Levenstein
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