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Motivation: let $A\in\mathbf{R}^{n\times n}$ be symmetric. Then by the method of Lagrange multipliers, a maximum of $x\mapsto x^tAx$ on the compact unit sphere $\mathbf{S}^{n-1}$ must be an eigenvector of $A$. In particular, we have that

($\star$) $\det(tI-A)$ has a real root if $A$ is a real symmetric matrix.

Now this is trivial using the fundamental theorem of algebra (which we did not) and also seems pretty strong. Thus:

Is there a simple way to deduce the fundamental theorem of algebra from ($\star$)?

(Note: I asked this on the Mathematics Stack Exchange site and did not get an answer.)

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    $\begingroup$ Derksen gave a proof using linear algebra and characteristic polynomials. See jstor.org/stable/3647746 or kconrad.math.uconn.edu/blurbs/fundthmalg/fundthmalglinear.pdf. $\endgroup$
    – KConrad
    Commented Mar 12, 2022 at 15:58
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    $\begingroup$ Try proving this way that $x^2+1$ has a root. $\endgroup$
    – markvs
    Commented Mar 12, 2022 at 15:59
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    $\begingroup$ Derksen's paper is here: sites.lsa.umich.edu/hderksen/wp-content/uploads/sites/614/2018/…. Of course necessary preliminaries are that one can take square-roots in $\mathbb{C}$ and that every odd-degree polynomial has a root. $\endgroup$
    – Mark Wildon
    Commented Mar 12, 2022 at 15:59
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    $\begingroup$ @markvs I don't want an "easy" proof, just wondering if we can somehow extend the argument I gave to nonsymmetric matrices (with complex eigenvalues). of course we can prove FTA in many other ways... $\endgroup$
    – tomm
    Commented Mar 12, 2022 at 16:11
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    $\begingroup$ @markvs $x^2+1$ has a root just by construction, namely $i$. But whatever you have in mind, the following fact is an easy exercise: let $K$ be a field of char $\neq 2$ in which $K^*/{K^*}^2$ has order 2 (e.g., the reals). Then for every quadratic extension $L$ of $K$, every element of $K$ has a root in $L$. $\endgroup$
    – YCor
    Commented Mar 12, 2022 at 16:38

2 Answers 2

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I don't think so. Indeed the result that every symmetric matrix is diagonalizable is true for some orderable non-real-closed field $K$ (see this answer by Will Sawin to Over which fields are symmetric matrices diagonalizable ? ). Hence you can't deduce that every nonconstant polynomial over $K[i]$ has a root, since it's false.

NB: about the terminology: orderable means that it admits a total ordering for which the set of positive elements is closed under addition and multiplication.

A real-closed field can be defined as an orderable field $K$ such that the field $K[i]$ is algebraically closed. Here $K[i]$ means the analogue of obtaining complex numbers from real numbers, starting from $K$ instead, namely $K^2$ with elements written as $x+iy$ with multiplication $(x+iy)(x'+iy')=(xx'-yy')+i(xy'+x'y)$. This is a field because $-1$ is not a square in $K$. Of course with basic background in elementary commutative algebra this is the same as $K[t]/(t^2+1)$.

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    $\begingroup$ Moreover, Will's answer could be adapted to apply to the field $K:=\bigcup_k \mathbb{R}((x^{1/2^k}))$. Then $K[i]$ is closed under square root, but still not algebraically closed. So we can't do it even using diagonalizability and the fact that $\mathbb{C}$ has square roots. $\endgroup$
    – David E Speyer
    Commented Mar 12, 2022 at 16:22
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(Expansion of my comment above.) This does not answer the diagonalisability of all matrices. It just extends it from symmetric to normal.

Assume that we have an ordered field $R$ for which one can prove that (a) any symmetric matrix over $R$ is diagonalisable and (b) that every positive number in $R$ has a square root. Here is a proof that any matrix $N$ such that $N$ and $N^{t}$ commute is diagonalisable.

  1. If $H$ is a skew-symmetric matrix with entries in $R$ then $\begin{pmatrix} 0 & H \\ -H & 0\end{pmatrix}$ is symmetric and thus diagonalisable. This means that there is a basis $x_i,y_i,z_j$ and numbers $a_i$ in $R$ such that $Hx_i=a_iy_i$, $Hy_i=-a_ix_i$, $Hz_j=0$, for some sequence of indices $i$ and $j$.

  2. If $K$ is a orthogonal matrix with entries in $R$ such that neither of $\pm 1$ is an eigenvalue, then $H=(K+1)(K-1)^{-1}$ is an invertible skew-symmetric matrix. Applying the above, one can show that $K$ is diagonalisable over $R[\sqrt{-1}]$.

  3. If $K$ is any orthogonal matrix with entries in $R$, then $K$ is diagonalisable over $R[\sqrt{-1}]$. (Apply the above to the perpendicular of the eigenspaces for $\pm 1$.)

  4. Given a normal matrix $N$ (meaning $N$ commutes with $N^t$), we write $N^{t}N=P^2$ where $P$ is positive definite symmetric by using diagonalisation and the existence of square-roots of positive elements. It follows that $K=NP^{-1}$ is orthogonal and $K$ commutes with $P$. Since $K$ and $P$ are diagonalisable (over $R[\sqrt{-1}]$ and commute, so is $N=KP$.

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