Can one deduce the fundamental theorem of algebra from real calculus and linear algebra? 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.)
 A: 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 $K[t]/(t^2+1)$.
A: (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.

*

*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 $j$.


*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 $R[\sqrt{-1}]$.


*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 $\pm 1$.)


*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 $N=KP$.
