Preparing my Linear Algebra lecture I like the determinant free approach of Axler because the proof that operators $T$ on an $n$-dimensional complex vector space have eigenvalues is so simple:


Fix any non-zero vector $x$, observe that $x,T(x),\ldots,T^n(x)$ are linearly dependent to get a non-trivial linear combination $\sum\limits\_{k=0}^nc\_k T^k(x)=p(T)=0$, factorize the polynomial, and conclude that at least one factor $\lambda\_k-T$ is not injective just because compositions of injective maps are injective.


The existence of a single eigenvalue is enough to prove the spectral theorem for normal operators by induction.
However, I also try to mention the algorithmic aspects, and to make this proof an algorithm you really have to find *all* zeros of $p$. This is in contrast to the usual determinant approach where you only need *one* zero of the characteristic polynomial to get an eigenvalue.


Assume you have an oracle telling you *one* zero of a complex polynomial each time you ask. Is there a determinant free argument similar to Axler's which would lead to an eigenvalue?


Of course, without asking the oracle $n$ times, but if one quesion isn't enough perhaps at least considerably less than $n$ calls.



How to find eigenvalues following Axler?
Using minimality can help.


Without using the polynomial-root oracle you can find the *minimal* polynomial of $T$ by looking for the first linear dependence among $I, T, T^2, \dotsc, T^r$ (for the minimal $r$—starting with just $I,T$ and increasing $r$ until you find a linear dependence). Alternatively, find the first linear dependence among $x, T(x), \dotsc, T^r(x)$ for a fixed (arbitrary) nonzero $x$. Either way, the resulting polynomial will have the property that all its roots are eigenvalues of $T$. So, only one call to the oracle is needed at that point.


(In practice, it seems to me that it probably goes the other way: to find a root of a polynomial, one looks for an eigenvalue of the companion matrix. But pedagogically it makes sense.)


