# Eigenvalues of a Linear Operator over Finite-Dimensional, Non-Zero, Real Vector Space

We know that every operator on a finite-dimensional, non-zero, complex vector space has an eigenvalue. The proof of this theorem, as done in Axler's Linear Algebra Done Right book, depends on the fundamental theorem of algebra, where a complex polynomial $$p(z)$$ is constructed and applied over the operator, i.e., $$p(T)$$.

Suppose $$T \in L(V)$$ where $$V$$ is a real vector space of dimension $$n>0$$. If the corresponding (real) polynomial $$p(x)$$ happens to have $$n$$ roots (including multiplicities), then we can use the proof above to show that $$T$$ has at least one eigenvalue, which is one of the (real) roots. Here $$p(x)$$ is constructed the same way as in Axler's proof.

Question:

1. If $$p(x)$$ has no root, can we conclude that $$T$$ has no eigenvalue?

2. If $$p(x)$$ has less than $$n$$ roots, can we say anything about the number of eigenvalues of $$T$$?

The number of real roots of $$p(x)$$ is the number of eigenvalues of $$T$$. In particular, if $$p(x)$$ has no real root, then $$T$$ has no eigenvalue.
• The proof for the complex case, which is in the link in the question, works because we can factor a complex polynomial of degree $n$ into $n$ factors (but, still, that proof guarantees one eigenvalue only). In the real case, a real polynomial of degree $n$ might not have $n$ roots so the proof would not even apply in this case in the first place. Thus, can you explain more why the number of real roots of $p(x)$ is the number of eigenvalues of $T$? – A Slow Learner Nov 10 '19 at 11:43
• Directly from Axler's proof, no, I can't. But if $p(x)$ has degree $n$ and $m$ real roots $\lambda_1,\ldots,\lambda_m$, then you can write $p(x)$ as $(x-\lambda_1)\cdots(x-\lambda_m)q(x)$, where $q(x)$ has no real roots. But then $\lambda_1,\ldots,\lambda_m$ are eigenvalues of $T$. So, $T$ has at least $m$ eigenvalues. – José Carlos Santos Nov 10 '19 at 11:48
• So $\lambda_1, ...,\lambda_m$ are eigenvalues of $T$ but this result requires a proof different from Axler's proof? – A Slow Learner Nov 10 '19 at 11:53