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.


  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 $T$?


1 Answer 1


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.

  • $\begingroup$ 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 $T$? $\endgroup$ Nov 10, 2019 at 11:43
  • $\begingroup$ 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. $\endgroup$ Nov 10, 2019 at 11:48
  • $\begingroup$ So $\lambda_1, ...,\lambda_m$ are eigenvalues of $T$ but this result requires a proof different from Axler's proof? $\endgroup$ Nov 10, 2019 at 11:53
  • $\begingroup$ No. Why should it? $\endgroup$ Nov 10, 2019 at 12:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .