# Eigenvalue of a linear operator

Let $T$ be a linear operator on a finite dimensional vector space $V$ with dimension $n$. Then, is it necessary that $T$ has $n$ eigen values or could it be less than $n$?

• The other answers are more worth reading, but just to quickly answer your question: there could be anywhere between $0$ and $n$ eigenvalues. – Callus Aug 4 '16 at 19:10
• It's always $n$... if you define "eigenvalue" and count them the right way. – Hurkyl Aug 4 '16 at 23:53

Depends on your field. For an algebraically closed field, all eigenvalues can be found and the linear transformation will have $n$ eigenvalues (of course, including multiplicity). But you may not be able to "find" all eigenvalues, if your field is, say $\mathbb R$.

It could be less than $n.$

Consider $\begin{bmatrix}1&1\\0&1\end{bmatrix}$

1 is an eigenvalue, and $\begin{bmatrix} 1\\0 \end{bmatrix}$ is an eigenvector, but there is not a second independent eigenvector.

• However, there is a 2nd independent generalized eigenvector. – Paul Sinclair Aug 4 '16 at 17:47
• @PaulSinclair The question was about eigenvalues, not eigenvectors. – Marc van Leeuwen Aug 4 '16 at 19:34
• Does the matrix I provided have a second eigenvalue? Nonetheless, it is easy to show that the characteristic polynomial could have multiple roots for any single eigenvalue. It is relevant, that there is not necessarily a space of eigenvectors associated with that value of dimension equal to the multiplicity. – Doug M Aug 4 '16 at 19:37
• @MarcvanLeeuwen - Yes, but I was not answering the question. I am not contradicting Doug M's post, just adding some relevant information. While the eigenspaces of an operator do not necessarily span the entire complex vector space, the generalized eigenspaces always will. So the number of eigenvalues counted by dimension of the eigenspaces may be less than $n$, but when counted by dimension of the generalized eigenspaces (which match the algebraic multiplicity of the eigenvalue as a root of the characteristic polynomial), it will always be $n$. – Paul Sinclair Aug 4 '16 at 20:42
• @Doug M. It doesn't have another independent eigenvector. However, while an eigenvector associated with the eigenvalue $\lambda$ satisfies $$(A - \lambda I)v = 0$$ a generalized eigenvector satisfies the equation $$(A - \lambda I)^kv = 0$$ for some integer $k > 0$. If $A$ is your matrix, then $(A-1\cdot I)^2 = 0$, so every vector is a generalized eigenvector associated with the eigenvalue 1. – Paul Sinclair Aug 4 '16 at 20:47

If $T:\, V \rightarrow V$ then it admits the matrix representation with respect to some base of $V$. The eigenvalues are then computable (and defined) as the roots of the monic polynomial $p(\lambda )$ $$p(\lambda)=\text{det}(A-\lambda I),$$ where $A$ is the matrix representation of $T$ in the given base. As the Fundamental Theorem of Algebra states, any polynomial in $\mathbb{C}$ (and hence also in $\mathbb{R}$) of degree $n$ has exactly $n$ complex roots. Hence the answer is that there are always exactly $n$ eigenvalues, however not necesserily real.

Note that the finite-dimension of $V$ is necessery, for infinite-dimensional case the operators can have only finitely many eigenvalues.

• Actually I think you need the root-factor-theorem also. But other than that really good answer – mathreadler Aug 4 '16 at 15:37
• Thanks for reminding. That is true of course, however for the clasic fields $\mathbb{R}$ or $\mathbb{C}$ this is somewhat automatically assumed (at least as I have seen). Nonetheless, you are right. – michalOut Aug 4 '16 at 15:42
• Note that this is counting the eigenvalues by (algebraic) multiplicity. There need not be $n$ distinct roots. – Robert Israel Aug 4 '16 at 15:48