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 $n$?
 A: 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. 
A: 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.
A: 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 $\mathbb R$.
