Suppose that $T\in \scr L(V)$ where $\scr L(V)$ denotes the set of all linear operators over $V$ and $\dim \mathbb {range} T=k$.
Prove that $T$ has at most $k+1$ distinct eigenvalues.
Let the distinct eigen values of $T$ be $\lambda_1,\lambda_2,….\lambda_m$ and the corresponding eigen vectors be $v_1,v_2,…,v_m$ .Then $\{v_1,v_2,…,v_m\}$ is linearly independent.
But how should I show that $m\le k+1$? Please give some hints.
Best Answer
Note that if $v$ is an eigenvector with non-zero eigenvalue, then $v$ is in the range of $T$ (do you see why?). Hence we can have at most $k$ distinct non-zero eigenvalues because of the linear independence you already noted.
Then we get $k + 1$ because $0$ might be an eigenvalue.