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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.

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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.

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  • $\begingroup$ For the first step I found that $T(v_j)=\lambda_j v_j\implies T(\dfrac{v_j}{\lambda_j})=v_j \implies v_j\in \Bbb {range T}$ if $\lambda_j\neq 0$ $\endgroup$ – Learnmore Mar 23 '16 at 17:04
  • $\begingroup$ Okay ,it's clear now $\endgroup$ – Learnmore Mar 23 '16 at 17:10
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The hint is that $v_1,\ldots,v_m$ are linearly independent and belong to the range of $T$.

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