TITLE: If $\dim V=v$ and $\dim(\ker T)=n$, prove that $T$ has at most $v-n+1$ distinct eigenvalues
QUESTION [6 upvotes]: Let $T:V\to V$ be a linear operator. If $\dim V=v$ and $\dim(\ker T)=n$, prove that $T$ has at most $v-n+1$ distinct eigenvalues.

I have been working on this proof for a few days and I am not sure what direction to really go with it? I feel like starting with the rank nullity theorem is correct and relating that to the sum of eigenspaces may be my next move. Though I cant think of how to bring these two ideas together to create a fluid proof? Thank you for your help...

REPLY [0 votes]: We know that $T$ has $v$ eigenvalues (counted according to multiplicity).
Since $\dim \ker T = n$, there is a basis $x_1,...,x_n$ for $\ker T$, and $T v_i = 0. v_i$, hence $T$ has an eigenvalue at $0$ of multiplicity $n$. Since there are $v-n$ remaining eigenvalues, there can be at most $v-n+1$ distinct eigenvalues (the $+1$ to account for $0$ eigevalue).
To show that this can be attained, let $x_1,...,x_v$ be a basis for $V$, and define $T x_i = i x_i$ for $i = 1,...,v-n$, and $T x_i = 0$ for $i=v-n+1,...,v$. Then $T$ has distinct eigenvalues $0,1,...,v-n$ (that is, $v-n+1$ of them).
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