If $\dim V=v$ and $\dim(\ker T)=n$, prove that $T$ has at most $v-n+1$ distinct eigenvalues 
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...
 A: I will assume that $V$ is a vector space over some field $k$ because you do not mention the ground field. 
Suppose $T$ has the set of eigenvalues $\lambda_0$, $\lambda_1$, ..., $\lambda_m$. Since distinct eigenspaces are in a direct sum, write $E_i$ for the eigenspace corresponding to $\lambda_i$. Then
$$
v = \dim(V) \ge \dim E_0 + \dim E_1 + \dots + \dim E_m \ge n+m.
$$
where $m$ is the number of non-zero eigenvalues. (The first inequality is because the direct sum of the eigenspaces is a subspace of $V$, and the second one is because the eigenspaces have dimension at least $1$). This means that $m \le v-n$, but the number of eigenvalues is at most $m+1$, hence if $0$ is an eigenvalue we have $m + 1\le v-n+1$ and if not, we still have $m \le v-n \le v-n+1$. 
Hope that helps,
A: Hint: Yes, rank nullity theorem is the way to go.
Hint: The reason for the extra $+1$, is because 0 is a possible eigenvalue.
A: So we need a few pieces for this problem:


*

*The rank nullity theorem tells us that the dimension of the image ($T(V)$) will have dimension $v-n$.

*Every vector in the kernel corresponds to an eigenvector with eigenvalue $0$ (why?)

*Each non-zero eigenvalue $\lambda_1,\lambda_2,...,\lambda_k$ of $T$ has at least one associated respective eigenvector, $v_1,v_2,...,v_k$.  These eigenvectors are linearly independent, as are $T(v_1),T(v_2),...,T(v_k)$ (why?).  That is, $T(v_1),T(v_2),...,T(v_k)$ forms a basis of some subspace of $T(V)$
Now, we can use that last fact to bring it all home.  We know that $T(V)$ has dimension $v-n$, which means $\{T(v_1),T(v_2),...,T(v_k)\}$ has at most $v-n$ elements, which means there are at most $v-n$ distinct non-zero eigenvectors, which means there are at most $v-n+1$ distinct eigenvectors total.
A: Suppose $V$ is a $n$-dimensional vector space over a field $F$. $\ker T$ has dimension $k>0$ (treat $k=0$ separately), so 0 is an eigenvalue of multiplicity $k$. $T$ can have at most $n$ eigenvalues counting multiplicity. We know that 0 accounts for exactly $k$ eigenvalues counting multiplicity. (keep in mind that $T$ may not have any other eigenvalues. Example: suppose $F=\mathbb{R} $ and the characteristic polynomial for $F$ is $p(x)=x(x^2+1)$).
A: 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).
