# Prove $T$ has at most two distinct eigenvalues

The question is from Axler's Linear Algebra text. The $\mathcal{L}(V)$ stands for the space of linear operators on the vector space $V$.

Suppose that V is a complex vector space with dim $V=n$ and $T \in \mathcal{L}(V)$ is such that $$\text{null} \ T^{n-2} \neq \text{null} \ T^{n-1}$$ Prove that $T$ has at most two distinct eigenvalues.

I fist thought of solving this by contradiction. That is, I thought, suppose there were three distinct eigenvalues. Then, there would be an equation like $$(x-\lambda_1I)^{d_1}(x-\lambda_2I)^{d_2}(x-\lambda_3I)^{d_3}$$ where the $d_i$'s are positive integers that sum to dim $V$. Call this polynomial $q(x)$ the characteristic poly. Thus, by Cayley's theorem, $$q(T)=(T-\lambda_1I)^{d_1}(T-\lambda_2I)^{d_2}(T-\lambda_3I)^{d_3}=0$$ Then multiplying out and setting dim $V = d_1+d_2+d_3 = n$, I could then get a poly with the various powers of $T$ (this is a little tricky to write down). In particular, I wanted to see the powers $n-2$ and $n-1$. I thought, ok, so now, rewrite the poly in terms of each of these and using the fact that $\text{null} \ T^{n-2} \neq \text{null} \ T^{n-1}$ you get some vector $v \in V$ such that, $$T^{n-1}v = (\text{poly}_1)v = 0$$ but $$T^{n-2}v = (\text{poly}_2)v = k \neq 0$$ I can think of some interesting things about $(\text{poly})_1$ and $(\text{poly})_2$, in particular, each has the monic term $T^n$. At this point, I'm not sure any of this helped.

Well, anyways, I'm sure someone has a much better approach! Thanks in advance to anyone who read this.

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Hints: consider the chain of subspaces $\{0\}=\ker T^0\subseteq\ker T\subseteq \ker T^2\subseteq \ldots...$ and think about what happens if $\ker T^{k-1}=\ker T^k$ at some point. Then prove that the assumption $\mbox{null} T^{n-2}\neq \mbox{null} T^{n-1}$ yields $\mbox{null} T^{n-1}=n-1$ or $\mbox{null} T^{n-1}=n$. In the latter case $T^{n-1}=0$, so it is easy to conclude. In the former case, take a basis of $\ker T^{n-1}$, complete it into a basis of $V$, and consider the matrix of $T$ in the latter.

Note: there is nothing special about $\mathbb{C}$ with this approach. It could be over any field.

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thanks a lot. quick question, the dim$\mbox{null}T^{n-1} = n-1 \text{or} n$? I was trying to figure that out at first but kept tripping up on this point. Is it because if, as you said, $\ker T^{k-1} = \ker T^k$ at some point, then the dim of the kernel of $T^n$ will equal also... said differently then since $\ker T^{k-1} = \ker T^k = \cdots = \ker T^n$, then dim$V = \text{dim}ker T + \text{dim} im T$, you can deduce that the above equals $n-1$ or $n$? – IQ472 Jun 5 '13 at 0:48
@IQ472 As you observed, if it is equal at some point, it is stationary. So the sequence can only become stationary after $n-1$. And before that, the inclusions were strict, the dimension was increasing by $1$ at least at each step. – 1015 Jun 5 '13 at 0:50

Suppose $T(v) = \lambda v$ where $\lambda \ne 0$.

Then $v$ is not part of the nullspace of $T^{n-1}$. Take a basis $\mathscr B_1$ for the nullspace of $T^{n-1}$ and add $v$ to that basis. Call this linearly independent set $\mathscr B$.

We want to prove that $\mathscr B$ is a basis for $V$.

$T$ is nilpotent on $\operatorname{span}( \mathscr B_1 )$, so $T^{|\mathscr B_1|} = 0$ But $T^{n-2} \ne 0$, so $|\mathscr B_1| > n-2$.

So $\mathscr B$ is a linearly independent set of order $n$ and is thus a basis. The only eigenvalue for a nilpotent matrix is $0$ (use a change of basis to make the matrix upper triangular), so the only eigenvalues for $T$ are $0$ and $\lambda$.

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