Question from Linear Algebra Done Right again.
Suppose $T \in \mathcal{L}(V)$ and $\operatorname{dim}\operatorname{range}T=k$. Prove that T has at most $k+1$ eigenvalues.

Question

Question from Linear Algebra Done Right again.
Suppose $T \in \mathcal{L}(V)$ and $\operatorname{dim}\operatorname{range}T=k$.  Prove that T has at most $k+1$ eigenvalues.

thomas5267 · Answer

I don't see why this is true.  Clearly if T is full rank then T has n eigenvalues, n being the dimension of V.  Other than this case I don't see why this is true.

thomas5267 · Answer

Wait.  I might have an idea how to do it.

thomas5267 · Answer

Actually I think I have a proof.  Will type up after dinner.

thomas5267 · Answer

$$
\begin{align*}
\left(\sum_{n=0}^ka_nT^n\right)v&=0\quad a_n \neq 0\text{ since }\operatorname{rank}(T)=k<k+1\
\sum_{n=0}^ka_nx^n&=c\prod_{n=0}^k(x-\lambda_n)\
\sum_{n=0}^ka_nT^n&=c\prod_{n=0}^k(T-\lambda_nI)
\end{align*}
$$
The map from $p \to p(T)$ is linear.  That is $p(x)+q(x)\to p(T)+q(T)$ and $ap(x)\to ap(T)$.  How can I conclude that $c\prod_{n=0}^k(T-\lambda_nI)$ is injective?  The best case is all $\lambda_n$ are distinct and hence k+1 eigenvalues at best.

thomas5267 · Answer

More precisely, in the first line, there exists some $a_j\neq 0,\,0\leq j \leq n$.

thomas5267 · Answer

I think I got a proper proof.  Will type it out next morning.

thomas5267 · Answer

Pick $v\in V$$$
\left(\sum_{n=0}^ka_nT^n\right)v=0\quad \text{some }a_n \neq 0\text{ since }\operatorname{rank}(T)=k<k+1
$$Making the $a_n$ the coefficient of a polynomial, we can write the following.
$$
\sum_{n=0}^ka_nx^n=c\prod_{n=0}^k(x-\lambda_n)
$$Such factorisation is guaranteed to exist by the fundamental theorem of algebra.  We can transfer this polynomial to $\mathcal{L}(V)$ by evaluating $x$ at $T$.
$$
\begin{align*}
\left(\sum_{n=0}^ka_nT^n\right)v&=0\
\left(c\prod_{n=0}^k(T-\lambda_nI)\right)v&=0
\end{align*}
$$This map from $p\to p(T)$ is linear and respects multiplication, hence $p(T)$ can be factored in the exact same way as $p$.  Since $\displaystyle \left(c\prod_{n=0}^k(T-\lambda_nI)\right)v=0$, one of the $T-\lambda_nI$ must be injective and hence $\lambda_n$ is an eigenvalue of $T$.  There are $n+1$ terms in the product, therefore the upper bound of (distinct) eigenvalues are $n+1$.

thomas5267 · Answer

This however raises a question.  Suppose $v\in V$, $T\in \mathcal{L}(V)$ and let $k$ be the largest value such that $\displaystyle \left(\sum_{n=0}^ka_nT^n\right)v=0$, so a linear dependency relationship with the largest index.  Construct the polynomial $\displaystyle \sum_{n=0}^ka_nT^n=c\prod_{n=0}^k(T-\lambda_nI)$ in the same manner as above.  Are all the $\lambda_n$ eigenvalues of $T$?

thomas5267 · Answer

Actually now I think about it my proof is incomplete.  If we choose $k'>k=\operatorname{rank}(T)$ the factorisation might contain more than $k+1$ terms and the whole argument falls apart.

Help!

thomas5267 · Answer

@hartnn Want to try?

thomas5267 · Answer

@ganeshie8 I thought I had the proof but I don't...

ganeshie8 · Answer

This looks hard for me
I'll go through this and give it a good try after breakfast, in 1 hour.. .

thomas5267 · Answer

Thanks a lot.  I should buy you a (virtual) breakfast XD!

ganeshie8 · Answer

Observation1 :

range(T) contains all the linear combinations of the columns of T,
and since eigenvectors of T are a subset of range(T), 
we can say that the dimension of range(T) cannot be less than the number of independent eigenvectors of T.

ganeshie8 · Answer

Observation2 : 
eigenvectors of different eigenvalues are linearly independent.

ganeshie8 · Answer

Observation3 : 
If a matrix is singular, then 0 is an eigenvalue of it.

thomas5267 · Answer

Can we salvage the proof by choosing the minimum $k$ such that $\left(\sum_{n=0}^ka_nT^n\right)v=0\quad \text{not all }a_n = 0$ aka the minimum linearly dependent set of vectors?

thomas5267 · Answer

So $k+1$ would be the maximum number of factors and some linear operators need fewer powers to achieve linear dependence?

thomas5267 · Answer

Like this, third answer: https://math.stackexchange.com/questions/655634/question-about-axlers-proof-that-every-linear-operator-has-an-eigenvalue

reemii · Answer

I don't handle that much the $z$ polynomials. so i'll present it that way (it's actually no more than what ganishie8 said):
By "contradiction": T has at least k+2 eigenvalues:
- take k+2 of them: $\lambda_1,\dots,\lambda_{k+1},\lambda_{k+2}$. Even if 0 is one of them, the dimension of the range of T will be at least k+1 because (as ganeshie8 said), the eigenvectors corresponding to different eigenvalues are linearly independent. Contradiction.

-> no more than k+1 eigenvalues. That's for sure. Now, is k+1 the upperbound (included)?
Yes, take the  non-zero $\lambda_i$'s: there are $k$ of them -> dim(range(T))=k.
If the dimension of ker(T) is 0, then there are no more eigenvalues.
If the dimension of ker(T) is > 0, then there is another eigenvalue, but the set of T(w) for w in ker(T) is just {$0_V$}. -> no dimension added.