If $\dim V=v$ and $\dim(\ker T)=n$, prove that $T$ has at most $v-n+1$ distinct eigenvalues

Question

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...

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... — Joe, Jun 11 '13 at 01:26
I added your comment to the question body. It is appropriate to give such background as part of the post, and easier for comments to be overlooked (especially if there end up being many of them). — Jonas Meyer, Jun 11 '13 at 01:31
Please correct your title. It uses different variables than your questions (in fact the variable $n$ is used in 2 different ways). It will get very confusing easily. — Calvin Lin, Jun 11 '13 at 01:35
@Joe : I see that you didn't accept an answer yet. If you have trouble understanding the details, please ask for them in the comments :) — Patrick Da Silva, Jun 11 '13 at 02:29

score 2 · Accepted Answer · answered Jun 11 '13 at 01:33

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,

After a little more thought and really thinking about it this made a lot more sense. So thank you it did help. — Joe, Jun 12 '13 at 20:55
@Joe : Those comments are the reason why I post. All the pleasure is mine! — Patrick Da Silva, Jun 13 '13 at 00:35

score 1 · Answer 2 · answered Jun 11 '13 at 01:33

1

Hint: Yes, rank nullity theorem is the way to go.

Hint: The reason for the extra $+1$, is because 0 is a possible eigenvalue.

answered Jun 11 '13 at 01:33

Calvin Lin

68,864

To add to this hint, $0$ is an eigenvalue whenever $n≠0$. – Ben Grossmann Jun 11 '13 at 01:41

score 1 · Answer 3 · answered Jun 11 '13 at 02:08

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.

score 0 · Answer 4 · answered Jun 11 '13 at 01:36

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 $p(x)=x(x^2+1)$).

score 0 · Answer 5 · answered Jun 11 '13 at 04:46

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).

If $\dim V=v$ and $\dim(\ker T)=n$, prove that $T$ has at most $v-n+1$ distinct eigenvalues

5 Answers5