# Number of Possible Distinct eigenvalues of $T$ when $\dim\operatorname{range}T = k$.

Is my attempt at the following Proof Correct?

PRELIMINATRY NOTATION AND THEOREMS

• $T\in\mathcal{L}(V)$ denotes the set of all linear operator on the vector space $V$.
• $(5.10)$ Given that $T\in\mathcal{L}(V)$ such that $\lambda_1,\lambda_2,...,\lambda_n$ are distinct eignenvalues of $T$ then the corresponding list of eigenvectors is linearly independent.

Theorem. Given that $T\in\mathcal{L}(V)$ such that $\dim\operatorname{range}T = k$ then $T$ has at most $k+1$ distinct eigenvalues.

Proof. We first prove the weaker claim that $T$ has at most $k$ non-zero distinct eigenvalues.

Assume on the contrary that $T$ has $n$ eigenvalues where $n>k$ namely $\lambda_1,\lambda_2,...,\lambda_n$ then using theorem $5.10$ we may establish that the corresponing list of vectors $v_1,v_2,...,v_n$ is linearly independent.

We now show that the list of vectors $Tv_1 = \lambda_1v_1,Tv_2 = \lambda_2v_2,...,Tv_n = \lambda_nv_n$ is linearly independent assume that $$c_1\lambda_1v_1+c_2\lambda_2v_2+\cdot\cdot\cdot+c_n\lambda_nv_n = 0$$ for some $c_1,c_2,...,c_n\in\mathbf{F}$ the linear independece of $v_1,v_2,v...,v_n$ implies that $c_j\lambda_j = 0$ for all $j\in\{1,2,...,n\}$ and since $\lambda_j\neq 0$ it follows that $c_j = 0$, but this implies that we have a linearly independent list of vectors namely $Tv_1,Tv_2,...,Tv_n$ with length larger than $\dim\operatorname{range}T$ an obvious contradiction.

Now returning to our original claim $0$ may or may not be an eigenvalue of $T$ in the event that it is we have $k+1$ distinct eigenvalues if not then we have $k$ distinct eigenvalues in both instances the number of distinct eigenvalues never exceeds $k+1$.

$\blacksquare$

• I find it nice. But notice that if $0$ is an eigenvalue of an operator $T\in\mathcal L(V)$ then $T$ is not injective, so $\text{rank } T<\dim V$. Roughly speaking, zero eigenvalues don't contribute to the rank of $T$, because the corresponding eigenvectors are precisely the kernel of $T$.(This does not invalidate your proof, but you can tune the last paragraph). Feb 14, 2018 at 16:16
• Thankyou @ajotatxe for you response though i fail to appreaciate how this concerns the above proof could you care to elaborate further Feb 14, 2018 at 16:36

• $(5.10)$ Given that $T\in\mathcal{L}(V)$ such that $\lambda_1,\lambda_2,...,\lambda_n$ are distinct eignenvalues of $T$ then the corresponding list of eigenvectors is linearly independent.
• Assume on the contrary that $T$ has $n$ distinct nonzero eigenvalues where $n>k$ namely $\lambda_1,\lambda_2,...,\lambda_n$ then using theorem $5.10$ we may establish that the corresponing list of vectors $v_1,v_2,...,v_n$ is linearly independent.
Remark about correction $2$, we need them to be distinct to use Theorem $5.10$ and we need them to be non-zero, since later in the proof, we uses $\lambda_j \ne 0$.