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Let $A\in \mathbb{C}^{n\times n}$ and $\lambda \in \mathbb{C}.$ $\lambda$ only eigenvalue of $A \iff A-\lambda E_n$ is nilpotent.

So the first direction: let $\lambda$ be the only eigenvalue of $A$ then the characteristic polynomial of $A$ has the form $\chi_A(\lambda)=\left(x-\lambda\right)^n$ and so does the minimal polynomial $\mu_A(\lambda)=\left(x-\lambda\right)^m, m\le n$ then by Cayley Hamilton theorem, the matrix satisfies $(A-\lambda E_n)^n=0$ and also $(A-\lambda E_n)^m=0$ so $A$ is nilpotent of degree $m.$ Any help with the other implication would be appreciated.

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    $\begingroup$ You don't need to use the minimal polynomial for the argument you gave. $\endgroup$ – lhf Jul 23 '17 at 15:18
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    $\begingroup$ Think about the Jordan normal form of $A$, if you know it already $\endgroup$ – tiefi Jul 23 '17 at 15:18
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Suppose $(A-\lambda I)^m =0$ and $Av = \mu v$ with $v\neq 0$.

Then $(A-\lambda I)v = (\mu-\lambda) v$ and so $(A-\lambda I)^m = (\mu-\lambda)^m v$. Since $(A-\lambda I)^m =0$ we must have $\mu=\lambda$.

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"Going the other way":

Suppose $T$ has at least two eigenvalues $\lambda_1$ and $\lambda_2$.

I claim $T - \lambda I$ is non-nilpotent for every $\lambda \in \Bbb C$.

For if $T$ has at least two eigenvalues $\lambda_1$ and $\lambda_2$, there are nonzero vectors $v_1$ and $v_2$ with

$Tv_1 = \lambda_1 v_1 \tag{1}$

and

$Tv_2 = \lambda_2 v_2; \tag{2}$

then for any $\lambda \in \Bbb C$ we have

$(T - \lambda)v_1 = (\lambda_1 - \lambda)v_1, \tag{3}$

and

$(T - \lambda)v_2 = (\lambda_2 - \lambda)v_2. \tag{4}$

Now if $\lambda \notin \{\lambda_1, \lambda_2 \}$, then

$\lambda_1 - \lambda \ne 0 \ne \lambda_2 - \lambda, \tag{5}$

whence for any positive integer $n$,

$(\lambda_1 - \lambda)^n \ne 0 \ne (\lambda_2 - \lambda)^n. \tag{6}$

Now it is easy to see that (3), (4) and (6) in concert yield

$(T - \lambda)^n v_1 = (\lambda_1 - \lambda)^n v_1 \ne 0 \tag{7}$

and

$(T - \lambda)^n v_2 = (\lambda_2 - \lambda)^n v_2 \ne 0; \tag{8}$

by virtue of (7) and (8) we must have

$(T - \lambda)^n \ne 0 \tag{9}$

for all positive $n \in \Bbb Z$; hence, $T - \lambda$ is not nilpotent. If $\lambda \in \{\lambda_1, \lambda_2 \}$, then at least one of (7), (8) still binds, and so the conclusion (9) binds in this case as well.

We have just seen that if $T$ has more than one eigevalue, $T - \lambda$ cannot be nilpotent. Thus $T - \lambda$ nilpotent forces $T$ to have precisely one eigenvalue.

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  • $\begingroup$ @copper.hat: me too! $\endgroup$ – Robert Lewis Jul 23 '17 at 16:23
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Hint: If $T$ is a nilpotent operator on a vector space of dimension $n$, then $T^n=0$.

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My previous answer to this question "came the other way", as it were, and proved that the matrix $A - \lambda I$ cannot be nilpotent for any $\lambda \in \Bbb C$ if $A$ has two or more distinct eigenvalues.

Whilst thinking on this problem it occurred to me that I would like to be able to "come this way", and prove the converse, that is, prove that if $A - \lambda I$ is nilpotent for some $\lambda \in \Bbb C$, then $\lambda$ is the unique eigenvalue or $A$. Furthermore, I wanted to do this if possible by elementary means, avoiding if I could complex results such as the Cayley-Hamilton Theorem, which I see as sort of a "big gun" in matrix theory. I couldn't find a proof which satisfied me until this morning, when I realized the required assertion may be quite simply demonstrated:

Suppose $A - \lambda I$ is nilpotent for some $\lambda \in \Bbb C$; then there is some positive $k \in \Bbb Z$ such that

$(A - \lambda I)^k = 0; \tag{1}$

let us consider the smallest of all such $k$. If $k = 1$, then

$A = \lambda I, \tag{2}$

and $\lambda$ is clearly the unique eigenvalue of $A$. If $k \ge 2$, then

$(A - \lambda I)^{k - 1} \ne 0, \tag{3}$

and so there must be some vector $w$ such that

$y = (A - \lambda I)^{k - 1} w \ne 0; \tag{4}$

then

$(A - \lambda I) y = (A - \lambda I)(A - \lambda I)^{k - 1}w = (A - \lambda I)^k w = 0, \tag{5}$

whence

$Ay = \lambda y, \tag{6}$

i.e. $\lambda$ is an eigenvalue or $A$. We may now invoke "going the other way", that is, the assertion proved in my other answer, that $A - \lambda I$ nilpotent only allows for one eigenvalue of $A$; thus $\lambda$ is sole eigenvalue of $A$.

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