A problem in linear algebra about diagonalizable operators 
Let $T$ be an operator in a vector space $V$ of finite dimension over $F$ such that
  $$(T−\lambda \cdot\text{Id}_V)^k=0$$ for some $λ∈F$ and $k>0.$ 
  Prove that $T$ is diagonalizable if and only if $T = \lambda \cdot\text{Id}_V $.

The proof of the inverse is trivial, but when we know $T$ is diagonalizable, how to get $T = \lambda \cdot\text{Id}_V$?
Use "$(T−\lambda \cdot\text{Id}_V)^k=0$"?
 A: Let $n = \dim V$.  If $T$ is diagonalizable, then represent $T$ by the matrix
$$
\begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{pmatrix}.
$$
Then what is the matrix that represents $T - \lambda \, \text{Id}$?  And so what is the matrix that represents $(T - \lambda \, \text{Id})^k$?  You can write this out very explicitly and then show that $(T - \lambda \, \text{Id})^k = 0$ implies that $\lambda_1 = \lambda_2 = \cdots = \lambda_n = \lambda$.
A: More generally, suppose that $\displaystyle \sum_{n=0}^{n}a_n x^n=f(x)\in F[x]$. Then, there is a natural way to interpret $f(A)$ for $A\in\text{Mat}_n(A)$. Namely, $\displaystyle f(A)=\sum_{n=0}^{n}a_n A^n$ where, by convention, $A^0=I$. It is not hard to prove that if the eigenvalues of $A$ with multiplicity are $\lambda_1,\cdots,\lambda_n$ then the eigenvalues for $f(A)$ are $f(\lambda_1),\cdots,f(\lambda_n)$. This fact is absolutely invaluable, and will allow you to kill many problems.
For example, here you have the polynomial $f(x)=(x-\lambda)^k$ and you know that $0=f(T)=(T-\lambda I)^k$. Thus, applying the previous paragraph you know that if $\lambda_1,\cdots,\lambda_n$ are the eigenvalues of $T$ then $f(\lambda_1),\cdots,f(\lambda_n)$ since these are the eigenvalues of $f(T)$, which is the zero matrix. But, $f(\lambda_i)=(\lambda_i-\lambda)^k=0$ implies that $\lambda=\lambda_i$. Thus, we must necessarily have that $\lambda=\lambda_i$ for all $i$. 
Here's where the second trick comes into play. Namely, the only matrix conjugate to a scalar matrix (a multiple of the identity) is that multiple of the identity. Indeed, if $A$ is conjugate to $\lambda I$ then $A=B\lambda IB^{-1}$ for some invertible $B$. But, $\lambda I$ commutes with every matrix in $\text{Mat}_n(F)$ and so $B\lambda IB^{-1}=\lambda I$ so that $A=\lambda I$. 
Now, as Michael Joyce points out, if $T$ were invertible it would be conjugate to the matrix with the eigenvalues of $T$ on the diagonal. But, since every eigenvalue of $T$ is $\lambda$ this implies that $T$ would be conjugate to $\lambda I$ which implies, by what we have just said, that $T=\lambda I$!
