Prove $A$ is diagonalizable while $A^{2}=kA$ 
Let $A$ be an $n$-order real matrix, with:
  $$A^{2}=kA$$ 
  Prove that $A$ is diagonalizable, i.e., there exists an invertible matrix $P$ such that  $P^{−1}AP$ is a diagonal matrix.

My thoughts: I assume that $$A=P^{-1}\mathrm{diag}(a_{1},...,a_{n})P,$$ so $$A^{2}=P^{-1}\mathrm{diag}(a_{1}^{2},...,a_{n}^{2})P=kP^{-1}\mathrm{diag}(a_{1},...,a_{n})P$$ which leads to $$\mathrm{diag}(a_{1}^{2},...,a_{n}^{2})=\mathrm{diag}(ka_{1},...,ka_{n})$$
so we have a series of equations:$$a_{1}^2=ka_{1},...$$
OK, I stopped here and want some help because the conclusion form my assumption seems rather weird. Anyone have better idea with it?
 A: This is false; take $k=0$ and $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Assuming $k\ne0$, reduce $A$ to Jordan form; let $\lambda E+N$ be a Jordan block; then
$\lambda^2 E+2\lambda N +N^2=kE+kN$, whence it follows that $N^2=0$, $(2\lambda-k)N=0$ and $\lambda^2=k\lambda$. If $N\ne0$, then $\lambda=k/2$, and hence $k^2/4=k^2/2\implies k=0$, a contradiction. Thus, $N=0$; i.e., there are no notrivial Jordan blocks and the matrix is diagonalizable.
A: I suppose $k > 0 $ otherwise there are counterexamples.
There is the following lemma:

A square matrix $A$ with coefficient in $\mathbb{R}$  is diagonalizable if the mininal polynomial has the form $$(x - \lambda_1 ) \cdot \ \ \ldots  \ \cdot (x - \lambda_r)$$ where the $\lambda_i \in \mathbb{R}$ are all different.

In your case the minimal polynomial is a divisor of $x^2 -k  = (x - \sqrt{k})(x + \sqrt{k})$
and then we can apply the lemma.
A: You must assume $k \ne 0$, otherwise $\pmatrix{0 & 1\cr 0 & 0\cr}$ is a counterexample.  The point is that if the minimal polynomial of $A$ has no 
repeated roots, $A$ is diagonalizable.
A: If $A$ is invertible than it can be only a multiple if identity.
$$ A^2 = k A
$$
$$
A = A^2 A^{-1} = k A A^{-1} = k I
$$
