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?
This is false; take $k=0$ and $A=\begin{pmatrix}0&amp;1\\0&amp;0\end{pmatrix}$. 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.
If $A$ is invertible than it can be only a multiple if identity.

$$ A^2 = k A
$$

$$
A = A^2 A^{-1} = k A A^{-1} = k I
$$