Q:

Find values of $k$ where $A$ is diagonalizable

I have the matrix 
A= $\begin{pmatrix}1&1&k\\ 1&1&k\\ 1&1&k\end{pmatrix}$
And need to find the values of $k$ where $A$ is diagonalizable. 
I'm not too sure how to go about this. Do I first row reduce it and find the eigenvalues?

A:

It is easily seen that the rank of the matrix $A$ is $1$. So $0$ must be an eigenvalue of $A$ for any value of $k$. Also since sum of each rows of $A$ is $k+2$, so $k+2$ is also an eigenvalue of $A$. Now consider two separated cases:
Case:I [$k+2=0$] 
Here $0$ is the only  eigenvalue of $A$, since $A^2=0$. Thus $A$ is not diagonalizable (minimal polynomial of $A$ is $m_a(t)=t^2$, not a product of distinct factors).
Case:II [$k+2\neq0$] Here $A$ has two different eigenvalue, namely $0$ and $k+2$. Using $dimE_\lambda=n-rank(A-\lambda I)$, for a $n\times n$ matrix $A$; $dimE_0=3-rankA=2$ which is equal to algebraic multiplicity of $0$. Also since algebraic multiplicity of $k+2$ is $1$, so $A$ is diagonalizable.
Thus for any real value  $k\neq-2$, $A$ is diagonalizable over $\mathbb{R}.$

