For what values of k is this singular matrix diagonalizable? So the matrix is the following:
\begin{bmatrix}
1 &1  &k \\ 
1&1  &k \\ 
 1&1  &k 
\end{bmatrix}
I've found the eigan values which are $0$ with an algebraic multiplicity of $2$ and $k+2$ with an algebraic multiplicity of $1$.
However I'm having trouble with the eigan vector of the eigan value 0. I've put this through Wolfram and I should've got: \begin{pmatrix}
-k\\ 
0\\ 
1
\end{pmatrix} and \begin{pmatrix}
-1\\ 
1\\ 
0
\end{pmatrix}
But when finding the eigan vectors what I end up (Having x_{1}=r and x_{2}=t) is \begin{Bmatrix}
r\\ 
t\\ 
\frac{-r-t}{k}
\end{Bmatrix} which even if I separate it wouldnt be the vectors Wolfram gave me. Is there something I'm missing? Also is there a better way to find out the initial answer?

Hint: Substitute $r = -k$ and $t = 0$ to get the first eigenvector Wolfram alpha is giving you, and $r = -1$ and $t = 1$ to get the second one. Wolfram alpha isn't giving you the entire two-dimensional eigenspace, but only an eigenbasis.
