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?
 A: 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.
A: A matrix is diagonalisable if its eigen values are distinct, $0$ and $k+2$ are the only eigen values..Now $0$ has multiplicity of $2$ and $k+2$ has $1$. If $k=-2$ then eigen value $0$ will have multiplicity of $3$ and now we can just check its eigen vector corresponding to $0$, if it produces $3$ linearly independent eigen vectors then the matrix is diagonalisable otherwise...Not diagonalisable.                                                                If $k$ is not equal to $-2$ then there will be another eigen value (say) $a$ of multiplicity $1$, hence it produces a single eigen vector and eigen value $0$ of multiplicity $2$ should produce $2$ linearly independent vectors in order to make matrix diagonalisable..if it doesnt the matrix is not diagonalisable.       
