For what values of k is this singular matrix diagonalizable?

Question

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?

Answer 1

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.

Answer 2

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.