# Finding the eigenvectors of a matrix that has one eigenvalue of multiplicity three

This is a simple question, which hopefully has a quick answer. I have a given matrix A, such that

$$A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$$

Since it's fairly straightforward, I'll just state that the eigenvalue of this matrix is $\lambda = 1$ with algebraic multiplicity $3$. To find the eigenvectors of this matrix, all I have to do is find the kernel of $(A-\lambda I)$. Thus,

$$(A-\lambda I) = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} -(1)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$$

The kernel of this matrix, according to my work and Wolfram Alpha, is $Ker(A-\lambda I) = \{(-1, 0, 1)^T, (0, 1, 0)^T\}$.

However, MATLAB and my calculator say that the eigenvectors are $(0, 1, 0)^T, (0,-1,0)^T, (0, -1, 0)^T$.

My question, then, is where did I go wrong? I looked through my book, and it does indeed cover eigenvalues with multiplicity, but it doesn't treat them any differently than the case with no multiplicity. Did I commit an algebraic error somewhere?