Finding Eigenvector of repeated eigenvalue

Question

How to find Eigenvector of following matrix: \begin{bmatrix} 1&0&1\\ 0&1&0\\ 0&0&1\\ \end{bmatrix}

I solved the characteristic of the the above matrix is \begin{equation} (1-\lambda)^{3}=0 \end{equation}

Answer 1

If $\lambda$ is an eigenvalue of a matrix $A$ then the eigenspace is given by the kernel of $A-\lambda I$ where $I$ is the identity matrix.

In your case, you have just the eigenvalue $\lambda=1$ and you compute the kernel of $$ \begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix}-1\cdot \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}= \begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix} $$