Finding eigenspaces of a matrix with parameter I have the following matrix:
$$
\begin{pmatrix}
3&0&0\\
k+2&1&0\\
5&k-1&1
\end{pmatrix}
$$
The exercise asks to find the eigenvalues of the matrix and, for all $k\in\mathbb{R}$, determine a basis of each eigenspace of the matrix.
Since ths is a lower triangular matrix, the eigenvalues are the values on the diagonals: $\lambda=3$ and $\lambda=1$.
Now, for $\lambda=3$ I need to find the null space of this matrix:
$$
\begin{pmatrix}
0&0&0\\
k+2&-2&0\\
5&k-1&-2
\end{pmatrix}
\rightarrow
\begin{pmatrix}
5&k-1&-2\\
k+2&-2&0\\
0&0&0
\end{pmatrix}
$$
By reducing in row echelon form ($R_2\leftarrow5R_2-kR_1$ and then $R_2\leftarrow R_2-2R_1$):
$$
\begin{pmatrix}
5&k-1&-2\\
0&-k^2-k-8&2k+4\\
0&0&0
\end{pmatrix}
$$
I'm stuck here because $-k^2-k-8$ doesn't have real roots. Any hints?

Since $-k^2-k-8$ can’t be zero, you can safely divide by it and continue merrily on your way with the row-reduction. However, because you’re working in $\mathbb R^3$ there’s a much simpler way to find a basis for this null space. The row space of $A-3I$ is clearly spanned by its two nonzero rows, and the null space of a matrix is the orthogonal complement of its row space, so the null space of $A-3I$ is spanned by $$(k+2,-2,0)\times(5,k-1,-2) = (4,2k+4,k^2+k+8).$$
