Finding eigenspaces of a matrix with parameter

Question

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?

Answer 1

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 $$(k+2,-2,0)\times(5,k-1,-2) = (4,2k+4,k^2+k+8).$$

Answer 2

Since $-k^2-k-8$ has no real roots, you can divide the second line by it (for every $k\in\mathbb R$); you will get$$\begin{bmatrix}5&k-1&-2\\0&1&-\frac{2k+4}{k^2+k+8}\\0&0&0\end{bmatrix}.$$And now you can use that $1$ from the middle of the matrix to eliminate the $k-1$ from the first row.