How to find a basis for the eigenspace of a $3 \times 3$ matrix?

I'm stuck on this linear algebra problem and I need some help. The problem is:

$$B=\left[\begin{array}{rrr} 5 & -2 & -6 \\ -2 & 2 & 3 \\ 2 & -1 & -2 \end{array}\right]$$ has eigenvalues 1 and 3, find the basis to the eigenspace for the corresponding eigenvalue.

I need to find the eigenvectors of B that correspond to each eigenvalue, and then use them as a basis for the eigenspace. But I don't know how to do that.

Here is what I have tried so far:

• I subtracted the eigenvalue from the diagonal entries of B and got two matrices: $$B-1I=\left[\begin{array}{rrr} 4 & -2 & -6 \\ -2 & 1 & 3 \\ 2 & -1 & -3 \end{array}\right]$$ and $$B-3I=\left[\begin{array}{rrr} 2 & -2 & -6 \\ -2 & -1 & 3 \\ 2 & -1 & -5 \end{array}\right]$$
• I tried to row reduce these matrices to find the null space, but I got stuck on both of them. For example, for the first matrix, I swapped the first and second rows and added the first row to the third row, and got: $$\left[\begin{array}{rrr} -2 & 1 & 3 \\ 4 & -2 & -6 \\ 0 & 0 & 0 \end{array}\right]$$ But then I don't know how to proceed from here. How do I find the eigenvectors from this matrix?

Can someone please explain how to solve this problem step by step? I would really appreciate it. Thank you!

2 Answers

By inspection is not difficult to find for example

• $$v_1=(1,-1,1)$$ for $$\lambda =1$$

• $$v_2=(1,2,0)$$ for $$\lambda =1$$

• $$v_3=(-2,1,-1)$$ for $$\lambda =3$$

or if we prefer by row operation, for the first one

$$\left[\begin{array}{rrr} 4 & -2 & -6 \\ -2 & 1 & 3 \\ 2 & -1 & -3 \end{array}\right]\stackrel{r_3+r_2}\to \left[\begin{array}{rrr} 4 & -2 & -6 \\ -2 & 1 & 3 \\ 0 & 0 & 0 \end{array}\right]\stackrel{2r_2+r_1}\to \left[\begin{array}{rrr} 4 & -2 & -6 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\stackrel{\frac12 r_1}\to \left[\begin{array}{rrr} 2 & -1 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$$

which leads to $$v_1$$ and $$v_2$$.

In a similar way we can proceed for the second one

$$\left[\begin{array}{rrr} 2 & -2 & -6 \\ -2 & -1 & 3 \\ 2 & -1 & -5 \end{array}\right]\stackrel{r_3+r_2}\to \left[\begin{array}{rrr} 2 & -2 & -6 \\ -2 & -1 & 3 \\ 0 & -2 & -2 \end{array}\right]\stackrel{r_2+r_1}\to \left[\begin{array}{rrr} 2 & -2 & -6 \\ 0 & -3 & -3 \\ 0 & -2 & -2 \end{array}\right]\stackrel{3r_3-2r_2}\to \left[\begin{array}{rrr} 2 & -2 & -6 \\ 0 & -3 & -3 \\ 0 & 0 & 0 \end{array}\right]$$

$$\stackrel{-\frac13r_3, \frac12 2r_1}\to \left[\begin{array}{rrr} 1 & -1 & -3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right]$$

which leads to $$v_3$$.

Let us take your example $$B-1I,$$ and finish the row-reduction by adding twice the first line to the second. You end up with $$\left[\begin{array}{rrr} -2 & 1 & 3 \\ 0&0&0\\ 0 & 0 & 0 \end{array}\right].$$ Now, the eigenspace of $$B$$ for the eigenvalue $$1,$$ i.e. the null-space of $$B-1I,$$ is the same as the null-space of that matrix. It is the subspace $$E_1=\left\{\begin{pmatrix}x\\y\\z\end{pmatrix}\in\Bbb R^3\mid-2x+y+3z=0\right\}$$ $$=\left\{\begin{pmatrix}x\\2x-3z\\z\end{pmatrix}\mid x,z\in\Bbb R\right\}$$ $$=\operatorname{span}\left(\begin{pmatrix}1\\2\\0\end{pmatrix}\begin{pmatrix}0\\-3\\1\end{pmatrix}\right).$$ You can train similarly on the second eigenspace (this time you will get 2 equations, and 1 spanning vector).