# Question about eigenvector, basis for the solution

I'm confused with some question currently I'm trying to solve. If you help that will be grateful.

Given the matrix find eigenvalues and eigenvectors $$A = \begin{bmatrix} 4 & -2 & 1 \\ 2 & 0 & 1 \\ 2 & -2 & 3 \\ \end{bmatrix}$$

$$|A - \lambda I| = 0$$ $$|A - \lambda I| = (\lambda-2)^2(\lambda-3)$$ $$\lambda_1 = 2, \lambda_2 = 3$$

Now in Case 1: $\lambda = 2$ book states that:

• $\begin{bmatrix} 2 & -2 & 1 \\ 2 & -2 & 1 \\ 2 & -2 & 1 \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}$ ,

which reduces the single equation $2x-2y+z=0$. This equation obviously has a 2-dimensional solution space. With $y = 1$ and $z = 0$, we get $x = 1$ and, hence, obtain the basis eigenvector $v_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix}$ . With $y = 0$ and $z=2$ we get $x = -1$ and hence the basis eigenvector $v_2 = \begin{bmatrix} -1 \\ 0 \\ 2 \\ \end{bmatrix}$ . The 2-dimensional eigenspace of $\mathbf A$ associated with the repeated eigenvalue $\lambda = 2$ has basis $\{v1, v2\}$

Isn't $v = \begin{bmatrix} 0 \\ -1 \\ -2 \\ \end{bmatrix}$ is also an eigenvector? Cause it also fits to equation $2x-2y+z=0$

• That's true. But $v=-v_1-v_2$. Jan 4 '15 at 2:20

Your third vector is a linear combination of the two vectors given in the solution, and hence does not produce anything new in the eigenspace. A two-dimensional eigenspace has many possible bases. (Your $v$ is equal to $-v_1-v_2$.)