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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 $2x-2y+z=0$

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  • $\begingroup$ That's true. But $v=-v_1-v_2$. $\endgroup$
    – Bernard
    Jan 4 '15 at 2:20
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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 $-v_1-v_2$.)

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