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$A$ is the matrix:

$$A = \begin{pmatrix} 4 & -5 & 2 \\ 5 & -7 & 3 \\ 6 & -9 & 4 \end{pmatrix}$$

As far As I know to find root subspace we have to find eigenvalue of linear operator, so:

$$\mathrm{det}|A-I{\lambda}| = \begin{vmatrix}4-\lambda & -5 & 2 \\ 5 & -7-\lambda & 3 \\ 6 & -9 & 4-\lambda \end{vmatrix} = \lambda^3-\lambda^2 $$

So:

$$\lambda^3-\lambda^2 = 0 \Rightarrow \lambda(\lambda^2-\lambda) = 0$$

Hence:

$\lambda_1 = 0$ and $\lambda_2 = 1$

How should I proceed to find root subspace?

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  • $\begingroup$ You have to solve the linear system $(A - \lambda I) \mathbf{x}=\mathbf{0}$ $\endgroup$ – Crostul Jun 8 '17 at 8:26
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Solve the equations $A.\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)=\lambda\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)$ when $\lambda=0$ and $\lambda=1$.

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