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Question:

How to find the eigenspace of $A$ corresponding to all the different real eigenvalues.

This matrix only three real eigenvalues, $\lambda = 5, 1, 1$. Step by step, how would I go about finding the eigenspace corresponding to an eigenvalue, say, $\lambda = 1$?

A = $\begin{bmatrix} 2 & 2 & -1\\ 1 & 3 & -1\\ -1 & -2 & 2\\ \end{bmatrix}$

Thanks!

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    $\begingroup$ It has two real eigenvalues with one repeated, $\lambda_{1, 2, 3} = 1, 1, 5$ (unless it is written incorrectly). Also, you can find three linearly independent eigenvectors, no generalized or chaining required. $\endgroup$ – Moo Apr 11 '16 at 2:19
  • $\begingroup$ I will correct it. $\endgroup$ – jackskis Apr 11 '16 at 2:21
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The equation $(A-I)x=0$ yields \begin{align}x_1+2x_2-x_3&=0\\ x_1+2x_2-x_3&=0\\-x_1-2x_2+x_3&=0, \end{align} and hence $$x_1+2x_2=x_3. $$ It follows that the eigenspace corresponding to $\lambda=1$ is the span of $$\begin{bmatrix}2\\-1\\0\end{bmatrix},\begin{bmatrix}1\\0\\1\end{bmatrix}. $$ Similarly, $(A-5I)x=0$ yields \begin{align} -3x_1+2x_2-x_3&=0\\ x_1-2x_2-x_3&=0\\ -x_1-2x_2-3x_3&=0, \end{align} and hence $$x_3=-x_1,\quad x_2=x_1. $$ It follows that the eigenspace corresponding to $\lambda=5$ is the span of $$\begin{bmatrix}1\\1\\-1\end{bmatrix}.$$

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  • $\begingroup$ How do you get from step 2 to the span corresponding to $\lambda = 1$? I can figure out how you did it for $\lambda = 5$ once you answer for $\lambda = 1$ $\endgroup$ – jackskis Apr 11 '16 at 16:07
  • $\begingroup$ A linear equation of the form $ax_1+bx_2+cx_3=0$ is a plane and so is the span of two linearly independent vectors in $\mathbb R^3$. I just let $x_2=0$ and $x_3=0$ to find two vectors on that plane. $\endgroup$ – Math1000 Apr 11 '16 at 16:51

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