# Linear Algebra: Eigenvectors and eigenvalues

Find a basis for the corresponding eigenspace to the listed eigenvalue: $$A=\left[\begin{matrix}4&0&1\\-2&1&0\\-2&0&1\end{matrix}\right], \lambda=1$$
This is what I've come up with: $$\lambda=1:\left[\begin{matrix}4&0&1\\-2&1&0\\-2&0&1\end{matrix}\right]-\left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\right]=\left[\begin{array}{ccc|c}3&0&1&0\\-2&0&0&0\\-2&0&0&0\end{array}\right]=\left[\begin{array}{ccc|c}1&0&\frac13&0\\-2&0&0&0\\-2&0&0&0\end{array}\right]=\left[\begin{array}{ccc|c}1&0&\frac13&0\\0&0&\frac23&0\\0&0&\frac23&0\end{array}\right]$$
$$x_1=-\frac13x_3$$
$$x_2=0$$ $$x_3=\text{free}$$ Which I would assume would be:$$\left[\begin{matrix}-\frac13\\0\\1\end{matrix}\right]$$ for the basis of $\lambda=1$, assuming that I've done my work correctly, which I'm sure is unlikely. Now the answer, according to the book, is $$\lambda=1:\left[\begin{matrix}0\\1\\0\end{matrix}\right]$$ Could someone explain what I've done wrong? Because I'm not coming out with that answer. Or maybe I am, I just don't know it. Regardless, some help would be much appreciated. Thanks!

• It's not $x_3$ that is free and $x_2=0$, it's the other way around. Mar 29 '14 at 17:54
• Remember, the third column tells you about $\ x_3 \ .$ Mar 29 '14 at 17:57
• Oh right! Alright yeah, I see why $x_2$ is the free variable. But what about $x_1$?
– Jc E
Mar 29 '14 at 17:57
• @JcE You know what $x_3$, so you can find $x_1$. Mar 29 '14 at 17:58
• @JcE There are infinite possible answers. From $x_3=0$, you can infer $x_1=0$. Since $x_2$ is free, the set of solutions to system is $\color{grey}{\left\{ \begin{bmatrix}x_1\\ x_2 \\ x_3 \end{bmatrix}\in \mathbb R^{3\times 1}\colon x_3=0\land x_1=-\frac 1 3 x_3\right\}=}\left\{ \begin{bmatrix}0\\ x_2 \\ 0 \end{bmatrix}\colon x_2\in \mathbb R\right\}$. Mar 29 '14 at 18:14

After you have $$\begin{bmatrix} 3&0&1\\-2&0&0\\-2&0&0 \end{bmatrix},$$ it is much easier if you reduce to the following: $$\begin{bmatrix}3&0&1\\1&0&0\\0&0&0\end{bmatrix}.$$ Now since the coefficient of $x_2$ is always $0$, $x_2$ can be any value. And by the second row we have $x_1=0$. But then by the first row $-x_3=3\cdot 0$, so that $x_3=0$.