# Finding eigenvectors of a $2\times 2$ matrix.

I need to find the eigenvectors of the following matrix. The eigenvalues are 0.8 and 1 (this was double checked with Wolfram Alpha).

$\begin{bmatrix}0.8 & 0\\ 2 & 1\end{bmatrix}$

To find the eigenvector of 1: $\begin{bmatrix}0.8 & 0\\ 2 & 1\end{bmatrix}$-$\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$=$\begin{bmatrix}-0.2 & 0\\ 2 & 0\end{bmatrix}$. Which implies that the vector is $\begin{bmatrix}0 \\ 0\end{bmatrix}$.

To find the eigenvector of 0.8: $\begin{bmatrix}0.8 & 0\\ 2 & 1\end{bmatrix}$-$\begin{bmatrix}0.8 & 0\\ 0 & 0.8\end{bmatrix}$=$\begin{bmatrix}0 & 0\\ 2 & 0.2\end{bmatrix}$, which implies the vector is $\begin{bmatrix}0.1\\ 1\end{bmatrix}$.

However, my answers are incorrect. Wolfram says the answers are respectively $\begin{bmatrix}0 \\ 1\end{bmatrix}$ and $\begin{bmatrix}0.0995\\ -0.995\end{bmatrix}$. (link here:http://www.wolframalpha.com/input/?i=eigenvectors+of+%7B%7B0.8%2C0%7D%2C+%7B2%2C1%7D%7D)

What did I do wrong? I've tried this problem many times without success.

• By definition, for an eigenvector $\vec x$ it holds $\vec x\neq \vec 0$. – thanasissdr Oct 4 '15 at 17:46
• @thanasissdr that's what rung a bell that told me something was off. – Grizzly0111 Oct 4 '15 at 17:47

for $\lambda = 0.8$ you need to solve

$$\begin{bmatrix}0.8 & 0\\ 2 & 1\end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix} = 0.8\begin{bmatrix}x \\ y \end{bmatrix}$$

so $$\begin{bmatrix}0.8 x \\ 2x+y \end{bmatrix} =\begin{bmatrix}0.8x \\0.8 y \end{bmatrix}$$

$0.8x=0.8x$ doesn't tell you anything but $2x+y=0.8y$ gives you $y=-10x$ so the eigenvector is any vector of the form $$k\begin{bmatrix} 1 \\ -10 \end{bmatrix}$$

Wolfram Alpha chooses $k=0.0995$ to make the eigenvector a unit vector.

Let's say that we want to find an eigenvector that corresponds to the eigenvalue $\lambda_1=1$. We can solve the equation: $$\begin{array}{l}A\cdot \mathbf{x} = 1\cdot \mathbf {x}\\[2ex] \begin{bmatrix} 0.8&0\\2&1\end{bmatrix}\cdot \begin{bmatrix}x_1\\x_2\end{bmatrix}= \begin{bmatrix}x_1\\x_2\end{bmatrix}\\[3ex] \left\{\begin{array}{l} 0.8x_1+0x_2=x_1\\ 2x_1+x_2=x_2 \end{array} \right. \end{array}$$ From the first equation, we have that $x_1=0$ and plugging that to the second equation we have that $x_2 = x_2$, which implies that $x_2$ can be any real number except for zero! Thus, we can select $x_2=1$. So, the first eigenvector (one of the ones that correspond to the eigenvalue $\lambda_1 =1$) is $\mathbf{x} = \begin{bmatrix} 0 \\1 \end{bmatrix}$.

Your first system of equations is:

$-0.2x + 0y = 0\\2x + 0y = 0.$

This tells us $x = 0$, but does not restrict $y$ in any way. Thus ANY non-zero value of $y$ lead to an eigenvector, $(0,y) = y(0,1)$.

Since any non-zero scalar multiple of an eigenvector is ALSO an eigenvector, we can choose ANY basis vector generating the subspace $\{(0,y):y \in \Bbb R\}$. It is traditonal to choose the eigenvector $(0,1)$, but any non-zero element of the subspace would do just as well.

Indeed we find that:

$\begin{bmatrix}0.8&0\\2&1\end{bmatrix}\begin{bmatrix}0\\y\end{bmatrix} = \begin{bmatrix}0\\y\end{bmatrix}$

which confirms $(0,y)$ is an eigenvector belonging to the eignevalue $1$ for any non-zero $y$.