1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ By definition, for an eigenvector $\vec x$ it holds $\vec x\neq \vec 0$. $\endgroup$ – thanasissdr Oct 4 '15 at 17:46
  • $\begingroup$ @thanasissdr that's what rung a bell that told me something was off. $\endgroup$ – Grizzly0111 Oct 4 '15 at 17:47
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\mathbf{x} = \begin{bmatrix} 0 \\1 \end{bmatrix}$.

$\endgroup$
0
$\begingroup$

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 $y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.