1. Mar 24, 2014

### 939

2. Mar 25, 2014

### Staff: Mentor

The equation he obtains is
$$\left( \begin{array}{cc} 8 & 1 \\ -8 & -1 \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) = 0$$
$(x,y) = (1, -8)$ is a solution, but not $(x,y) = (-8, 1)$. Try it for yourself: substitue the possible solutions in the above equation, and see what works.

By the way, there is an infinite number of possible solutions: any vector $a (1, -8)$ is also a solution (with $a$ a scalar). Which one to choose is arbitrary. In the video, he could have chosen $y=1$ instead, and found $(x,y) = (-1/8, 1)$ (you can check for yourself that solution also).

3. Mar 25, 2014

### HallsofIvy

Staff Emeritus
You talk here, and in a similar question in the homework section as if the "eigenvector" were just two numbers in arbitrary order. It is not- it is a vector which, in this case, can be represented[/b] as an ordered pair of numbers.

You should, yourself, have done the multiplication
$$\begin{pmatrix}8 & 1 \\ -8 & -1\end{pmatrix}\begin{pmatrix}-8 \\ 1\end{pmatrix}= \begin{pmatrix}-64+ 1 \\ 64- 1\end{pmatrix}= \begin{pmatrix}-63 \\ 63\end{pmatrix}$$
which is NOT
$$0\begin{pmatrix}-8 \\ 1 \end{pmatrix}$$