# Homework Help: Please check my Eigenvector solutions.

1. Aug 12, 2011

1. The problem statement, all variables and given/known data
Find the characteristic equations, eigenvalues and eigenvector of the following matrix

2. Relevant equations

3. The attempt at a solution

Somehow somewhere I think the solution is wrong, based on online Eigenvector calculator on the web. Please do provide me actual answers and solutions. Thanks in advance!

2. Aug 12, 2011

### HallsofIvy

Your first eigenvector corresponding to $\lambda= 3$ is correct.

However, you have the wrong characteristic equation and "1" is NOT an eigenvalue. You should have seen that when you wrote $(A- \lambda I)x= 0$:
$$\begin{bmatrix}2 & 0 \\ 8 & -2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
which gives
$$\begin{bmatrix}2x_1 \\ 8x_1- 2x_2\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$$
which then requires that $2x_1= 0$ and $8x_1- 2x_2= 0$.
From the second $x_2= 4x_1$ but the first says $x_1= 0$ so $x_2= 4(0)= 0$. There is NO non-trivial vector satisfying this. 1 is NOT an eigenvalue
(I have no idea where you got "$x_1+ x_2= 0$".)

The characteristic equation is given by
$$\left|\begin{array}{cc}3- \lambda & 0 \\ 8 & -1- \lambda\end{array}\right|= 0$$
Which is, of course, simply $(3-\lambda)(-1- \lambda)= 0$.

3. Aug 12, 2011

Take a look at this. I've corrected it. Please let this answer correct :)

Some mistake there;
Eigenvector for lambda = 1 is [0;1]

One more, on the second last line.
is that correct to state "Let x_2=t" or "Let x_1=t"?

Last edited: Aug 12, 2011
4. Aug 12, 2011

### stringy

If you compute what you think are an eigenvector and eigenvalue pair, stick them back in! They had better satisfy $Ax = \lambda x$ since that is, after all, the equation whose solutions you were looking for in the first place.

The variable $x_2$ should equal t, if that's what you're asking.

5. Aug 12, 2011

### HallsofIvy

You mean "for lambda= -1".

You had just shown that $x_1= 0$ so you can't say "let x_1=t".

6. Aug 13, 2011