# Please check my Eigenvector solutions.

Tags: check, eigenvector, solutions
 P: 15 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!
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 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 $(3-\lambda)(-1- \lambda)= 0$.
 P: 15 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"?
P: 90
Please check my Eigenvector solutions.

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.

 Quote by hadizainud One more, on the second last line. is that correct to state "Let x_2=t" or "Let x_1=t"?
The variable $x_2$ should equal t, if that's what you're asking.
Math
Emeritus
You had just shown that $x_1= 0$ so you can't say "let x_1=t".