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!
Your first eigenvector corresponding to [itex]\lambda= 3[/itex] is correct. However, you have the wrong characteristic equation and "1" is NOT an eigenvalue. You should have seen that when you wrote [itex](A- \lambda I)x= 0[/itex]: [tex]\begin{bmatrix}2 & 0 \\ 8 & -2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}[/tex] which gives [tex]\begin{bmatrix}2x_1 \\ 8x_1- 2x_2\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/tex] which then requires that [itex]2x_1= 0[/itex] and [itex]8x_1- 2x_2= 0[/itex]. From the second [itex]x_2= 4x_1[/itex] but the first says [itex]x_1= 0[/itex] so [itex]x_2= 4(0)= 0[/itex]. There is NO non-trivial vector satisfying this. 1 is NOT an eigenvalue (I have no idea where you got "[itex]x_1+ x_2= 0[/itex]".) The characteristic equation is given by [tex]\left|\begin{array}{cc}3- \lambda & 0 \\ 8 & -1- \lambda\end{array}\right|= 0[/tex] Which is, of course, simply [itex](3-\lambda)(-1- \lambda)= 0[/itex].
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"?
If you compute what you think are an eigenvector and eigenvalue pair, stick them back in! They had better satisfy [itex] Ax = \lambda x[/itex] since that is, after all, the equation whose solutions you were looking for in the first place. The variable [itex]x_2[/itex] should equal t, if that's what you're asking.