Hi
After after doing row reduction on matrix[1 3; 2 2]
what would you get?
Im checking if method is completely misguided.
Thanks.
this matrix was formed in order to find an eigenvector or basis using a given eigenvalue .
got[1 3;0 -4] after some reductions ...which involved ensuring theres a zero beneath the first leading entry..
row2/-4 then gives [ 1 3 ;0 1]
now assigning parameters y=t
x=-3t;y=t
then [x ;y]=t[-3;1]
eigenvector then in [-3;1]
Yes, that is correct for "row-reduction" but I don't know what you mean about "finding eigenvectors". You cannot find eigenvectors (or eigenvalues) by row reduction!
In this particular problem, it is easy to see that $\displaystyle \begin{bmatrix}1 & 3\\ 0 & 1\end{bmatrix}\begin{bmatrix}-3 \\ 1\end{bmatrix}= \begin{bmatrix}0 \\ 1\end{bmatrix}$ which is NOT a multiple of ([3, 1]. [-3, 1] is NOT an eigenvector for this matrix!
(Here, the eigenvalues are $\displaystyle 1+\sqrt{3}$ and $\displaystyle 1- \sqrt{3}$. Eigenvectors corresponding to eigenvalue $\displaystyle 1+ \sqrt{3}$ are multiples of $\displaystyle [3, \sqrt{3}]$ and eigenvectors corresponding to eigenvalue $\displaystyle 1- \sqrt{3}$ are multiples of $\displaystyle [3, -\sqrt{3}]$).