First question here. So I've been asked to find the eigenvectors for the following 3×3 matrix A:
\begin{bmatrix}
1 & 4 & 5 \\
0 & 2 & 6 \\
0 & 0 & 3 \\
\end{bmatrix}
I found the characteristic equation and found the eigenvalues to be $\lambda$=1, 2 and 3.
Just as an example of my issue, let's take $\lambda$=2. Plugging that into A-I$\lambda$ to find the eigenvectors, I can get the augmented matrix:
$$ \left[
\begin{array}{ccc|c}
-1 & 4 & 5 & 0 \\
0 & 0 & 6 & 0 \\
0 & 0 & 1 & 0 \\
\end{array}
\right] $$
This is where I get stuck. I honestly have no idea what the process is to pick a free variable out of the 3 (let's call them x, y and z). Could someone please run me through the process? I've found that I never seem to get these right. A general method to know which variable is the free variable would also be great if it's not too much to ask.
Thanks in advance!
Best Answer
Forget about free variables for the moment, and try to think about what the augmented matrix represents. With enough practice with examples of this kind, it will become clear which variables are going to be the free variables. In your working above, you got to the augmented matrix $$ \left[ \begin{array}{ccc|c} -1 & 4 & 5 & 0 \\ 0 & 0 & 6 & 0 \\ 0 &0 &1&0 \end{array} \right],$$ which is correct. With a few more elementary row operations (if you want the details, just post a comment and I will edit my answer to include these operations) we get $$ \left[ \begin{array}{ccc|c} -1 & 4 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 &0 &0&0 \end{array} \right].$$ Now, reading off this matrix, we see that $z=0$ and $-x + 4y =0.$ So we see that $$\left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 4y \\ y \\ 0 \end{array} \right) = y\left( \begin{array}{c} 4 \\ 1 \\ 0 \end{array} \right),$$ showing that $y$ is the free variable in this case. Further, we now see that $\left( \begin{array}{c} 4 \\ 1 \\ 0 \end{array} \right)$ is an eigenvector corresponding to the eigenvalue $\lambda = 2.$