so I have this matrix:
$$
\begin{matrix}
0.7 & 0 & 0 \\
0.1 & 0.6 & 0 \\
0 & 0.2 & 0.8 \\
\end{matrix}
$$
I managed to solve eigenvectors for the eigenvalues 0.7, 0.6 but for the eigenvalue 0.8, the first 2 terms are 0 and the 3rd term is not present.
I saw an answer in some university PDFs for this question, and they say the eigenvector for eigenvalue 0.8 is :
$$\pmatrix{0\\0\\1}$$
How can you get the eigenvector in a case like this when v1, v2 terms are 0 and v3 term is not included?
For eigenvalue 0.8 i got the following equations:
-0.1 * v1 = 0
0.1 * v1 – 0.2 * v2 = 0
0.2 * v2 = 0
There are no v3 terms included, so how can i figure it out?
Best Answer
Yes, there are no conditions on $x_3$. That means that $x_3$ can be anything, or rather any real (or complex...) number. Therefore you don't only have a single eigenvector, but a whole one-dimensional space of them! $x_3=1$ is one solution and all others are multiples of it.