I have this matrix which is corresponding to an eigenvalue $\lambda = 50$:
\begin{array}{ccc|c}
27 & 0 & 36&0\\
0&0&0&0 \\
36 & 0& -27&0
\end{array}
The eigenvectors for $\lambda$ are $(0, 1, 0)^T$ and $(1, 0, \frac{4}{3})^T$. But when I transform the matrix into reduced row echelon form I get:
\begin{array}{ccc|c}
1 & 0 & -7&0\\
0&0&1&0 \\
0 & 0& 0&0
\end{array}
which results only in the form $(0,a,0)^T$.
How do I see initially, that I can't use rref for finding eigenvectors?
Best Answer
When you're "finding the rref of a matrix", you're changing the matrix by applying row-operations. That is, you're multiplying on the left by an invertible matrix.
Now, suppose that $A$ has an eigenvector $v$ with eigenvalue $\lambda$, and suppose we've row reduced $A$ to the matrix $EA$, where $E$ is some invertible matrix. We then have $$ (EA)v = E(Av) = E (\lambda v) = \lambda Ev $$ so, whether $v$ is once more an eigenvector depends very much on the sequence of row-operations applied.
We also might not even have the same eigenvalues.