I am trying to obtain a basis for an eigenspace given the standard matrix of a linear operator over a space. I have done all of the work. I just need to confirm my results or find my mistake.
A=[F]=
\begin{array}{ccc}
3 & 2 & 1 \\
0 & 2 & 4 \\
0 & 0 & 4
\end{array}
After finding $|\lambda I – A|$ I get that the eigenvalues are $\lambda_{1}=2$, $\lambda_{2}=3$ and $\lambda_{3}=4$. I am having a problem with $\lambda=4$. When I compute $4I-A$, the computation yields that there is no basis for the nullspace, does this mean that there is no basis for this eigenspace?
4I-A=
\begin{array}{ccc}
1 & -2 & -1 \\
0 & 2 & -4 \\
0 & 0 & -4
\end{array}
So then upon row reducing to the canonical form $C$, I obtain the identity matrix.
I_3=
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}
This says that $x=y=z=0$.
So, is this the basis for the eigenspace? The basis for the nullspace does not exist since the only vector in the basis of the nullspace is the zero vector.
Best Answer
No. A basis is a linearly in-dependent set. And the set consisting of the zero vector is de-pendent, since there is a nontrivial solution to $c\vec{0}=\vec{0}$.
If a space only contains the zero vector, the empty set is a basis for it. This is consistent with interpreting an empty sum as zero.
But with respect to your computation, you are just not computing $4I-A$ correctly. Look again at the lower right entry.