Define $T \in \mathcal{L(F^2)}$ by $$ T(w,z) = (z,w)$$Find all eigenvalues and eigenvectors of $T$.
An eigenvalue is defined as $\lambda \in \mathcal{F}$ such that $Tv = \lambda v$ where $v \neq 0$. Expanding the definition, I get the following system of equations
$$z = \lambda w \\ w = \lambda z \tag 1.$$
Now I can simplify this system by substituting $w$ for $z$ values to get $$ z = \lambda^2 z \\
0 = \lambda^2 z – z \\
0 = z(\lambda^2 – 1)$$
Solving for $\lambda$ gives $ \lambda = \pm 1$ so the two eigenvalues are $1, -1$.
To find the corresponding eigenvectors, from (1) I think the vector I get is $(z, w) = (\lambda ^2 z, \lambda z) = (z,z)$ when $\lambda = 1)$ and
$(z, w) = (\lambda ^2 z, \lambda z) = (z,-z)$ when $\lambda = -1)$
However I am not sure if I found the corresponding eigenvectors. What would be a way I could check for myself that these are correct?
Best Answer
Just check. If $(x,y)$ is an eigenvector of $\textsf T$ with corresponding eigenvalue $\lambda=1$, then $(x,y)$ satisfies $$(y,x)=\textsf{T}(x,y)=1(x,y)=(x,y)$$ that is, $x=y$. The vectors $(x,y)$ that satisfies $x=y$ are exactly the vectors living in the span of $\{(1,1)\}$. Similarly, the eigenvectors with corresponding eigenvalue $\lambda=-1$ are the vectors in $\operatorname{span}\big(\{(1,-1)\}\big)$.