Let $V=\mathbb{C^2}$, vector space over the field $\mathbb{C}$,
and let $T:V \rightarrow V$ be defined by $T(z,w)=(z+iw, iz+w)$. I am asked to prove that T is diagonalizable (i.e., that there is a basis $\beta \subset V $
such that $[T]_\beta$ is a diagonal matrix).
I proposed the set $\delta=\{(1,0),(0,i)\}$ and calculated the solutions for the characteristic polynomial of $[T]_\delta$, in order to figure out a set of eigenvalues for T, and thus, a couple of linearly independent eigenvectors.
Since $[T]_\delta$ $= \begin{pmatrix}
1 & -1 \\
1 & 1
\end{pmatrix}$, then the characteristic polynomial of $[T]_\delta$ is $f(t)=det( [T]_\delta -tI)=-(1-t)^2$, and thus $\lambda=1$ is an eigenvalue of $T$. Therefore the left multiplication by $( [T]_\delta -1I)$ is a non-invertible linear transformation, which implies that there is some $v \in V$, with $v \neq 0$, such that $( [T]_\delta -1I)v=0$. However, when I try to find such a vector $v$, I am lead to the conclusion that $v=0$ (i. e.
$( [T]_\delta -1I)v=0$ implies that $v=0$). But that contradicts the fact that $f(1)=0$.
What is going on?
Best Answer
You computed the characteristic polynomial incorrectly. We have $$[T]_\delta -tI=\begin{pmatrix} 1-t & -1 \\ 1 & 1-t\end{pmatrix}$$ which has determinant $$(1-t)^2+1=t^2-2t+2,$$ not $-(1-t)^2$.