Suppose V is a finite-dimensional complex vector space and $T \in \mathcal{L}(V)$ Prove that if $T$ is diagonalizable then $V=\text{null}(T- \lambda I) \oplus \text{range}(T- \lambda I)$.
I already did the problem:if $T$ is diagonalizable then $V= \text{null}T \oplus \text{range}T$.
So would all I have to do to finish this problem, show that $T$ is diagonalizable $\implies$ $T- \lambda I$ is diagonalizable? I know the top implication goes both ways, but I do not have to prove the converse. Can someone verify if all I have to do to prove this direction is show $T$ is diagonalizable $\implies$ $T- \lambda I$ is diagonalizable? Also I am having trouble figuring out how to do this. Any suggestions?
Best Answer
Our OP user764658 has proved that for any diagonalizable $T$,
$V = \text{null}(T) \oplus \text{range}(T); \tag 1$
now $T$ is diagonalizable provided that there is some invertible matrix $P$ and some dialgonal matrix $D$ such that
$PTP^{-1} = D; \tag 2$
also, for any $\lambda$ we have
$P(T - \lambda I)P^{-1} = PTP^{-1} - P(\lambda I)P^{-1}$ $= PTP^{-1} - \lambda PIP^{-1}= PTP^{-1} - \lambda I = D - \lambda I; \tag 3$
since $D$ and $\lambda I$ are each diagonal, so is $D - \lambda I$; hance $T - \lambda I$ is also diagonalizable via (3). Hence we have
$V = \text{null}(T - \lambda I) \oplus \text{range}(T - \lambda I). \tag 4$
This shows that indeed all that is required is to show that $T - \lambda I$ is diagonalizable if $T$ is, since (1) holds for any diagonalizable operator.