Let $ V $ be a finite-dimensional inner product space over $ \mathbb C $, and let $ T: V \rightarrow V $ be a linear operator.
(a) Show that $ T=0 \Leftrightarrow T^*=0 \Leftrightarrow TT^* \Leftrightarrow T^*T =0. $
(b) Show that if $ T $ is self adjoint, and $ T^n=0 $ for some $ n $, then $ T=0 $.
(c) Show that if $ T $ is normal, and $ T^n=0 $ for some $ n $, then $ T=0 $. (Hint: Show that if $ T $ is normal then $ TT^* $ is self adjoint.)
(d) Give an example of $ V $ and $ T $ where $ T \neq 0 $ and $ T=0 $. Show directly that t is not normal.
My attempt:
(a) $ T=0 \Leftrightarrow Tv=0 \Leftrightarrow \lt Tv,Tv \gt = 0 \Leftrightarrow <T^*w,T^*w>=0 \Leftrightarrow <T^*w=0 \Leftrightarrow T^*=0 \Leftrightarrow TT^*=0 \Leftrightarrow T^*T=0 \Leftrightarrow TT^*=0. $
(b) Since T is self adjoint $ \ T=T^*$ So, $\ T^2 =T^*T=0 \Rightarrow \ T=0 $ and the base case holds. Suppose $ m $ is an integer for which the hypothesis holds. Then $\ T^m=0 \rightarrow \ T=0$
$ \ T^{m+1} = T^m . T^*=0\rightarrow T^*=T=0.$ by part (a). $(\ TT^*=0 \rightarrow \ T^*=0) $ Hence the results holds for all $\ n\in \mathbb N. $
(c) This is my first question. Since $ (ST)^* = T^*S^*$ we have $ (TT^*)^* = T^{**}T^* = TT^* $ which does not make sense. Where have I gone wrong?
(d) Let $ V= \mathbb{F}_2 = \mathbb{Z}/2 \mathbb{Z} $ and suppose $ T $ is the linear transformation given by $ \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}$.
How do I find the explicit linear transformation given by this matrix?
Are my solutions to part (a) and (b) correct?
Any help is greatly appreciated.
Quick correction to (a): you should emphasize that $T = 0 \iff Tv = 0$ for all $v \in V$. For all steps with $v,w$, it should be clear that the equality is meant to hold for all such vectors (or at least over a basis).
For (b), your proof by induction is faulty. In particular, it's not clear how you deduce $$ T^mT^* = 0 \implies T^* = T = 0 $$ this is not a clear application of part (a). Instead, I would note that if $k<m \leq 2k$, then $$ T^m = 0 \implies T^{2k} = 0 \implies (T^k)(T^k)^* = 0 \implies T^k = 0 $$ which allows us to set up a proof by strong induction. (If it helps, take $k = \lceil m/2 \rceil$).
For (c), you have not gone wrong at all: $TT^*$ is necessarily self-adjoint, regardless of whether $T$ is normal. The trick is to note that $$ T^m = 0 \iff (T^m)(T^m)^* = 0 \iff (TT^*)^m = 0 \iff TT^* = 0 \iff T = 0 $$
For (d), remember that $V$ should be an inner product space. Consider instead $$ A = \pmatrix{0&1\\0&0} $$ The linear transformation given by $A$ is $T(v) = Av$.