Let $U$ and $V$ denote two finite-dimensionel inner product vectorspaces over $\mathbb{C}$, where $\mathrm{dim}\hspace{1pt} U=6$ and $\mathrm{dim}\hspace{1pt} V=3$. Furthermore $R\in\mathcal{L}(U)$, $K\in\mathcal{L}(V)$ and $T\in\mathcal{L}(U,V)$, where $R$ and $K$ are bijective and $T$ is surjective. Let $S=KTR$. Prove that $SS^{*}$ is invertible.
My attempt at doing the proof:
$KT\in\mathcal{L}(U,V)$ is surjective since both $K$ and $T$ are surjective, however $KT$ can't be injective since $\mathrm{dim}\hspace{1pt} U=6>\mathrm{dim}\hspace{1pt} V=3$.
$TR\in\mathcal{L}(U,V)$ is also only surjective by the same argument as above.
Now observe that $S\in\mathcal{L}(U,V)$ is surjective since both $KT$ and $R$ are surjective (or since both $K$ and $TR$ are surjective) and not injective by the same argument as before. Thus $S^{*}\in\mathcal{L}(V,U)$ is injective.
That the linear operator $SS^{*}\in\mathcal{L}(V)$ is invertible is equivalent to it being surjective and injective. However $SS^{*}$ is not injective since $S$ is not injective; that $S$ is not injective implies that there exists a vector $v\neq0\in V$ such that $S^*(v)=u\neq0 \in U$ such that $SS^{*}(v)=S(u)=0$ . Hereby, it has been shown that $\mathrm{ker}\hspace{1pt}SS^{*}\neq\{0\}$ which implies that $SS^{*}$ is not invertible.
What am I doing wrong??
Best Answer
It doesn't follow that $\ SS^*\ $ isn't injective merely from the fact that $\ S\ $ isn't. All you need for $\ SS^*\ $ to be injective is that $\ S^*\ $ be injective, and that $\ \ker(S)\cap\mathcal{R}(S^*)=\{0\}\ $, because if that's the case, and $\ SS^*x=0\ $, then $\ S^*x\in \ker(S)\cap\mathcal{R}(S^*)\ $, so $\ S^*x=0\ $ from the second condition, and then $\ x=0\ $ from the first.
And in fact, it's not difficult to show that these conditions hold for your problem. You've already shown that the first one holds, and if $\ y=S^*x\in \ker(S)\cap\mathcal{R}(S^*)\ $, then $\ \|y\|^2=$$\langle y,y\rangle =$$ \langle S^* x,y\rangle =$$ \langle x,Sy\rangle=0\ $, because $\ y\in\ker(S)\ $. Therefore $\ y=0\ $, and the second condition holds as well.