User:J D Bowen/Math725 HW11

1a) Let $$M \ $$ have singular value decomposition $$M=UDV^H \ $$. Observe that $$M \ $$ is injective if and only if $$\text{ker}(M) = \left\{{0}\right\} \ $$.  But since $$\text{ker}(V^H)\subseteq\text{ker}{M} \ $$, we must have $$(M \ \text{injective}) \implies \text{ker}(V^H)=\left\{{0}\right\} \ $$