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Let $V$ and $W$ be two finite-dimensional inner product spaces over the same field and let $T\in \mathcal{L}(V,W)\ $ be a linear transformation. Show that $T$ is injective if $T^*$ is surjective.

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Start by proving that $\mbox{Ker}T^*=(\mbox{Im} T)^\perp$. This gives you the if. To get the only if, recall that in finite dimension $F^\perp=\{0\}$ if and only if $F$ is the whole space (replace by dense in infinite dimension). –  1015 Apr 10 '13 at 21:46
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$T \in L(V,W),\ T^* \in L(W,V) $ as for inner product spaces duals can be identified to itself. So if for $ v \in V $ we have $ T(v) = 0 $ then for all $ w\in W $ we have $$ \langle T^*(w),v\rangle = \langle w,T(v)\rangle = 0 $$ But if $T^*$ is surjective then there exists $ w \in W $ such that $ T^*(w) = v $, hence $ \|v\| = \langle v,v\rangle = 0 $ so $ v= 0 $. Hence $T$ is injective.

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