Let $V$ be a finite inner product space with $T\colon V\to V$ a linear transformation.
How can I prove that, $(\ker{T})^{\perp}\subseteq \operatorname{Im}T^{*}$ ?
Edit:
My purpose is to prove that:
$\operatorname{rank}(T)=\operatorname{rank}(T^{*})$
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5$\begingroup$ It is easy to prove $(\mbox{Im} T^*)^\perp=\mbox{Ker}T$ by writing $0=(Tx,y)=(x,T^*y)$ for every $y$, and every $x\in\mbox{Ker} T$. So $((\mbox{Im} T^*)^\perp)^\perp=(\mbox{Ker}T)^\perp$. Now for $F$ a subspace in general, $(F^\perp)^\perp=\overline{F}$. This follows from $H=\overline{F}\oplus \overline{F}^\perp=\overline{F}\oplus F^\perp$. Finally, in finite dimension, every subspace is closed: $\overline{F}=F$. $\endgroup$– JulienCommented May 6, 2013 at 10:36
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$\begingroup$ What does $T^\perp$ mean? $\endgroup$– Michael MoradCommented Sep 21, 2023 at 15:22
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1 Answer
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I thought of an alternative proof to what you originally wanted to show:
If $T = [A]$ in an orthonormal basis (exists because we're in a finite space) $[T^*] = \bar{A}^t = A^*$.
$rank A^* = rank \left(\bar{A}^t \right)$ and because the conjugate does not change the rank(tell me if you need a proof to this one as well) it's the same as:
$rank \left(A^t\right)$ but as you've probably proved that $\forall M_{n\times n} rank_C\left(M \right) = rank_R\left(M \right) $ so the transpose does not have any change the rank so it equals to $rank \left(A\right)$.
Which therefore means that given the isomorphism of matrices and the linear transformation they describe that $\operatorname{rank}(T)=\operatorname{rank}(T^{*})$.
