A question about linear algebra about orthogonal projection

Question

$T: \Bbb R^n \to \Bbb R^n$ a linear map. I want to prove that if the orthogonal complement of the image of $T$ intersects the subspace $E$ in only the $0$ subspace, then $P_E(T)$ is onto $E$. Here $P_E$ is the orthogonal projection to the subspace $E$. I tried going with checking dimensions but couldn't do anything.

Answer 1

The question can be stated in terms of subspaces. Denote $V=T(\mathbb{R}^n).$ We want to show that if a subspace $V$ satisfies $V^\perp\cap E =\{0\}$ then $P_E(V)=E.$ It can be solved analyzing the dimensions of subspaces (I leave this method to the asker). Another explanation can be done as follows. Let $V$ and $E$ be subspaces such that $V^\perp\cap E=\{0\}.$ Assume $x\in E$ and $x\perp P_E(V).$ We decompose $x$ as $$x=x_V+x_{V^\perp},\quad x_V\in V,\quad x_{V^\perp}\in V^\perp $$ Then $x=P_E(x_V)+P_E(x_{V^\perp}).$ We take the inner product with $x$ and get $$\langle x,x\rangle =\langle P_E(x_V),x\rangle +\langle P_E(x_{V^\perp}),x\rangle=\langle P_E(x_{V^\perp}),x\rangle$$ As $\|P_E(x_{V^\perp})\|\le \|x\|,$ we conclude that $P_E(x_{V^\perp})=x.$ The latter implies $x=x_{V^\perp},$ thus $x=0.$ Summarizing the subspace $E$ does not contain a nonzero vector orthogonal to $P_E(V).$ Hence $P_E(V)=E.$

Remark The same proof can be applied to Hilbert spaces and closed subspaces but the conclusion should be replaced by $P_E(V)$ is dense in $E,$ i.e. $\overline{P_E(V)}=E.$

Answer 2

Sticking with the finite-dimensional case, I will give a somewhat different suggestion. The key fact is that if $U$ and $V$ are subspaces of $\Bbb R^n$ (or any finite-dimensional inner product space), then $$U^\perp\cap V^\perp = (U+V)^\perp.$$ (I leave this to you to check.) Taking orthogonal complement again (here is where we use finite-dimensionality), we have $$(U^\perp\cap V^\perp)^\perp = U+V.$$ We apply this in your case with $U=\text{im}(T)$ and $V=E^\perp$. Since we are given $U^\perp\cap E = (0)$, we conclude that $U+E^\perp = \Bbb R^n$. From this it follows that $E=P_E(\Bbb R^n)=P_E(U+E^\perp)= P_E(U)$, as required.