Will $V+V^{\perp}$ always span $\mathbb{R}^n$?

Question

Let's say we have a subspace $V$, that is a subset of $\mathbb{R}^n$. Does $V + V^{\perp}$ always span $\mathbb{R}^n$?

Answer 1

What you mean is a linear subspace, not a linearly independent subspace, which does not make any sense.
Proof: Take a basis for $V$, $B_V=(v_1,...,v_k)$. The condition $w\in V^\perp$ is equivalent to $\langle v_i,w\rangle=0$ for $i=1,...,k$. Hence you have a system of $k$ linear equations. Since $B_V$ are linearly independent, those equations are independent as well. Hence the space of solutions to this system, which is exactly $V^\perp$, will be of dimension $n-k$. Take a basis $B_{V^\perp}=(u_1,...,u_{n-k})$. Check that $(v_1,...,v_k,u_1,...,u_{n-k})$ is linearly independent and thus is a basis for $\mathbb{R}^n$. So $V+V^\perp=\mathbb{R}^n$

Answer 2

Yes, every subspace of a normed linear space with finite dimension is closed. It is also true that if $X$ is a closed, linear subspace of a Hilbert space $H$ (in this case $\mathbb{R}^n$), then $H = X \bigoplus X^{\perp}$, where $\bigoplus$ denotes the direct sum of $X$ and $X^{\perp}$ (i.e., $X \cap X^{\perp} = \{0\}$ and $h = x + x'$, for all $h \in H$ with $x \in X$ and $x' \in X^{\perp}$).

Answer 3

For $w\in \mathbb R^n$, select $v\in V$ that minimizes $|w-v|$ (why is that possible?). Then $w-v\in V^\perp$ (why?).