Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
The phrase "linearly independent subspace" doesn't make sense. A set of vectors may be linearly independent (or not), but not a subspace, since the zero vector is contained in any subspace. – Shaun Ault Jan 2 '13 at 15:49
Yes, it happens. I'll give you an hint. Set an orthogonal base of $V$, complete it orthogonally and look at the matrix. – Ivan Jan 2 '13 at 15:51
I'd make that $+$ into a $\oplus$ to indicate a direct sum of the perp spaces. – JohnD Jan 2 '13 at 16:02
Although you are asking about $\Bbb R$, it's worth noting that this is not true for finite fields. There are, for example, proper subspaces such that $V=V^\perp$ using the "usual" inner product, so that $V+V^\perp$ is not the whole space. – rschwieb Jan 2 '13 at 17:41
up vote 1 down vote accepted

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 $V+V^\perp=\mathbb{R}^n$

share|cite|improve this answer

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?).

share|cite|improve this answer
I don't really understand what you mean? Does it have anything to do with the zero vector being in both subspaces? – JohnPhteven Jan 2 '13 at 16:07

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 $x' \in X^{\perp}$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.