# Orthogonal Complement and Direct Sum

I deal with the following problem:

Let $V$ be an inner product vector space and let $S$ and $T$ be two subspaces such that $V = S \oplus T$. Then, prove that $V = {S^ \bot } \oplus {T^ \bot }$ where ${S^ \bot }$ and ${T^ \bot }$ are orthogonal complement of $S$ and $T$, respectively. I don't know that $V$ is finite or infinite dimensional.

My attempt is here:

For finite dimensional inner product vector space $V$, let $\left\{ {{s_1}, \ldots ,{s_n}} \right\}$, $\left\{ {{s_{n+1}}, \ldots ,{s_{n+m}}} \right\}$, $\left\{ {{t_1}, \ldots ,{t_m}} \right\}$ and $\left\{ {{t_{m+1}}, \ldots ,{t_{m+n}}} \right\}$ be a basis for $S$, ${S^ \bot }$, $T$ and ${T^ \bot }$, respectively. Then, I have to show that the union of $\left\{ {{s_{n+1}}, \ldots ,{s_{n+m}}} \right\}$ and $\left\{ {{t_{m+1}}, \ldots ,{t_{m+n}}} \right\}$ is a basis for $V$. I know that $V = {S^ \bot } \oplus {S }={T^ \bot } \oplus {T }$. But I can not go on. I will pleause if one answers this question.

One can construct a counter-example as follows. Let $V= l_0({\bf N})$ be the vector space of sequences of reals numbers $(u_n)$ with $u_n=0$ if $n$ is large enough : $V$ is just the union of all finite dimensional euclidian spaces. Let $S$ the hyperplane defined by $\Sigma _n u_n =0$, and $T$ the line directed by the sequence $(1,0,...,0,..)$. Then this pair is a counter example to the question, as the orthogonal to $S$ is reduce to $0$ whereas the orthogonal to $T$ is the set of sequences with $u_1=0$.

If the vector space is finite dimensional, the result can be proved just by counting dimensions.

• I think it is not enough to count the dimension. You must also show that the union of the basis of the orthogonal complements of $S$ and $T$ is linearly independent. Can you prove it Thomas? Oct 26, 2015 at 14:15
• If a vector is orthogonal to $S$ and $T$, then it is orthogonal to $S+T=V$, and it is $0$. Oct 26, 2015 at 14:23
• You are right, Thomas. Thank you very much. Oct 27, 2015 at 6:56
• The counter example is very good. $V$ could also be considered as the free vector space over any infinite set. You might expand on why the orthogonal complement of $S$ is {0}: I think because firstly you prove all coefficients must be equal, and secondly that they must be zero ? Oct 27, 2015 at 8:05
• Let $x=(x_1,..,x_n,0,0,0....)$, and $s=\Sigma x_i$. Note that $y=(x_1,..,x_n, -s, 0,..)$ is in $S$ but not orthogonal to $x$ unless $x=0$. No non zero vector $x$ can be orthogonal to every vector in $S$. Oct 27, 2015 at 8:36

If $V=S\oplus T$, then $T=S^{\perp}$ and $S=T^{\perp}$. Therefore ...

(Note: I am assuming $\oplus$ means orthogonal decomposition.)

• I assumed it meant direct sum, and not necessarily orthogonal in which case it doesn't follow that , $T = S^{\perp}$ Oct 27, 2015 at 7:48
• @TomCollinge : Yes, based on what the OP was being asked to show, you have to assume orthogonal decomposition. Oct 27, 2015 at 13:15