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Let $V$ be a finite dimensional real vector space with inner product $\langle \, , \rangle$ and let $W$ be a subspace of $V$. The orthogonal complement of $W$ is defined as $$ W^\perp= \left\{ v \in V \,:\, \langle v,w \rangle = 0 \text{ for all } w \in W \right\}. $$ Prove the following: $\dim W + \dim W^\perp= \dim V$.

I'm not sure how to find the relationship between number of basis vectors in $W$ and $W^\perp$.

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5 Answers 5

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Let $\beta=\{w_1,w_2,\ldots,w_k\}$ and $\gamma=\{x_1,x_2,\ldots,x_m\}$ be the bases for $W$ and $W^\perp$, respectively. It suffices to show that $$\beta\cup\gamma=\{w_1,w_2,\ldots,w_k,x_1,x_2,\ldots,x_m\}$$ is a basis for $V$. Given $v\in V$, then it is well-known that $v=v_1+v_2$ for some $v_1\in W$ and $v_2\in W^\perp$. Also because $\beta$ and $\gamma$ are bases for $W$ and $W^\perp$, respectively, there exist scalars $a_1,a_2,\ldots,a_k,b_1,b_2,\ldots,b_m$ such that $v_1=\displaystyle\sum_{i=1}^ka_iw_i$ and $v_2=\displaystyle\sum_{j=1}^mb_jx_j$. Therefore $$v=v_1+v_2=\sum_{i=1}^ka_iw_i+\sum_{j=1}^mb_jx_j,$$ which follows that $\beta\cup\gamma$ generates $V$. Next, we show that $\beta\cup\gamma$ is linearly independent. Given $c_1,c_2,\ldots,c_k,d_1,d_2,\ldots,d_m$ such that $\displaystyle\sum_{i=1}^kc_iw_i+\sum_{j=1}^md_jx_j={\it 0}$, then $\displaystyle\sum_{i=1}^kc_iw_i=-\sum_{j=1}^md_jx_j$. It follows that $$\sum_{i=1}^kc_iw_i\in W\cap W^\perp\quad\mbox{and}\quad \sum_{j=1}^md_jx_j\in W\cap W^\perp.$$ But since $W\cap W^\perp=\{{\it 0}\,\}$ (gievn $x\in W\cap W^\perp$, we have $\langle x,x\rangle=0$ and thus $x={\it 0}\,$), we have $\displaystyle\sum_{i=1}^kc_iw_i=\sum_{j=1}^md_jx_j={\it 0}$. Therefore $c_i=0$ and $d_j=0$ for each $i,j$ becasue $\beta$ and $\gamma$ are bases for $W$ and $W^\perp$, respectively. Hence we conclude that $\beta\cup\gamma$ is linearly independent.

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    $\begingroup$ I cannot grab this part: $\sum_{i=1}^kc_iw_i=-\sum_{j=1}^md_jx_j$ implies $\sum_{i=1}^kc_iw_i\in W\cap W^\perp\quad\mbox{and}\quad \sum_{j=1}^md_jx_j\in W\cap W^\perp.$ Can you explain this further? $\endgroup$
    – mgus
    Jan 21, 2018 at 2:52
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    $\begingroup$ By assumption, $\displaystyle\sum_{i=1}^kc_iw_i\in W$ and $\displaystyle-\sum_{j=1}^md_jx_j\in W^\perp$. So if $\displaystyle\sum_{i=1}^kc_iw_i=-\sum_{j=1}^md_jx_j$, we naturally get $\displaystyle\sum_{i=1}^kc_iw_i\in W^\perp$ and $\displaystyle-\sum_{j=1}^md_jx_j\in W$ as well. $\endgroup$
    – Solumilkyu
    Jan 21, 2018 at 12:43
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Hint:

Take a basis $w_1,\dots,w_r$ of $W$, and consider the linear forms on $V$ defined by $w_i^*:v\mapsto\langle w_i,v\rangle$.

These linear forms are linearly independent, hence the solutions of the system of equations $w_i^*(v)=0,\ i=1,\dots r$ has codimension $r$ by the rank-nullity theorem. These solutions are precisely the orthogonal complement $\;U^{\bot}$.

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  • $\begingroup$ This proof Is better sumce it does not require the <> to ve an actual inner product. So this works also in the space F_q^n fór finite field F_q. $\endgroup$ May 2 at 18:50
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It is sufficient to show that $V=W\oplus W^{\perp}$. If $v\in W\cap W^{\perp}$, then $\left\langle v,v\right\rangle=0$. Hence it remains to show that any vector $v\in V$ can be written as $v=w+w'$ with $w\in W$ and $w'\in W^{\perp}$.

Choose an orthonormal basis $\left\{w_1,\dots , w_k\right\}$ of $W$ and extend to an orthonormal basis $\left\{w_1,\dots,w_k,v_{k+1},\dots ,v_n\right\}$ of $V$. By definition $v_i\in W^{\perp}$ for all $n\geq i\geq k+1$. Hence any $v\in V$ can be decomposed as we needed to show.

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  • $\begingroup$ I like Bernard's answer better, also, for my answer you need to know that you can take orthonormal bases of finite dimensional vector spaces, thus you need knowledge of the Gramm-Schmidt procedure. $\endgroup$ May 1, 2016 at 11:50
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    $\begingroup$ This is false as stated. You can't just extend to a basis of $V$ and then say "by definition." $\endgroup$ May 1, 2016 at 12:11
  • $\begingroup$ Hold on, take any $w\in W$, then $w=\sum_{i=1}^k\lambda_i w_i$. Hence $\left\langle w,v_j \right\rangle=\sum_{i=1}^k\lambda_i \left\langle w_i,v_j\right\rangle=0$, hence $v_j\in W^{\perp}$ for any $j\geq k+1$. Or am I completely wrong here? $\endgroup$ May 1, 2016 at 12:16
  • $\begingroup$ The $v_j$ are very, very unlikely to be in $W^\perp$. $\endgroup$ May 1, 2016 at 12:25
  • $\begingroup$ Ah ok, sorry, I meant that you extend to an orthonormal basis of $V$ as well! $\endgroup$ May 1, 2016 at 12:29
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There is something missing in @Solumilkyu 's answer.

In order to show that $W$ and $W^\perp$ have bases, we must show that $W, W^\perp \neq \{\vec{0}\}$, which cannot always be satisfied.

In fact, if $W=V$, then from $W \cap W^\perp=\{\vec{0}\}$ (To prove this, let $\vec{x} \in W \cap W^\perp$. Then $<\vec{x},\vec{x}>=0 \to \vec{x}=\vec{0}$) we can conclude that $W^\perp=\{\vec{0}\}$. Then $\dim W+\dim W^\perp=\dim W$. If $W\neq V$, then from the fact that $v \in V$ implies that $v=v_1+v_2$ for $v_1 \in W \wedge v_2 \in W^\perp$, we can conclude that $W^\perp \neq \{\vec{0}\}$ just by setting $v \in V \setminus W \wedge v \neq \{\vec{0}\}$ and $v_1 = \vec{0}$.

On the other hand, if $W = \{\vec{0}\}$, then obviously $W^\perp = V$.

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I have some qualms with @Solumilkyu’s answer. To prove that that a set of vectors is indeed a basis, one needs to prove prove both, spanning property and the independence. @Solumilkyu has demonstrated $\beta \cup \gamma$ is linearly independent, but has very conveniently assumed the spanning property.

To answer the question, though its quite old, one can use the rank-nullity theorem, coupled with the fact that $\text{rank}(A) = \text{rank}(A^{T})$ and $N(A^{T}) = V^{\bot}$.

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