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 $W^\perp$.


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.

  • $\begingroup$ 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 '18 at 2:52
  • 1
    $\begingroup$ 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 '18 at 12:43


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 $\;U^{\bot}$.


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.

  • $\begingroup$ 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. $\endgroup$ – Mathematician 42 May 1 '16 at 11:50
  • $\begingroup$ This is false as stated. You can't just extend to a basis of $V$ and then say "by definition." $\endgroup$ – Ted Shifrin May 1 '16 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? $\endgroup$ – Mathematician 42 May 1 '16 at 12:16
  • $\begingroup$ The $v_j$ are very, very unlikely to be in $W^\perp$. $\endgroup$ – Ted Shifrin May 1 '16 at 12:25
  • $\begingroup$ Ah ok, sorry, I meant that you extend to an orthonormal basis of $V$ as well! $\endgroup$ – Mathematician 42 May 1 '16 at 12:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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