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Hey I have a question:

Say I have a euclidian Vectorspace $(V,\langle\cdot,\cdot\rangle)$ (finite dimension, lets say $n$). And now say that I have an ordered subset of $V$, say $S = (v_1 , ..., v_k)$.

Then I can expand this subset using Steinitz into a Basis of $V$ and then using the Gram-Schmidt process I can turn this basis into a orthogonal and then a orthonormal basis $S = (v_1,...,v_k, v_{k+1},...,v_n)$ of $V$.

Now if have to prove that if the subspace $W=\operatorname{span}(v_1,...,v_k)$ then a $F = (v_{k+1},...,v_n)$

is a Basis of $W^⊥$ \begin{aligned}[]W^{\perp }=\{ u\in V\mid \forall v\in W:\langle u,v\rangle =0\} .\end{aligned}

Question 1: Since I have a Orthonormal Basis then all Elements in this Basis are Orthonormal to each other and only by defining the subspace $W$ can I then define the subspace $W^⊥$? I am pretty sure this is true but then something does not add up in question 2.

Question 2: idea of proof for $W^⊥ \subseteq\operatorname{span}(F)$: Since I have a Orthonormal Basis $S$ then

for all $v\in V: v = \sum_{i=1}^n \langle v_i,v\rangle v_i $

now if $v\in W^⊥: v = \sum_{i=k+1}^n \langle v_i,v\rangle v_i \in \operatorname{span}(F)$

For $1\leq i \leq k$ the inner product is 0 since they are Orthogonal and in $W$. But aren't my other vectors (in $W^⊥$ also orthogonal to each other) meaning that say $v$ the sum would only consist of one vector where $v$ is not orthogonal to it times a scalar? Or is this only true if I can't write $v$ as a linear combination of my elements in $F$ but only as a multiple of one element? And if choose a $v$ as a linear combination of elements in $F$ then all $k+1 \leq i \leq n$ dissapear in the inner product except those used in the linear combination?

Thank You!

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Regarding question 1: If I understand your question correctly, you are right: $W^\perp$ is defined in terms of $W$. Note that the subspace $W$ appears in the definition of $W^\perp$ that you give: $W^\perp=\{u\in V | \forall v \in W, \langle u, v \rangle = 0\}$.

Regarding your second question: Let's illustrate with an example. Let $V=\mathbb{R}^3$ with the usual basis $\{e_1,e_2,e_3\}$. Let $W=$ span$(e_3)$, so $W^\perp$ is the $xy$-plane. If $v=(a,b,c)\in W^\perp$, then $v=\sum_{i=1}^3\langle e_i,v \rangle e_i=ae_1+be_2$, i.e. $c=0$. So as you can see the sum consists of a linear combination of more than one vector. Unless $v$ is simply a scalar multiple of one of the basis vectors, in general more than one basis vector will appear in the linear combination.

The idea for your proof is correct. It shows that $W^\perp\subset$ span$(F)$, and since it is obvious that span$(F)\subset W^\perp$ and it is assumed that $F$ is linearly independent, you can conclude that $F$ is a basis for $W^\perp$. Hope this helps.

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  • $\begingroup$ Got it thank you! $\endgroup$
    – James
    Jun 15 '18 at 5:49

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