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Let $\{w_1,w_2,...,w_k\}$ be a basis for a subspace $W$ of $V$. Show that $W^⊥$ consists of all vectors in V that are orthogonal to every basis vector.

I know that the intersection of the two subspaces is a set containing only the zero vector.

the set is a basis so it's linearly independent but it's not orthogonal because then the intersection would have more than just the zero vector. and if the set is a basis for the subspace $W$ then it is also the $col(W)$ and I know that the orthogonal complement of $col(W)$ would be $null(W^T)$ This is where I am stuck

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2 Answers 2

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Go back to the definitions.

Let $S$ be the set of all vectors that are orthogonal to every basis vector. We need to show two things:

  • if $x \in W^\perp$, then $x \in S$
  • if $x \in S$, then $x \in W^\perp$

After proving that, we could say that $W^\perp \subseteq S$ and $S \subseteq W^\perp$, so that $W^\perp = S,$ which is what we're trying to show.

Proving the first statement is relatively easy, and uses only the definition of $W^\perp$. Proving the second statement requires that you use the definition of a basis and go through some algebra.

Remember that $x$ is orthogonal to $w$ means $\langle w,x \rangle = 0$.

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1)Assume $x \in W^{\bot}$ then $<x,w_i>=0$ where $w_i \in W, i=1,...,k$ thus $x \in S$ and $W^{\bot} \subseteq S$

2) suppose $x \in S$. Consider any $w \in W$. We may write $w = c_1w_1 + \cdots + c_k w_k$. It follows that $$ \langle x,w_i \rangle=0 \quad \text{for }i = 1,\dots,k \implies \\ c_1\langle x,w_1 \rangle + \cdots + c_k\langle x,w_k\rangle =0 \implies\\ \langle x, c_1w_1+c_2w_2+...+c_kw_k \rangle =0 \implies \\ \langle x,w \rangle = 0 $$ thus $x \in W^{\bot}$ and $W^{\bot} \subseteq S$

Therefore $W^{\bot} = S$

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  • $\begingroup$ Your proof looks great, well done! LaTeX note: you should use \langle and \rangle for brackets like these $\langle \rangle$ rather than $<>$. Also, I find that the \implies arrow ($\implies$) looks nicer than $\Rightarrow$. $\endgroup$ Commented Nov 12, 2014 at 13:32
  • $\begingroup$ There are also some parts of the proof that you should elaborate on, even if you do have the broad strokes. $\endgroup$ Commented Nov 12, 2014 at 13:34
  • $\begingroup$ will do! thanks! $\endgroup$ Commented Nov 13, 2014 at 19:36

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