So there was a question in my linear algebra textbook, that didn't have an answer in the "answer section" and there wasn't really any example on how to prove it, (google isn't really helping either!). The question was:
"Let R^4 have the euclidian inner product, and determine if a vector v = (2,-1,0,3) is orthogonal to the space spanned by w1 = (0,0,0,0), w2 = (2,4,5,0), w3 = (6,0,6,-4)"
After a bit of messing around, I thought that (at least in 2-space), a vector (a1) orthogonal to a another vector (b1), would automatically be orthogonal to every vector in the span of b1 and some other vector b2 (since that would make up a plane, and it would be orthogonal to the plane) so that by showing that v is/isn't orthogonal to w1,w2 and w3 seperatly, it would determine if it was orthogonal to their span.
As it turns out the inner product of (v,w1) (v,w2) (v,w3) are all zero, hence it's orthogonal to every vector in the span. Is this a reasonable conclusion/good way of answering the question?
Thanks in advance!
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$\begingroup$To say that $v$ is orthogonal to the space means that $v$ is orthogonal to each element of the space. Now an element of the space spanned by $w_1,w_2,$ and $w_3$ looks like \begin{equation*} w = aw_1 + bw_2 + cw_3 \end{equation*} for some real numbers $a,b,c\in\mathbb{R}$. Then, since $v$ is orthogonal to each of $w_1$, $w_2$, and $w_3$, we have \begin{equation*} v\cdot w = a(v\cdot w_1) + b(v\cdot w_2) + c(v\cdot w_3) = 0. \end{equation*} So $v$ is orthogonal to each element of the space spanned by $w_1$, $w_2$, and $w_3$, and we're done.
(Short answer: Yes, you've got the right idea.)
$\endgroup$ 2 $\begingroup$You are correct, If $(\vec v,\vec w_1) = (v,\vec w_2) = 0$, then by bilinearity of the scalar product $$(\vec v, x_1 \vec w_1 + x_2 \vec w_2) = x_1 (\vec v,\vec w_1) + x_2 (\vec v,\vec w_2) = 0$$ for any $x_1,x_2$. Because the span of $\vec w_1$ and $\vec w_2$ is exactly the set of alllinear combinations $x_1 \vec w_1 + x_2 \vec w_2$, you are done.
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