## anonymous 4 years ago prove that {vector a, perp of vector b on vector a} is an orthogonal set which spans S if {vector a, vector b} is a basis for a subspace S of R^n Not sure how to prove this. I tried proving it and I get that it is a false statement. Just need help how to start the proof, if its provable.

1. anonymous

basically my proof starts with c1(vector a)+c2(vector b)=0 c1(vector a)*(vector a)+c2(vector b)*(vector a)=0*(vector a) simplifying it I get c1/-c2=(vector b)*(vector a)/(vector a)^2 c1/-c2*(vector a)=projection of vector b on a am I doing anything wrong?

2. anonymous

Hmm... If I were to prove this, I'd probably start by giving a vector in S arbitrary coordinates with respect to a and b, and then show that they unambiguously transform into coordinates of a and b', where b' is your perpendicular vector.

3. anonymous

Something like.... let "v" be some arbitrary vector in S : $\vec{v} \in S$ Then, because the vectors a and b are a basis for S, the vector "v" has coordinates x and y with respect to a and b, i.e. $\vec{v} = x\cdot \vec{a} + y\cdot \vec{b}$ Note that a and b are not necessarily perpendicular. Since their dot product is not zero, we can define a new vector v' that is perpendicular to a like this: $\vec{b'} = \vec{b} - \left(\frac{\vec{a}\cdot \vec{b}}{a^2}\right)\vec{a}$ (Because a is a basis vector, it cannot be the zero vector, and so we're safe dividing by a^2)

4. anonymous

We can therefore rewrite b as $\vec{b} = \vec{b'} + \left(\frac{\vec{a}\cdot \vec{b}}{a^2}\right)\vec{a}$ plugging this in will yield new coordinates with respect to a and b'.

5. anonymous

Oops... in the first post there, that should be "a new vector b' " not v'...

6. anonymous

kk, i'm not sure how to solve for b' do I solve it with respect to x and y

7. anonymous

Just plug that into $\vec{v} = x\cdot \vec{a} + y\cdot \vec{b}$

8. anonymous

You don't need to solve for anything, per se. A set of vectors spans a space iff any vector in the space can be written as a linear combination of the vectors in the set. We're just directly showing that if the vector can be written as a linear combination of a and b, then it can definitely be written as a linear combination of a and b'.

9. anonymous

oh ic

10. anonymous

i.e. $\vec{v} = x\cdot \vec{a}+y\cdot \left( \vec{b'} + \left(\frac{\vec{a}\cdot \vec{b} }{a^2} \right) \right) = \left[x + y\left(\frac{\vec{a}+\vec{b}}{a^2} \right) \right]\cdot \vec{a} + y\cdot \vec{b'}$

11. anonymous

so a dot b' equals zero when you plug it so it satisfies both conditions

12. anonymous

Crap. Make that plus sign a dot product, and we're good. You see what I mean? We did not specify anything about the vector v except that it could be written as a linear combination of a and b (which is a given, since a and b span S). Then, we did some fairly simple vector manipulation to define a vector b' that is perpendicular to a, and showed that our arbitrary vector v can ALSO be written as a linear combination of a and b'.

13. anonymous

Yes, b' is perpendicular to a. We defined it as such, right?

14. anonymous

ok, thanks a lot, I get it now, would probably never have thought of it that way

15. anonymous

No problem. Also note that if b just so happens to be perpendicular to a in the first place, the dot product vanishes and the coordinates stay the same.