A community for students.
Here's the question you clicked on:
 0 viewing
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.
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.

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0basically 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm... 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Something 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We 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'.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oops... in the first post there, that should be "a new vector b' " not v'...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0kk, i'm not sure how to solve for b' do I solve it with respect to x and y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just plug that into \[\vec{v} = x\cdot \vec{a} + y\cdot \vec{b}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You 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'.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i.e. \[\vec{v} = x\cdot \vec{a}+y\cdot \left( \vec{b'} + \left(\frac{\vec{a}\cdot \vec{b} }{a^2} \right) \right) = \left[x + y\left(\frac{\vec{a}+\vec{b}}{a^2} \right) \right]\cdot \vec{a} + y\cdot \vec{b'} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so a dot b' equals zero when you plug it so it satisfies both conditions

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Crap. 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'.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, b' is perpendicular to a. We defined it as such, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, thanks a lot, I get it now, would probably never have thought of it that way

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.