Inner Product Space and Orthogonality proof question (is this the correct way?)

[ryan8642]

Homework Statement

Let V be the inner product space. Show that if w is orthogonal to each of the vectors u1, u2,...,ur, then it is orthogonal to every vector in the span{u1,u2,...,ur}.

Homework Equations

u.v=0 to be orthogonal
If u and v are vectors in an inner product space, then ||u+v||^2 = ||u||^2 + ||v||^2

The Attempt at a Solution

So we know ||u+v||^2 = ||u||^2 + ||v||^2 for ortho vectors

for u1: <u1,w>=0
||u1+w||^2 = <u1+w, u1+w>
= <u1,u1> + 2<u1,w> + <w,w>
= ||u1||^2 + ||w||^2

then i do the same for u2 and ur.

Is this the correct way to go about this?

Thanks

 

[Dick]

Homework Statement

Let V be the inner product space. Show that if w is orthogonal to each of the vectors u1, u2,...,ur, then it is orthogonal to every vector in the span{u1,u2,...,ur}.

Homework Equations

u.v=0 to be orthogonal
If u and v are vectors in an inner product space, then ||u+v||^2 = ||u||^2 + ||v||^2

The Attempt at a Solution

So we know ||u+v||^2 = ||u||^2 + ||v||^2 for ortho vectors

for u1: <u1,w>=0
||u1+w||^2 = <u1+w, u1+w>
= <u1,u1> + 2<u1,w> + <w,w>
= ||u1||^2 + ||w||^2

then i do the same for u2 and ur.

Is this the correct way to go about this?

Thanks

No, it isn't. What's the definition of span?

 

[ryan8642]

So w has to be orthogonal to all of the vectors in the span{u1,u2,...,ur}

k1u1+k2u2+...+krur=b to test for spanning where b is an arbitrary vector.
would 'b' be w in this case?

 

[Dick]

So w has to be orthogonal to all of the vectors in the span{u1,u2,...,ur}

k1u1+k2u2+...+krur=b to test for spanning where b is an arbitrary vector.
would 'b' be w in this case?

No, b would be an arbitrary vector in span{u1,u2,...,ur}. So you want to show b.w=0. What is (k1u1+k2u2+...+krur).w?

 

[ryan8642]

i am sooo confused.

b and w are both vectors.
so b=(b1,b2,...br)
w=(w1,w2,...wr) w vector is perpendicular to b vector (b lies in the span of u1,u2,...ur)

so b1w1+b2w2+...+brwr=0

*i noticed u edited what u asked me.

What is (k1u1+k2u2+...+krur).w?

k1u1w1+k2u2w2+...+krurwr = 0 (im assuming it has to equal zero to be orthogonal)

 

[Dick]

i am sooo confused.

b and w are both vectors.
so b=(b1,b2,...br)
w=(w1,w2,...wr) w vector is perpendicular to b vector (b lies in the span of u1,u2,...ur)

so b1w1+b2w2+...+brwr=0

Look b=(k1u1+k2u2+...+krur). That's an arbitrary vector in the span. The k's are numbers and the u's are vectors. w is just w. Compute (k1u1+k2u2+...+krur).w by using the distributive property. You know (a+b).c=a.c+b.c where a,b,c are vectors, yes? You also know (xa).c=x(a.c) where a and c are vectors and x is a number, yes? Use those properties.

 

[ryan8642]

Compute (k1u1+k2u2+...+krur).w
=w.k1u1 + w.k2u2 + ... + w.krur
=k1(w.u1) + k2(w.u2) + ... + kr(w.ur)

sorry about bein really dumb with this, I am horrible with proofs. but good with numbers...

 

[Dick]

Compute (k1u1+k2u2+...+krur).w
=w.k1u1 + w.k2u2 + ... + w.krur
=k1(w.u1) + k2(w.u2) + ... + kr(w.ur)

sorry about bein really dumb with this, I am horrible with proofs. but good with numbers...

I'm sure you are good with numbers. 'Abstract' throws some people. Just pretend they are numbers. Now what are w.u1, w.u2, etc?

 

[ryan8642]

w.u1=0
w.u2=0
..
w.ur=0

they are all orthogonal to each other.

=k1(w.u1) + k2(w.u2) + ... + kr(w.ur)
=k1(0) +k2(0) +...+ kr(0)
=0

and btw man i really appreciate u walkin me through this proof. This helps me sooo much for this proof and just solving proofs in general.

 

[ryan8642]

btw, u must know proofs well and understand them well.
Do you have any tips for studying them/remembering proofs for tests/exams?

 

[Dick]

btw, u must know proofs well and understand them well.
Do you have any tips for studying them/remembering proofs for tests/exams?

I'll give you one that's pretty important. Look up the definition of any terms you are a little vague on. That's why my first hint was "What's the definition of span?". Your first attempt didn't even use that. That's a pretty good hint you are missing something. Other than that, practice makes perfect.

 

[ryan8642]

ok, i will for sure do that while I am studying.
Thanks for all the help.