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

#### ryan8642

1. The problem statement, all variables and given/known data
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}.

2. Relevant 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

3. 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.

Thanks

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#### Dick

Homework Helper
1. The problem statement, all variables and given/known data
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}.

2. Relevant 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

3. 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.

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

Homework Helper
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

Homework Helper
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, im horrible with proofs. but good with numbers....

#### Dick

Homework Helper
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, im 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

Homework Helper
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 im studying.
Thanks for all the help.

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