Show that I have a basis, linear algebra

[MaxManus]

Homework Statement

(X,<,>) is a inner product space over R
{e_i}_{i in N} is an orthonormal set in X

Show that if every element u in X can be written as a linear combination
u = \sum_{i=1}^\infty a_i e_i then {e_i}_{i in N} is a basis for X

Homework Equations

Let {e_i}be a sequence of elements in a normed vector
space V . We say that {e_n} is a basis for V if for each x in V there is a
unique sequence {a_n} from K such that:
x = \sum_{n=1}^{\infty} a_i e_i

The Attempt at a Solution

I'm not sure what I have to prove. Is not the definition of a basis given in the question?

 

[Klockan3]

It is true that they have given you that there is one sequence, what is left to prove is that it is unique. The proofs for linear algebra are usually quite trivial like this, but remember to read the conditions and what you are given thoroughly since it is easy like in this case to mistakenly take some details for given when in fact they are not.

 

[MaxManus]

Thanks. Can I say that it is unique because it is an orthonormal set and therfor lineraly independent? And then the proof is complete?

 

[Klockan3]

Thanks. Can I say that it is unique because it is an orthonormal set and therfor lineraly independent? And then the proof is complete?

I would say that you need more than that. Is the definition of an orthonormal set that all linear combinations are unique? Otherwise you need to show that link as well, as far as I know the definition of an orthonormal set is not that all linear combinations are unique.

 

[MaxManus]

But I am on the right track?

If
a_1 e_1 + a_2 e_2 + ... e_n e_n = 0
Taking the norm.
|| a_1 e_1 + a_2 e_2 + ... a_n e_n ||^2 = 0
e_i are orthonormal so the norm is 1
a_1^2 + ... + a_n^2 = 0
Which means that all the as are zero which means that the {e_i} are linearly independent.

 

[Klockan3]

But I am on the right track?

If
a_1 e_1 + a_2 e_2 + ... e_n e_n = 0
Taking the norm.
|| a_1 e_1 + a_2 e_2 + ... a_n e_n ||^2 = 0
e_i are orthonormal so the norm is 1
a_1^2 + ... + a_n^2 = 0
Which means that all the as are zero which means that the {e_i} are linearly independent.

You are using way too many secondary results, as I said use the definitions. You can prove it like that, yes, but it is ugly and questionable since you don't mention why a_1 e_1 + a_2 e_2 + ... e_n e_n = 0 implies || a_1 e_1 + a_2 e_2 + ... a_n e_n ||^2 = 0, personally I wouldn't give points for that. The definition of an orthonormal set is not that if their linear combination is zero then their norm will also be zero, look up the definition if you have to instead of guessing like this.

 

[Citan Uzuki]

You're also proving the wrong thing. In order to show that the representation is unique, you need a little more than linear independence, since L.I. only says that nontrivial finite linear combinations cannot be zero, and nothing about infinite sums.

I think you're on the right track though, since the result you're being asked to prove is equivalent to saying that no nontrivial infinite linear combination converges to zero (note that you have to prove this, not just assume it). And if you can show that the squared norm of \sum_{i=1}^{\infty}a_ie_i is \sum_{i=1}^{\infty} a_i^2, then you're done, by the approach you've suggested. (Hint: take limits of finite sums, and use the continuity of the norm).

You are using way too many secondary results, as I said use the definitions. You can prove it like that, yes, but it is ugly and questionable since you don't mention why LaTeX Code: a_1 e_1 + a_2 e_2 + ... e_n e_n = 0 implies LaTeX Code: || a_1 e_1 + a_2 e_2 + ... a_n e_n ||^2 = 0 , personally I wouldn't give points for that. The definition of an orthonormal set is not that if their linear combination is zero then their norm will also be zero, look up the definition if you have to instead of guessing like this.

He's not invoking the definition of an orthonormal set, he's using something much more basic, namely the fact that the norm of the zero vector is zero. The orthonormality is used in the next step, when he asserts that the norm of a_1e_1+\cdots +a_ne_n is a_1^2+\cdots a_n^2, which is by the Pythagorean theorem for inner product spaces. Now, this is a derived result, but frankly I would use it on a problem like this. Since the problem is precisely to show that nontrivial infinite linear combinations of orthonormal vectors can't add up to zero, and the fastest way to do that is going through the norm.

 

[Klockan3]

He's not invoking the definition of an orthonormal set, he's using something much more basic, namely the fact that the norm of the zero vector is zero. The orthonormality is used in the next step, when he asserts that the norm of a_1e_1+\cdots +a_ne_n is a_1^2+\cdots a_n^2, which is by the Pythagorean theorem for inner product spaces. Now, this is a derived result, but frankly I would use it on a problem like this. Since the problem is precisely to show that nontrivial infinite linear combinations of orthonormal vectors can't add up to zero, and the fastest way to do that is going through the norm.

Oh, I missed that this was an infinite dimensional linear space! I just assumed that it was finite dimensional since the topic title says linear algebra, since when you get to infinite spaces you get a lot of problems about convergence etc. Yes, then some things have to be done differently and there should be no problem with using more advanced results, if this was about the finite case I hope that you can agree that it would be strange to use a result like that in the proof.

 

[MaxManus]

You're also proving the wrong thing. In order to show that the representation is unique, you need a little more than linear independence, since L.I. only says that nontrivial finite linear combinations cannot be zero, and nothing about infinite sums.

I think you're on the right track though, since the result you're being asked to prove is equivalent to saying that no nontrivial infinite linear combination converges to zero (note that you have to prove this, not just assume it). And if you can show that the squared norm of \sum_{i=1}^{\infty}a_ie_i is \sum_{i=1}^{\infty} a_i^2, then you're done, by the approach you've suggested. (Hint: take limits of finite sums, and use the continuity of the norm).

Thanks to both.
First Part: "no nontrivial infinite linear combination converges to zero "
Here I'm not sure where to start
Second Part:
e_i e_i = 1
e_i e_j = 0
&lt;\sum_{i=1}^{\infty}a_i e_i,\sum_{i=1}^{\infty}a_i e_i&gt;^2 = \lim_{N-&gt; \infty} &lt;\sum_{i=0}^{N}a_i e_i,\sum_{i=1}^{N}a_i e_i&gt;^2 = \lim_{N-&gt; \infty} \sum_{i=1}^N a_i^2 = \sum_{i=1}^{\infty}a_i^2
Right?

 

[Citan Uzuki]

That's right for the second part. For the first part, you just note that if \sum_{i=1}^{\infty} a_ie_i = 0, then \sum_{i=1}^{\infty} a_i^2=0, and so each a_i is zero. Now show that that indeed implies the representation is unique.

 

[MaxManus]

Thanks, but I'm not sure how to show that this implies the representation is unique. I'm not supposed to use linearly independence, but what can I use.

 

[Citan Uzuki]

Thanks, but I'm not sure how to show that this implies the representation is unique. I'm not supposed to use linearly independence, but what can I use.

Well, you've now proven the infinite-dimensional analog of linear independence, so you can derive unique representation from that using the same method that you would derive unique representation with respect to a finite basis from finite linear independence. If you still need a hint, see below:

Spoiler

If you have two distinct representations of the same vector, then the difference between them is a nontrivial representation of zero

 

[MaxManus]

Had to use the hint:-)

If u can be represented with another infinite linear combination of e_i called v then u-v = 0
u-v = \sum_{i= 1}^{\infty} a_i e_i - \sum_{i= 1}^{\infty} b_i e_i =( a_1 e_1 + a_2 e_2 + ... + a_n e_n + ...) -( b_1 e_1 + b_2 e_2 + ... + b_n e_n + ...) = (a_1 - b_1)e_1 + (a_2-b_2)e_2 + ... (a_n -b_n)e_n + ... = \sum_{i= 0}^{\infty} (a_i -b_i )e_i

\sum_{i= 0}^{\infty} (a_i -b_i )e_i = 0
Taking the norm on both sides.

\sqrt{ \sum_{i=1}^{\infty}(a_i-b_i)^2} = 0

So a_i = b_i, the representation is unique. Is this the whole proof (or the proof at all)?

 

[Citan Uzuki]

Yeah, that's it

 

[MaxManus]

Super, thanks for all the help.