# Abstract Linear Algebra

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1. Dec 8, 2014

### teme92

1. The problem statement, all variables and given/known data
Let V be a finite-dimensional real vector space with inner product <⋅,⋅> and L: V → R a linear transformation. Show that there exists a unique vector a ∈ V such that L(x) = <a,x>.

2. Relevant equations
Hey everyone, so I'm a physics student who had to choose a few electives in the maths department. Unfortunately I picked a proof based module without realizing it as I thought it would've been a carry on from first year Linear Algebra. I decided to soldier on through as I think it might be useful down the line. I am however having difficulty unlike the maths students as they do this sort of question answering a lot. So I hope by explaining what I know someone will be able to help me to answer this question correctly.

3. The attempt at a solution
I know that the inner product for real numbers is the dot product. I know the properties the inner product satisfy (associative and commutative). Its the L(x) = <a,x> part that I don't fully understand. If anyone could help me I'd appreciate it. Thanks in advance.

2. Dec 8, 2014

### PeroK

You have two things to do: you have to find a vector a and you have to show that it is unique.

Hint: what's the most important thing to find in a Vector Space?

3. Dec 8, 2014

### Fredrik

Staff Emeritus

Once you have proved uniqueness, you can try to show existence. This is where PeroK's hint comes in.

I would prefer a notation like $z$ or $x_0$ instead of $a$, because I find it convenient to use letters at the start of the alphabet for numbers, and letters at the end of the alphabet for vectors. Are you using Greek letters for numbers? That's OK too.

4. Dec 8, 2014

### HallsofIvy

Staff Emeritus
If V is a finite dimensional vector space with given inner product, < , >, then there exist an "orthonormal basis". That is, there exist a basis $\{v_1, v_2, \cdot\cdot\cdot, v_m\}$ such that $<v_i, v_j>$ is 1 if i= j, 0 if i is not equal to j. Apply A to each of those basis vectors in turn: let $a_i= A(v_i)$. Show that $\sum a_iv_i$ is the desired vector.

5. Dec 8, 2014

### teme92

Hey everyone thanks for the replies. I understand uniqueness in the circumstance where if a and b are numbers than there exists one number x such that a+x=b. I just take a from both sides so:

(a+x)-a=b-a

Therefore x=b-a. How do I apply that to the situation I'm given?

6. Dec 8, 2014

### Fredrik

Staff Emeritus
What does it mean to say that the vector you denoted by $a$ is not unique? Write down that statement, and then find some of its implications.

7. Dec 8, 2014

### teme92

If a is not unique I think this is how I say it?
If v1,...,vn is a basis for V then since every v ∈ V has a unique expression:

v = ∑ni=1rivi, where r1,....r2 ∈ ℝ

8. Dec 8, 2014

### Staff: Mentor

The order here seems backwards to me. You would have a real problem if you showed that the vector was unique, but then were not able to find such a vector.

@teme92, the usual approach in showing uniqueness is to a proof by contradiction. That is, you assume that there are two such vectors, and show that can't possibly happen. By reaching a contradiction, you are showing that your initial assumption is not true. IOW, there can be only one such vector.

9. Dec 8, 2014

### teme92

I don't know how to show by contradiction here. Its the L(x) = <a,x> that I don't understand. Currently this is how I'm seeing it:

a = (a1,.....,an)T and x = (x1,....xn)T, therefore:
<a,x> = a1x1 + .......+ anxn = aTx

From this I have no idea how to show uniqueness and I can't find anything in my notes or book about it. Would anyone be able to clear this up? Thanks for the help again guys.

10. Dec 8, 2014

### Fredrik

Staff Emeritus
The order is irrelevant. To prove uniqueness is to prove that there's at most one. If the actual number is zero, then perhaps time would have been better spent trying to prove that there's exactly zero, but there would be no real problem.

Right, you can assume that there are two different vectors $u$ and $v$ that both have the given property, and then you show that these assumptions lead to a result that's known to be false. Typically, we would prove that $u=v$, which contradicts our assumptions. But that suggests a slightly simpler way to prove uniqueness. You assume that $u$ and $v$ are vectors with the given property, and then you prove that $u=v$. Now you can conclude that there's at most one vector with the given property.

11. Dec 8, 2014

### Fredrik

Staff Emeritus
That's not it. This statement doesn't really make sense. The word "since" suggests that you're stating an implication, i.e. a statement of the form "if p then q", which can be equivalently stated as "since p, q". But your p and q seem to be the same statement: There's a unique n-tuple of real numbers $(r_1,\dots,r_n)$ such that $v=\sum_{i=1}^n r_i v_i$.

Also, your statement doesn't say anything about $a$ or $L$.

The statement you want to prove is this: There's a unique $a\in V$ such that $L(x)=\langle a,x\rangle$ for all $x\in V$. The statement I was hoping you would find is this: There exist vectors $a,b\in V$ such that $L(x)=\langle a,x\rangle$ for all $x\in V$ and $L(x)=\langle b,x\rangle$ for all $x\in V$. Now you need to think about what this statement implies.

The elements of V aren't necessarily n-tuples of real numbers. There are lots of other things they can be, e.g. polynomials or matrices.

12. Dec 8, 2014

### teme92

That statement implies that a=b and therefore a is unique. But is that answer the question satisfactorily?

13. Dec 8, 2014

### vela

Staff Emeritus
When you say "L: V → R a linear transformation," it just means that $L$ is linear and that it maps vectors in V to real numbers. It doesn't tell you much else. In particular, the statement doesn't specify how to calculate the real number that a vector $\vec{x}$ maps to. There are a bunch of different ways you could map vectors to real numbers, some linear and some not. If the mapping is linear, it turns out you can always express it in the form $\langle \vec{a},\vec{x}\rangle$ where $\vec{a}$ is unique.

By the way, the concept of the inner product is a generalization of the dot product. The dot product is an inner product, but an inner product isn't necessarily the dot product. So you shouldn't assume that you're working with the usual dot product.

14. Dec 8, 2014

### Fredrik

Staff Emeritus
You also have to explain how it implies that $a=b$. You have to find a statement that obviously follows from the one I wrote down, and then you find a statement that obviously follows from that, and so on, until you find the result $a=b$.

You should probably start by writing down the definition of "inner product", because you will need it.

15. Dec 8, 2014

### teme92

Thanks vela is does make it a bit clearer

16. Dec 8, 2014

### teme92

So my current attempt is as follows:

Let v1,....,vn be unit vectors in V. If x=x1v1+....+xnvn is any vector then:

L(x)=L(x1v1+....+xnvn)
=x1L(v1) +....+ xnL(vn)

If we now let ai=L(vi) we see that:

L(x)=x1a1+....+xnan=<x,a>=<a,x>

This seems correct, am I right in saying this? Thanks for all the help especially Fredrik

17. Dec 9, 2014

### Fredrik

Staff Emeritus
Your assumption that $v_1,\dots,v_n$ are unit vectors in V is not strong enough to ensure that the equality $x_1a_1+\cdots+x_na_n=\langle x,a\rangle$ holds. But you are at least doing the sort of calculations that are involved in the existence part of the proof.

What about the uniqueness? Did you figure out how to prove it? It's the sort of proof that's very easy for someone who's experienced with proofs, but can be difficult for someone who isn't.

18. Dec 9, 2014

### teme92

No I haven't figured out how to do it. I understand its concept but not how to apply it here.

19. Dec 9, 2014

### Fredrik

Staff Emeritus
I see. This is our assumption:

There exist vectors $a,b\in V$ such that $L(x)=\langle a,x\rangle$ for all $x\in V$ and $L(x)=\langle b,x\rangle$ for all $x\in V$.​

Let's just try to find a logical consequence of this assumption. Let $x$ be an arbitrary element of $V$. What does the assumption tell us about $x$? In particular, can we make a statement about $x$ that also involves $a$ and $b$?

20. Dec 9, 2014

### teme92

Can I say a+x=b as b is any element of V?