# Linear Algebra proof

1. Mar 3, 2014

### tiger2030

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

If dim(X)=n, show that the vector space of k-linear forms on X is of dimension nk.

2. Relevant equations

3. The attempt at a solution

So I know we need to let x1, x2,...xn be a basis for X. My professor then said to "show that the function fj1,...,jk, 1≤jl≤n defined by fj1,...,jk(xi1,...xik) = δi1j1,...,δikjk and then extend multilinearly." This is where I am lost on what to do. Any help would be much appreciated.

2. Mar 3, 2014

### micromass

Staff Emeritus
In one variable. Let's say you have a basis $e_1,...,e_n$. You can define a unique linear map $T$ by saying that $T(e_i) = y_i$.
Indeed, the actual definition of $T$ is

$$T(\sum \alpha_i e_i) = \sum \alpha_i y_i$$

This is what it means to extend a function linearly: you start by defining it on a basis, and then use linear combinations to define the function on the entire space.

Does that make sense?

3. Mar 3, 2014

### tiger2030

Yes, this makes sense because to be linear, you must be able to pull the constant out and still yield the same answer.

4. Mar 3, 2014

### micromass

Staff Emeritus
Good, so does that explain the professor's hint:

It's just the same but in multiple dimensions.

5. Mar 3, 2014

### tiger2030

Ok so first I need to check that they are well defined k-linear forms, correct?

6. Mar 3, 2014

### micromass

Staff Emeritus
That's one thing you need to verify, yes.

7. Mar 3, 2014

### tiger2030

so all fjk map to either 1(for fjk(xik)) or 0(for fjk(xim), where k≠m.

also, if xik=yik, then fjk(xik)=1=fjk(yik)

Therefore all f are well defined.

8. Mar 3, 2014

### micromass

Staff Emeritus
How are the $f_{i_1....i_k}$ defined on non-basis elements?

9. Mar 3, 2014

### tiger2030

As a linear combination of basis elements?

10. Mar 3, 2014

### micromass

Staff Emeritus
Can you give an exact definition like I did in my post 2?

11. Mar 3, 2014

### tiger2030

so fjk(∑αiei)=∑aiδi=∑ai?

12. Mar 3, 2014

### tiger2030

Say x31=αe1+βe2+γe3.
Then fj1(x31)=fj1(αe1+βe2+γe3)= αfj1(e1)+βfj1(e2)+γfj1(e3)=α+β+γ?

13. Mar 3, 2014

### micromass

Staff Emeritus
So going back to the original problem. Take $x_1,...,x_n$ a basis for $X$. Take $y_1,...,y_k$ any $k$ elements in $X$. How do we define

$$f_{i_1,....,i_k}(y_1,....,y_k)$$

14. Mar 3, 2014

### tiger2030

That would equal ∑ai, where yi=∑aixi

15. Mar 3, 2014

### micromass

Staff Emeritus
No, that's not correct.

16. Mar 3, 2014

### tiger2030

ok so I am just going to try and use an example to see where my thought process is wrong and then use that to apply to a general case.
Say y2=3x1+2x2+4x3. Then fj2(y)=fj2(3x1+2x2+4x3)=fj2(3x1)+fj2(2x2)+fj2(4x3)=2

17. Mar 3, 2014

### micromass

Staff Emeritus
It seems you misunderstand the indices $j_1$ and so on.
Let's work in one variable. In that case, you are given an index $j_1$ and you know that $1\leq j_1 \leq n$. So $j_1$ could be anything from $1$ to $n$. And for each value of this $j_1$, you have a map.

So you have maps

$$f_1,~f_2,~f_3,~f_4,...$$

What $f_4$ (for example) simply does is take out the fourth basis vector and give its coordinate. So, for example

$$f_4(3x_1 + 2x_2 + 4x_3 + 6x_4) = 6$$

Now, in the case of two variables, you have two indices $j_1$ and $j_2$ which can take on values anything from $1$ to $n$. Let's say $n=3$, then you have maps

$$f_{1,1},~f_{1,2},~f_{1,3},~f_{2,1},~f_{2,2},~f_{2,3},~f_{3,1},~f_{3,2},~f_{3,3}$$

So there are $9$ maps. (with general $n$, there are $n^2$ maps).

Let's look at a specific map like $f_{2,1}$. This map is a biliniear map, meaning it takes in two elements of $X$. And what it does is select the 2nd coordinate of the first element and the first coordinate of the second element and multiply them. So

$$f_{2,1}(3x_1 + 6x_2 + 4x_3,x_1 + 2x_2 + 5x_3) = 6\cdot 1 = 6$$

Likewise for example,

$$f_{1,3}(3x_1 + 6x_2 + 4x_3,x_1 + 2x_2 + 5x_3) = 3\cdot 5 = 15$$

18. Mar 3, 2014

### tiger2030

Ok, that clears things up a lot more. So instead I would get the coefficient with the first basis vector from y1 multiplied by the coefficient with the second basis vector from y2, and so on until it it multiplied by the coefficient with the nth basis vector from yn

19. Mar 3, 2014

### micromass

Staff Emeritus
Right, so what we do is decompose each $y_j$ in the basis $x_1,....,x_n$. Let's write

$$y_j = \alpha_{1,j} x_1 + .... + \alpha_{n,j} x_n$$

Then

$$f_{i_1,....,i_k}(y_1,....,y_k) = \alpha_{i_1,1}\cdot .... \cdot \alpha_{i_k,k}$$

I know the notation is very awkward, but you need to get used to it.

20. Mar 3, 2014

### tiger2030

The explanation of the notation cleared a lot of stuff up for me. So I get know that we have taken apart each yj and are multiplying the coefficients together but how does this show the dimension is nk?