Linear Algebra Proof: Dim(X)=n, Show nk Dim Vector Space

[tiger2030]

Homework Statement

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

Homework Equations

The Attempt at a Solution

So I know we need to let x₁, x_2,...x_n be a basis for X. My professor then said to "show that the function f_j1,...,jk, 1≤j_l≤n defined by f_j1,...,jk(x_i1,...x_ik) = δ_i1j1,...,δ_ikjk and then extend multilinearly." This is where I am lost on what to do. Any help would be much appreciated.

 

[micromass]

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?

 

[tiger2030]

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

 

[micromass]

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

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

"show that the function $f_{j_1,...,j_k}$, $1\leq j_l\leq n$ defined by $f_{j_1,...,j_k} = \delta_{i_1j_1}...\delta_{i_kj_k}$ and then extend multilinearly."

It's just the same but in multiple dimensions.

 

[tiger2030]

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

 

[micromass]

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

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

 

[tiger2030]

so all f_jk map to either 1(for f_jk(x_ik)) or 0(for f_jk(x_im), where k≠m.

also, if x_ik=y_ik, then f_jk(x_ik)=1=f_jk(y_ik)

Therefore all f are well defined.

 

[micromass]

so all f_jk map to either 1(for f_jk(x_ik)) or 0(for f_jk(x_im), where k≠m.

also, if x_ik=y_ik, then f_jk(x_ik)=1=f_jk(y_ik)

Therefore all f are well defined.

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

 

[tiger2030]

As a linear combination of basis elements?

 

[micromass]

As a linear combination of basis elements?

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

 

[tiger2030]

so f_jk(∑α_ie_i)=∑a_iδi=∑a_i?

 

[tiger2030]

Say x₃₁=αe₁+βe₂+γe₃.
Then f_j1(x₃₁)=f_j1(αe₁+βe₂+γe₃)= αf_j1(e₁)+βf_j1(e₂)+γf_j1(e₃)=α+β+γ?

 

[micromass]

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)

 

[tiger2030]

That would equal ∑a_i, where y_i=∑a_ix_i

 

[micromass]

No, that's not correct.

 

[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 f_j2(y)=f_j2(3x₁+2x₂+4x₃)=f_j2(3x₁)+f_j2(2x₂)+f_j2(4x₃)=2

 

[micromass]

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 f_j2(y)=f_j2(3x₁+2x₂+4x₃)=f_j2(3x₁)+f_j2(2x₂)+f_j2(4x₃)=2

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

 

[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 y_n

 

[micromass]

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.

 

[tiger2030]

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

 

[tiger2030]

I understand there would be n^k different permutations of the coefficients indices but is this enough to prove the dim?

 

[micromass]

The idea is to show that the set of all $f_{i_1,...,i_k}$ forms a basis for the vector space of all $k$-forms.