Understand Linear Functionals & Vector Space X

[wurth_skidder_23]

Here is the problem I have been asked to solve:

Assume that m < n and l1, l2, . . . , lm are linear functionals on an n-dimensional vector space X.
(a) Prove there exists a non-zero vector x in X such that the scalar product < x, lj >= 0 for 1 <= j <= m. What does this say about the solution of systems of linear equations?
(b) Under what conditions on the scalars b1, . . . , bm is it true that there exists a vector x in X such that the scalar product < x, lj >= bj for 1 <= j <= m? What does this say about the solution of systems of linear equations?I am having trouble understanding the concept of a linear functional and how it relates to the vector space X.

 

[AKG]

x is a vector in X, and l_j is a linear functional on X, so what is < x, l_j >? Do you mean l_j(x)? Linear functionals are linear functions which map vectors to scalars. By linearity, they are uniquely determined by their behaviour on the basis vectors of the vector space. Let f be a linear functional on a vector space V, and let v₁, ..., v_n be basis vectors for V. If x is any vector in X, then there exist scalars x₁, ..., x_n such that

x = x₁v₁ + ... + x_nv_n

Then:

f(x)
= f(x₁v₁ + ... + x_nv_n)
= f(x₁v₁) + ... + f(x_nv_n)
= x₁f(v₁) + ... + x_nf(v_n)
= (x₁, ..., x_n).(f(v₁), ..., f(v_n))

If l₁, ..., l_m are linear functionals, consider the row vectors (l₁(v₁), ..., l₁(v_n)), ..., (l_m(v₁), ..., l_m(v_n)). Do you see how the question is reduced to the following:

If m < n, and {l_i(v_j) | i = 1, ..., m; j = 1, ..., n} is any set of scalars, do there exists scalars x₁, ..., x_n such that for each i, it holds that:

(x₁, ..., x_n).(l_i(v₁), ..., l_i(v_n))

Consider:

(\mbox{Span}\{(l_i(v_1),\, \dots ,\, l_i (v_n))\ :\ 1 \leq i \leq m\})^{\perp}

 

[wurth_skidder_23]

< x, lj > is the scalar/dot product

 

[AKG]

I know that. Suppose v is a vector and x is toenail. What is <v,x>?

 

[wurth_skidder_23]

Yeah, I still have no idea what I'm proving for this problem, nor do I understand the final question, though I imagine if I could do the proof, the question would make sense. A hint would be appreciated.

 

[AKG]

The dot product takes two vectors of the same space and gives you a scalar. A linear functional on V is not a vector in V. There is a strong relation between the value of a linear functional at a vector and the dot product of that vector with a particular other vector, but that's still a little off in the distance for us. First, you need to get the basics.

7 is a real number. So is -\pi. It makes sense to talk about, say, 7 x -\pi. You can multiply numbers. Can you multiply a number by a function? What is 7 x sine? It doesn't really make sense (okay, you can make sense of it, but try to understand the basic idea). Similarly, v and x are both vectors, say. Then it makes sense to take their dot product, v.x If f is a linear functional, then what sense does it make to take v.f? You can multiply two numbers together, but you can't multiply a number by a toenail, it's just absurd. Likewise, you can take the dot product of a vector with another vector, but you can't take the dot product of a vector with a functional, it's just absurd. It makes as much sense as taken the dot product of a vector with a toenail.

If V is a vector space over a field F, then a linear functional is a function: f : V -> F that is linear. So a linear functional (I'll just call it a functional) is a function, which is just something that maps elements of one set to elements of another set. In addition, it is a special kind of function, it is a linear function. That means f(v+w) = f(v)+f(w), and f(av) = af(v), where v and w are vectors in V, and a is a scalar in F. Do you understand this so far?

Suppose you have an n-dimensional vector space V (over a field F), and m (with m < n) linear functionals f₁, ..., f_m. You want to prove that there is a non-zero x such that f_i(x) = 0 for 1 < i < m.

Let me give you a related problem. Let V be an n-dimensional vector space. Let {v₁, ..., v_m} be any set of m vectors from V, with m < n. Prove that there exists a non-zero x such that <x,v_i> = 0 for 1 < i < m.

 

[matt grime]

Duals always seem to cause problem's, conceptually, and I don't know why.

If V and W are vector spaces you are really happy with what the space of linear maps between V and W is. If we are the kind of person who likes bases then it is the set of dim(W) by dim(V) matrices. Well, all we've done is take a general case you're happy with and looked at one particular example, when W is one dimensional. Surely that is even easier to understand than the general case?

 

[abhi_kirk]

Hi.
I have the same question as wurthskidder23 and I've read the reply by AKG. Can anybody give some further hint as how to solve the question. I am not sure how to prove that x is non-zero?