Linear algebra assumption

Linear algebra assumption

November 17, 2023

what we have #

(P1) From linear dependent \( n \) vectors, we can have \(n-1\) vectors which span the same set. #

This is true because “dependent” and “span” actually means the same thing.

(P2) # of independent vectors < # of spanning vector (finite dimensional) #

For a spanning set, we could add a vector from the independent set. This act will plug out a dependent vector because of P1

What we want to prove #

Every subspace of a finite-dimensional vector space is finite-dimensional. #

  1. A: is it infinite? B: No

  2. A: Does it have finite spanning list?

    1. A: Does it have a vector? B: yes

      A: Let’s call it \(v_1\).

      1. B: Does it mean we have a way to select \( v_1 \)? A: No

      2. A: How about assuming we are given \(v_1\) B: Ok

      3. A: Can we ask if it spans?

        B: What if we can’t tell?

        A: How about assuming the answer was given?

        B: Are we assuming there is no inbetween span & non-span?

        A: Yes

        B: Can we be sure? A: Yes

        B: How? A: That’s our assumption. That \( \text{span}(v_1, … v_n) ?= \text{ subspace} \) is well defined. ie it has a truth value.

        B: So, we can say, a mathematical construct is assumed to have property of deterministic and identity ? (Let’s call this, A1 our assumption)

        e.g. (P3) A subspace has a property set_member where set_member(subspace) is deterministic and doesn’t change.

        Ok. let’s go back

  3. B: Let’s go back to vector selection, if we have a set of vectors \( V^s \), we can test spanness. (eqality to a vector space)

    1. B: if our set of vector spans, we are good. otherwise, add other vector v

      1. B: P2 says, if I have spanning set \(V^s \), \( \text{len}(V^s) \leq n \), but it doesn’t say there will be such \(V^s\).

        Wait, we can use A1 here? How do we invoke A1 here?

        1. We can think of all \( V^s \) and hence we can say “existance of finite \( \text{span}(V^s)\)”
        2. “existance of finite \( \text{span}(V^s)\)” is a mathematical statement and is assumed to have the property A1

      2. A: By P3, we can always say our \(V^s\) is either span or not-span. If not-span, we are saying there exists a vector which is in subspace but not in \( \text{span}(V^s) \)

        So we bring in this vector and add to our \( V^s \)

        B: So A1 (which seems a static property) gives us existential construction process, interesting.

        B: and adding a vector can’t go over the subspace, because ‘closedness’ of vector space forbits “splitting over” by adding a new vector. So we again end up where we are either spanning or non-spanning.

        So all we needed was, that a set can be compared. Comparison here means, we can test if there’s a vector in one set not in another.

QUESTION But then, how do we know it’s safe to assume P3 is true? #

Or are we just saying, let’s assume P3 is true?

For a number, I have no problem with saying \( n1 = n2 \) or \( n1 \neq n2 \).

P3 just asserts a mathematical object ‘set’ has the same property.

I guess I’m just not sure if we can plugin in any ‘set’ such as all polynomials and be sure if P3 applies to this set.