# Vectors make sense

## Description

Check out Brilliant: https://brilliant.org/LookingGlassUniverse/

(if you click through this link then the Vector Spaces chapter of their Linear Algebra course is free for 2 weeks from upload)

Vectors may seem very difficult when you're first introduced to them, but I hope this video helps you see they're not that scary! This is the start of a whole series of linear algebra, and I will cover vectors, the scalar product, matrices, eigenvalues/ eigenvectors and Dirac notation.

DON'T FORGET TO DO YOUR HOMEWORK:

Prove these 2 statements about bases

1. If you have two different bases for the same space, then they must have the same number of basis elements in them. (E.g, there are many different choices of basis for the plane, but no matter what basis you choose, there are only 2 vectors in each basis.)

2. Once you pick a basis (say {v_1, v_2}), there's only one correct way to write another vector as a linear combination of the basis vectors. Eg, say v=a v_1+ b v_2. Then you can't also write v=a' v_1+ b' v_2, where a' and b' are different from a and b.

The multiple choice questions from Brilliant:

Q1. Which of these vectors is redundant (i.e. can be written as a linear combination of the other 2):

i) (1 2 3)

ii) (1 3 5)

iii) (2 5 8)

iv) Each of the above

The answer is available here: brilliant.org/practice/linear-independence/?p=2

Q2. Consider the following 3 vector spaces:

A= The vector space spanned by {(1 2)}

B= The vector space spanned by {(1 2), (2 3)}

C= The vector space spanned by {(1 2), (2 4), (3 6)}

Question: which of the following is true?

i) A is a subspace of B, which is a subspace of C

ii) C is a subspace of A, which is a subspace of B

iii) B is a subspace of C, which is a subspace of A

iv) A is a subspace of B and C, which are not subspaces of each other

The answer is here: brilliant.org/practice/subspaces-and-span/?p=6

HINTS FOR PROOF QUESTIONS:

2 is easier, so let's do that first

Hint 2.1 Let's do the case with just 2 basis vectors first. If there are 2 basis vectors v_1 and v_2, the one thing you know about them is that they are not just a multiple of each other (otherwise it wouldn't be a basis). Try and get a contradiction with this fact.

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Hint 2.2 Assume v= a v_1+ b v_2 = a' v_1+ b' v_2, but a and a' aren't equal, and b and b' aren't equal.

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Hint 2.3 Use the above equation to write a relationship between v_1 and v_2. Oh no, that looks like they are multiples of each other!

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Hint 2.4 Now do the case where there are n basis vectors. What you know about them is that you can't write them as linear combinations of the others. Try and get a contradiction with this fact.

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Hint 1.1. Imagine you had 3 vectors and they span 2D space. Doesn't that one of them is redundant? The following in this case first:

B1={u_1,u_2}

B2={v_1, v_2,v_3}

Write each u_i in terms of B1.

Remember that there is a redundancy in B2 if you can write c u_1+ d u_2= u_3. So write this, and let's see if we can find a solution for c and d.

Plug in your equations for u_i into c u_1+ d u_2= u_3

You now have a vector on the right hand side in B1 and a vector on the left hand side in B1

Using the result from question 2 (dammit, I really should have swapped the order of these questions), you know that the coefficient in front of v_1 and v_2 must be the same on both sides (since there is only one unique way to write a vector in B1)

So now you have 2 linear equations with 2 unknowns (c and d- everything else is 'known').

Show that they only don't have a solution for c and d if B2 was actually linearly dependent all along. (Yes this will require you to know some linear algebra to do efficiently (although technically possible without)).

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When you assume that the two bases can be any size it's most efficient to do this with linear algebra (sorry!!)

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The solution: yutsumura.com/if-there-are-more-vectors-than-a-spanning-set-then-vectors-are-linearly-dependent/

Music: Epidemic sound, Summer nights 2