Basis and Dimension

Now we know about vector spaces, so it's time to learn how to form something called a basis for that vector space. This is a set of linearly independent vectors that can be used as building blocks to make any other vector in the space. Let's take a closer look at this, as well as the dimension of a vector space, and what that means.

Script by Howard Whittle

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[michaelpisciarino]

0:27 A Basis
1:31 Check Linear Combination
2:26 Span
2:52 Satisfying Linear Independence

3:21 A more complicated example (R 2x2)
3:54 Span check
4:23 Distribute the Scalars. Add up the new matrix.
4:45 Make sure a solution exists
5:43 Check Linear Independence

6:57 Row Echelon Form:
- No Free Variables
- All Scalars must be = 0

7:34 Both conditions verified ✔️

Basis With N Elements= Dimension N

9:05 Check Comprehension

[quantumleap]

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[ChocolateMilkCultLeader]

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[christycaroline]

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[Abdulrahman-hbfy]

I have a sweet information
When n (number of columns) is not equal to m (number of rows) then the set is always not a basis (if the question ask if a specific set of vectors is a basis or not)

but when n = m, then you have two possibilities depending on the det(A), in other words:
det(A) is equal to 0 ==> the set is not a basis
det(A) isn't equal to 0 ==> the set is a basis

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[Mark-nhvh]

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[dter]

The reduced row echelon form isn't finished yet at 7:22, you can still do R3+R4, R2-R4 and after that R1-R3 which doesn't require you to solve the remaining set of equations.

[perspicacity]

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[mmbpaja]

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[immortalspiritualbeing]

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[thesoccerful]

5:05 since you’re taking the determinate of the square matrix and it’s a none zero number, isn’t also linearly independent too?

[Soyokaze-ifkp]

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