Linear Algebra 6d: The Equivalence of the Two Definitions of Linear Dependence

[lukedavis]

these videos are getting really good. I love how you fixed the definition and played with it instead of just saying it. that really helps with understanding the essence of the definition

[wawangsf]

it's unfortunate a huge drop off of viewership occur in these videos as they are meaningful and concise

[maxpercer]

I love this video (watched it again). Personally I find it easier than reading a math book (there are advantages to reading math books too).
Watching you gives me hope for humanity.
If i may critique, the alpha, beta, gamma, delta are not necessarily the same in the → part and the ← part, since they are arbitrary. This is not really a critique because it is obvious to anyone who has sophistication in proofs. Also, as someone once said, if you give too many details to a proof you can lose your audience. One must strike a balance between detail and conveying the 'gist of it'.

[khuldoonusman]

When you explain the equivalence from the 1st to the 2nd definition, in the first definition: 'a' can be a linear combination of trivial coefficients but if that is the case then 'a' itself would be the zero vector. However if we then move 'a' to the other side to equate it to zero, it cannot be equal to the second definition because of the trivial coefficients and the fact that 'a' is a zero vector. So then they would not be equal, right?

Or does 'a' being a zero vector mean it can have any coefficient?

[maxpercer]

How do we go about showing the two definitions are equivalent in the 'extreme' or boundary cases? In the case that the set contains a single vector, and the case when the set contains no vectors at all - the so called empty set Ø . I guess there would have to be some kind of vacuous proof.

[pinazo]

Why without subtitles? Subtitles greatly help non-English speakers understand the content. Add please! There are many videos that lack subtitles!!