Linear Algebra 4c: Impossible Decomposition in ℝⁿ

[antonellomascarello]

0:22 IMPORTANT point : Treating different vector spaces in their own terms
1:20 IMPORTANT point : Do not automatically associate geometric vectors with Rn vectors
2:40 Example #1
4:06 Example #2
5:55 Example #3

[Hythloday]

It is a great course and all so far. I like how you are making great effort to introduce concepts in a manner that allow for internalising of larger ideas rather than rote memorisation. This is the way I would advocate all tuition be done. I myself have always had a passion for understandings and trouble with memory, thus, and somewhat unfortunately, I have had to struggle and find understandings for myself. A huge task. I find myself with a first in mathematics attempting to redo / relearn all my knowledge. As is known, knowledge learned by rote is knowledge quicker lost.

One criticism. The constant advertisements are annoying with such bite sized lessons. I preferred your longer lectures on tensors, as far as I got anyhow... Many thanks though.

[ivanzagar]

Oh my god!
I just realized that this is really similar to how we can define a set in set theory. We must choose a property that has sense and we get another subset. We could also say that the elements of a set hold that property.
In LA there are linear properties which are the only ones that make sense. Vectors hold properties and if any is linear (survives under linear combinations) then it follows that the target vector also has that property.
But now instead a subset we get a subspace that is a subset of vectors that are by all rules of algebra a vector space.

[MrCrazyShock]

In the last example, there is also another property that is common to the decomposition factors but which is not present in the target vectors; in each decomposition vector, there is a difference of 1 between each entry. For example, in the last vector 7+1=8, 8+1=9. Same thing for the two other decomposition vectors. However, it is not the case with the target vector.

I try with a vector that as a bigger diffrence between entries but for which the middle entry is still the mean of the two others (1 3 5). Guess what? Any linear combination still keeps this proprety...

[invictusdomini]

Two columns are also linearly dependent, once the left-most column is subtracted from the middle and right, in the bottom example.

[agh]

For the second example would it be sufficient to say that the second entries are all multiples of 5 therefore any linear combination would also have a second entry that is a multiple of 5?

[agh]

For the third example would it also be valid to say that the first and second entries of the decomposition vectors differ by the same amount as the second and third entries, and therefore any linear combination should also have this property?

[revanthkalavala]

In third example we can also see that middle vector is average of other two
I have seen it it column wise, you explained it in row wise but my question is this useless to think column wise ?