Linear Algebra 4b: Impossible Decomposition with Polynomials

[IgorKuts]

Actually writing down your extremely precise explanations word-by-word. Learning to speak math, so to speak. It is clear that you were emphasizing the correct wording and this subtle detail says a lot to me about the quality of your videos. Thank you prof. Grinfeld.

[rd-tkjs]

Illuminating lectures for a 56 yr greenhorn ... Thanks professor !

[OtterMorrisDance]

Another wonderful video. Thank you Pavel!

[antonellomascarello]

1:17 Example #1
2:35 9:02 Example #2
4:25 IMPORTANT NOTE : Different vector spaces but strong analogy in terms of LA framework
5:43 Example #3

[TheDutLinx]

Very cool subject and explanation. Thanks for the video!

[Jnglfvr]

There's a SINGLE fundamental reason why these three vectors cannot sum to the given totals and that is simply because, in each case they are linearly dependent and the vector x^2 + 1 is not in the 2 dimensional subspace spanned by any of these three vectors. Eg., in case 1 we have 7*( 7*x^2 - x) = 37*( x^2 + x) + 4*( 3*x^2 - 11*x). In case #2 we have x^2 - 2*x +1 = (x^2 - 1) - 2*( x -1) and in case 3 we have x^2 - 4*x + 4 = (x^2 - 4 ) - (4/3)*(3*x - 6).

[MrDizzy]

You draw a cubic polynomial generated from the quadratic decomposition vectors :)

[xoppa]

2:10 'free term' refers to the constant term

[karanbirsidhu]

For the second example, can we instead also say that the target polynomial has a complex root, and the basis polynomials chosen have real roots instead. Would that be an equivalent argument?