Linear Algebra 3b1: Decomposition with Geometric Vectors 1

[budokan]

The most enjoyable linear algebra course I have ever watched. Excellent examples, great explanations. Thank you for all your efforts Dr. Pavel Grinfeld.

[michaelszymczak]

Decomposition of C here is finding its components A and B. An A component of C can determined by applying an A projection transformation to C. The A projection transformation has a null space (not defined yet in this course) that takes any vector in this space to zero. As it happens, vector B lives in this space. Geometrically, this manifests itself as a perpendicular line, and this is I think the reason why it is 'special'.

[alojzybabel]

I think that what makes the decomposition easier when we have the right angle, is that the dot product works out of the box. The dot product makes the orthogonal projection of one vector upon another, that is, it projects it at the right angle. So when the vector b is also at right angle to a, then all you need is to use the dot product. However, if the angle between a and b is different, you need to use a slanted projection along the direction of b upon the vector a, and this is not what the dot product gives you out of the box. You will have to compensate for that by some additional tricks, which makes it more complex.

[antonellomascarello]

0:25 : IMPORTANT : Significantly different objects, but Lin decomposition applies equally well to all of them
0:46 : Example #1
3:33 : Example #1 / IMPORTANT NOTE
6:42

[mukov]

No need for orthogonal and guessing: let basis (OA, OB), where OA is vec from point O to point A. Vec OC = (|OA'|/|OA|)*OA + (|OB'|/|OB|)*OB, where A' = intersection of (line at C, parallel to OB) and (line OA); B' = intersection of (line at C, parallel to OA) and (line OB).

[michaellewis]

5:31 Mate “much richer” they are in the same linear sub space, they are scalar multiplications of each other so there is a relational invariant but still an extra degree of freedom for specifying each quantity so there is an infinite amount of combinations possibilities.

[Cookstein]

take a line parallel to b and move it to the end of c. Where it intersects the extension of a is coefficient alpha. And then vice versa for coefficient beta

[Cookstein]

the money example is difficult because whilst the dollar and cents vectors are different vectors they really only differ in magnitude i.e. they are vectors in 1D and subsequently share direction i.e. are linearly dependent?

[Mahmood]

how to get practice with those ?? any website or a book with answers to the problems at the end of it so to check my answers plz will be helpful