0:00
Hello there.
00:01
Okay, so for this exercise we got two linear maps, f and g, defined as follows.
00:11
We also got the composition of these two maps.
00:14
So g -o -f is defined from v to dov.
00:22
So we need to show that first the rank of the composition of the functions is less than the rank of g.
00:33
So let's prove this.
00:35
Let's first consider the image of the mapping f.
00:40
It's going to be a super space of u.
00:46
And then if we take the map g, we got that g -o -f of the whole set v.
01:00
It's going to be contained on g of u.
01:04
So they say that the dimension of g of g.
01:11
Of of b is less or equal than the dimension of g of u and the dimension this definition of dimension is equal to say is equal to the rank so basically what we got here is that the rank of g o f is less or equal than the rank of g you can gain some intuition in this because you got here for example the three sets, v, u on w.
01:55
Here you got the whole set v, and when you take the map f, you can either have a subspace of u or the whole space u, right? so that's why we got this relation here...