00:01
Okay, so for this exercise we need to prove a theorem that say that, well, let's consider two vector spaces over the field k.
00:11
You got some basis on the vector v, and you got any vectors on u.
00:18
So these are not necessary a basis on you.
00:23
So we need to show that there exists a map from b to u such that f evaluated at vi, that corresponds to the basis of the vector b is equal to these vectors ui on u for all i from 1 to n.
00:45
Okay, so we need to show this, so how to start.
00:48
Well, first we need to define these, the first step is to define the mapping from u to b and show that this well defined.
01:00
Okay, so that means that we know that f applied to vi is equal to ui, okay, for all these i's from one to end.
01:19
So let's consider any vector v on v.
01:26
Here i got a mistake.
01:30
Here we need to define the mapping from v to you.
01:36
Okay, so we choose any vector on the vector space v.
01:42
So that means this is equivalent to say that v is written as a linear combination of the basis.
01:53
The basis is given by v1, v2 up to vn.
01:57
So here we obtain a1 v1 plus a2 v2 plus dot dot dot dot.
02:09
A, vn.
02:17
And now we're going to apply the mapping to this vector b, based on the definition of the mapping.
02:29
So if we apply f to b, what we're going to obtain is a1, u1, plus a2, u2, plus d, dot, dot, plus un here, and vm, a1, and a scalar and these skillers are unique so this map is well defined okay so that's the first step even more we can for example we can define a1 for example for f b1 is just taking 0 plus 0 plus 0 plus in this case the constant that multiplies to this is just one and the rest of the coefficients are going to be zero okay so based on this definition we got that yes f of b i is equal to ui and the mapping is well defined so that's the first thing now we need to show that this map is linear okay so far what we have shown is that there exists this map we can define this kind of mapping is well defined but now we need to show if this is linear or not okay so the second part is show that f from u sorry from b to u is linear okay so and we know how to do this so let's take any vector v on the vector space v we know that this to say that v is equal to a1.
05:22
I'm going to short this notation so we can write this just as the summation from i equals to 1 up to n of a .i.
05:36
And we're going to also consider a vector w, also on this vector space.
05:47
And w is also written as a linear combination of the basis.
05:56
So that means that the only change our the coefficient.
06:00
So, b -i -b -i.
06:02
Okay, so we got these two vectors.
06:06
And in order to show that f is a linear map, we need to first verify if the summation of these two vectors is equal to the map applied to the summation of the two vectors is equal to the sum of the map of these two vectors.
06:29
This equality we need to show.
06:33
So let's start with the left -hand side, from the left -hand side...
