00:01
This exercise we need to prove one of the theorems.
00:02
The terms say that if we got some finite dimensional space v here, and also we got that the dimension of b and u are the same, and the map that goes from v to u, then f is an isomorphism, even if f is non -singular.
00:25
So let's first prove this direction of the double implication.
00:32
So let's suppose that f is an isomorphism.
00:42
So if f is an isomorphism, that means that f of zero only maps to the zero vector.
00:53
So this implies that the kernel of f is just the vector zero.
01:01
And this implies that f is non -singular.
01:10
So this direction is really straightforward.
01:12
However, the other direction is not that easy.
01:17
So now we need to show the other direction.
01:22
That means that if f is non -singular, then that implies that f is an isomorphism.
01:39
So we know that f is non -singular, that means that the kernel of f is equal to zero.
01:50
The dimension, sorry, the dimension of, well, the kernel of f is just the zero vector, and this implies that the dimension of the kernel...