00:01
Is one to one and prove that t is on to okay so you want to show that t is on to so just apply rank nality theorem to t so you will get that dimension of v is equal to dimension of kernel of t but we are assuming t is one to one right so kernel of t is just a zero subspace and dimension of that is zero so dimension of v is equal to dimension of range of t which is also known as rank of t.
00:55
And you also know that it's given in the proposition that dimension of v is equal to dimension of w.
01:04
And from a previous problem, you know that if dimension of w is equal to dimension of range of t, then t is on to.
01:17
This is simply because range of t is always a subspace of w.
01:23
And if you have a subspace inside a vector space whose dimension is equal to the, full space then it is equal to the full space so t is on to okay now let us prove the other equivalence assume that t is on to and you want to show that t is one to one here again i'm going to apply rank nullity theorem so you have dimension of v equal to dimension of w which is equal to which is equal to dimension of kernel of t plus dimension of range of t now, since t is onto, range of t is equal to w.
02:51
So, this is equal to dimension of w.
02:55
So this thing will cancel with this thing.
02:58
And you will get that dimension of kernel of t is zero...