00:01
Let's start the solution.
00:02
Part a, in part a we have given t is r square from r square and we have t x, y is equal to y, x.
00:16
Now kernel of t is equal to x, y belongs to r square such that t x, y is 0, 0.
00:28
So that is x, y such that t x, y is y, x which is equal to 0, 0.
00:38
So from here we get that implies y is 0 and x is 0 so this is 0, 0.
00:48
So dimension of kernel t is 0 that implies t is 1 to next part b and in that we given t is r square to r cube and t x, y that is equal to x, y, x plus y.
01:21
So that means kernel of t that is equal to x, y such that t x, y which is equal to 0, 0, 0...