Let A 3 3 9 2 1 6 and L be the transformation defined by LX

Let A = [-3 3 -9 2 1 6] and L be the transformation defined by: L(X) = AX, L: R^2 rightarrow R^3. 1) Find the kernel of L. Is the transformation L injective? 2) Find the image of L. Is the transformation L surjective?

Solution

Since L(X) = AX, therefore, A is the standard matrix of L and hence Ker(L) = Null(A)and Im(L) = Col(A).

To determine Null(A) and Col)(A), we ill reduce A to its RREF as under:

Multiply the 1st row by -1/3

Add -3 times the 1st row to the 2nd row

Add 9 times the 1st row to the 3rd row

Multiply the 2nd row by 1/3

Add 2/3 times the 2nd row to the 1st row

Then the RREF of A is

1

0

0

1

0

0

Hence Im (T) = Col(A) = { (-3,3,-9)^T,(2,1,6)^T} or, { (1,0,0)^T, (0,1,0)^T}.

Also, Ker(T) = Null(A) is the set of solutions to the equation AX = 0. If X = (x,y)^T, then this equation is equivalent to x = 0 and y = 0. Hence Ker (T) = {(0,0)^T}.

Since L(X)= AX= 0 has only the trivial solution X = (0,0)^T, therefore L is injective.

Since the columns of A do not span R³, hence T is not surjective.

1	0
0	1
0	0