Let A 9 12 4 6 1 1 Find bases of the kernel and image of A

Let A = [-9 -12 -4 -6 -1 -1] Find bases of the kernel and image of A (or the linear transformation T(x) = Ax).

Solution

To find the kernel of a matrix A is the same as to solve the system AX = 0, and one usually does this by changing A to its RREF B. The matrix A and B have exactly the same kernel. In both cases, the kernel is the set of solutions of the corresponding homogeneous linear equations, AX = 0 or BX = 0.

Here A = [ -9    -4     -1]

[ -12 -6     -1]

B, the RREF of A is the matrix B =    [ 1   0   1/3]

[ 0   1 -1/2 ]

The solution to the equation AX = 0 is the same as the solution to x +1/3 z = 0 and y – (1/2)z = 0., where X = ( x , y , z)^T.Now, if z = t, then x = (- 1/3) t and y =( 1/ 2) t so that X = ( x , y , z)^T = t ( - 1/3 ,1/ 2 , 1)^T. The vectors so determined span the kernel of A and a basis for the kernel of A is the set { ( - 1/3 ,1/ 2 , 1)^T}.

The basis set of the transformation T is the set { T(e₁) ,T(e₂), T (e₃) } . Now, T ( e₁) = A(e₁) = [ -9    -4     -1] [ 1 ]

[ -12 -6     -1] [ 0 ]

[ 0 ]

                

Thus T (e₁ ) = (- 9, -12)^T = -3( 3, 4)^T . Similarly, T( e₂ ) = ( -4 – 6)^T = -2 ( 2, 3)^T and T( e₃ ) = (-1, -1)^T =      -1(1, 1)^T . However , these 3 vectors are not linearly independent as ( 1, 1)^T = ( 3 , 4)^T - (2, 3)^T . Therefore, a basis for the image of A is { ( 3, 4)^T , ( 2, 3)^T }