Let L be the linear operator in R2 defined by Lx4x10x28x10x2

Let L be the linear operator in R2 defined by L(x)=(4x1+0x2,8x1+0x2)T

Find bases of the kernel and image of L .

Solution

The matrix of the transformation L is A = [ - 4     0]

[ 8      0]

The RREF of A is [ 1 0 ]

[ 1 0 ]

The Kernel of L is the set of solutions to the equations x₁ = 0. x₂ is arbitrary, say t. Then x = ( x₁ , x₂ )^T = t( 0, 1)^T Thus, a basis for the kernel of L is { ( 0, 1)^T }

The basis for the image of L is { L (e₁ ), L(e₂ )} where e₁ =   ( 1, 0)^T and e₂ = (0, 1)^T . Now, L (e₁ ) = ( -4, 8) = -4( 1,2) and L (e₂ ) = (0, 0). Therefore, a basis for the image of L is { ( 1,2)^T }.