Image and nullspace bases of a linear transformation

[featheredteap]

Homework Statement

Let T be the linear transformation T: M_2x2-->M_2x2 given by

T([a,b;c,d]) = [a,b;c,d][0,0;1,1] = [b,b;d,d]

Find bases (consisting of 2x2 matrices) for the image of T and the nullspace of T.

Homework Equations

Standard basis of a 2x2 matrix: {[1,0;0,0],[0,1;0,0],[0,0;1,0],[0,0;0,1]}

rank(T) + nullity (T) = n (number of columns of T)

The Attempt at a Solution

I multiplied the elements of the standard basis by T to find the image points of the transformation. I then put those image points in the form [a,b;c,d] in a matrix, which equalled the matrix T relative to the standard basis S:

[T]_S=[0,0,0,0;0,1,0,0;0,0,0,1;0,0,0,0]

To find a basis for the image, I took the columns with leading entries, but I'm not completely sure it's correct:

basis: {[0,1;0,0],[0,0;1,0]}

As for the basis for the nullspace, wouldn't it just be {0} because there is only one solution to the system of equations (i.e. they are linearly independent)? Or is it [0,0;0,0]? However, rank + nullity = 4 and (assuming I got the basis for the image right) rank = 2, so nullity should = 2. Does the nullity of a 2x2 matrix = 2?

 

[lanedance]

woops looked at it again - ok, so
$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} b & b\\ d & d \end{pmatrix}$$

as you have sort of done, you can imagine the matricies as vectors in $$\mathbb{R}^4$$

if should be clear that any matrix with b=d=0, will be mapped to the zero matrix, giving the null space

if either b or d is non-zero, then the resultant transformation will be non-zero, giving the basis of vectors that will be mapped to non-zero vectors. input these individulaly to find a basis for the image space... (i think the one you presented is incorrect)