Cfrgtkky

Let $$A=
begin{bmatrix}
0 & 0 & 6 & -18 \
0 & 0 & -1 & 3 \
0 & 0 & -2 & 6 \
end{bmatrix}
$$
Find a basis for the kernel and image of the linear transformation $T$ defined by $T(x) =Ax$.

My question is how do you handle the first 2 columns of A which are all zero's? Does that mean the top 2 values of the basis of the kernel are variables? both zero? The methods I know for calculating the basis for the kernel and image of a transformation matrix are not producing the correct answers in WebWork.

By elimination, you will end up with
$$
begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & -1 & 3 \
0 & 0 & 0 & 0 \
end{bmatrix}
$$
which gives you the dimension of the image of $T$ and the kernel of $T$.

Now, you can read from the matrix a basis for the image of $T$. On the other hand
$$
-x_3+3x_4=0
$$
tells you how to find a basis for the kernel of $T$.

Image is spanned on linearly independent columns of matrix $A$, which follows from definition of matrix of linear transformation. To find it do the row reduction of $A^T$ and its nonzero columns span the image, or equivalently, if you row reduce $A$ then columns with pivots will correspond to columns of $A$ that form image of the transformation.
Kernel is set of vectors $x$ such that $Ax=0$. To find it's base you need to solve system of equations given by $Ax=0$.

The kernel basis is:

$$ left[
begin{array}{cc}
0&0&1&1/3\
end{array}
right]^T, left[
begin{array}{cc}
1&0&0&0\
end{array}
right]^T, left[
begin{array}{cc}
0&1&0&0\
end{array}
right]^T$$

The image basis is:

$$left[
begin{array}{cc}
6&-1&-2\
end{array}
right]^T$$