How to find a basis for the kernel and image of a linear transformation matrix Let $$A=
        \begin{bmatrix}
        0 &amp; 0 &amp; 6 &amp; -18 \\
        0 &amp; 0 &amp; -1 &amp; 3 \\
        0 &amp; 0 &amp; -2 &amp; 6 \\
        \end{bmatrix}
$$
  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 
$$
        \begin{bmatrix}
        0 &amp; 0 &amp; 0 &amp; 0 \\
        0 &amp; 0 &amp; -1 &amp; 3 \\
        0 &amp; 0 &amp; 0 &amp; 0 \\
        \end{bmatrix}
$$
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 $T$.
The kernel basis is:

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

The image basis is:

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