# How to find a basis for the kernel and image of a linear transformation matrix

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:

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