Ufyukyu

up vote 1 down vote favorite 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. linear-algebra linear-transformations share | cite | improve this question ...