1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $T$.

$\endgroup$
0
$\begingroup$

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 $Ax=0$.

$\endgroup$
-1
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.