Define the linear transformation T by T($\vec x$) = A$\vec x$. Find (1) the kernel of T and (2) the range of T
$A = \begin{bmatrix}1&-2&1\\0&2&1\end{bmatrix}$
1) Finding the Kernel(T)
$\begin{bmatrix}1&-2&1\\0&2&1\end{bmatrix}$ => RREF $\begin{bmatrix}1&0&2\\0&1&.5\end{bmatrix}$
$\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$ = $\begin{bmatrix}-2x_3\\-.5x_3\\t\end{bmatrix}$ =
$\begin{bmatrix}-2t\\-.5t\\t\end{bmatrix}$ =
t$\begin{bmatrix}-2\\-.5\\1\end{bmatrix}$
My Answer – Kernel(T) = {t(-2, -.5, -1)}, t is any real number
2) Finding the Range(T)
$\begin{bmatrix}1&-2&1\\0&2&1\end{bmatrix}$ => RREF $\begin{bmatrix}1&0&2\\0&1&.5\end{bmatrix}$
My Answer – Basis for range(T) = {(1.0), (0,1)}
This is what I got for the kernel and range, however the answer words it in an odd way. It says "Kernel of A is null space of A which is span{(4,1,-2)} and the Range of A is $R^2$.
Is my answer equivalent with the book's answer? Did I do anything wrong?
Best Answer
Your answers match those of the book; they're just written differently. Note that if you set $t = -2$, you will see that the set of vectors spanned by $\begin{bmatrix}-2 \\ -0.5 \\ 1\end{bmatrix}$ is the same as that spanned by $\begin{bmatrix} 4 \\ 1 \\ -2\end{bmatrix}$. Similarly, $\left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1\end{bmatrix} \right\}$ is a basis for $\mathbb{R}^2$.