We are given:
Find $\ker(T)$, and $\textrm{rng}(T)$, where $T$ is the linear transformation given by
$$T:\mathbb{R^3} \rightarrow \mathbb{R^3}$$
with standard matrix
$$ A = \left[\begin{array}{rrr} 1 & -1 & 3\\ 5 & 6 & -4\\ 7 & 4 & 2\\ \end{array}\right]\textrm{.} $$
The kernel can be found in a $2 \times 2$ matrix as follows:
$$ L = \left[\begin{array}{rrr} a & b\\ c & d\\ \end{array}\right] = (a+d) + (b+c)t $$
Then to find the kernel of $L$ we set
$$(a+d) + (b+c)t = 0$$ $$d = -a$$ $$c = -b$$
so that the kernel of $L$ is the set of all matrices of the form $$ A = \left[\begin{array}{rrr} a & b\\ -b & -a\\ \end{array}\right] $$
but I do not know how to apply that to this problem.
$\endgroup$ 13 Answers
$\begingroup$$$ A = \left[\begin{array}{rrr} 1 & -1 & 3\\ 5 & 6 & -4\\ 7 & 4 & 2\\ \end{array}\right] $$ Consider a linear map represented as a $m × n$ matrix $A$ . The kernel of this linear map is the set of solutions to the equation $Ax = 0$ $$ ker(A)=\{x \in R^n|Ax=0\} $$ $$ det(A)=1(12+16)-(-1)(10+28)+3(20-42)=0 $$ Since $det(A)=0$ , $x\ne0$ and $0$ is a vector here. $$ \left[\begin{array}{rrr} 1 & -1 & 3\\ 5 & 6 & -4\\ 7 & 4 & 2\\ \end{array}\right] \left[\begin{array}{r} a\\b\\c \end{array}\right] =\left[\begin{array}{r} 0\\0\\0 \end{array}\right] $$ In row-reduced form, $$ A = \left[\begin{array}{rrr} 1 & 0 & \frac{14}{11}\\ 0 & 1 & \frac{-19}{11}\\ 0 & 0 & 0\\ \end{array}\right] $$ $$x=\frac{-14}{11}z$$ $$y=\frac{19}{11}z$$ $$ \left[\begin{array}{r} a\\b\\c \end{array}\right] =\left[\begin{array}{r} -14\\19\\11 \end{array}\right]z $$ Similarly for $2×2$ matrix .
$\endgroup$ $\begingroup$For range (T), just row reduce A to Echelon form, the remaining non-zero vectors are basis for Range space of T.
$\endgroup$ $\begingroup$To find the range(image) of T, find the transpose of the matrix first and then reduce the transposed matrix to an echelon form, the remaining non zero matrix becomes the basis for the range and the dimension becomes the rank
$\endgroup$ 2
