Define the linear transformation $T$ by $T(x) = Ax$, where
$A=\left(\begin{matrix} \frac{9}{10} & \frac{3}{10}\\ \frac{3}{10} & \frac{1}{10} \end{matrix}\right)$.
Find (a) $\ker(T)$, (b) $\text{nullity}(T)$, (c) $\text{range}(T)$ and (d) $\text{rk}(T)$.
Apparantly, $\ker(T) = \{(t,-3t) : t \in\mathbb{R}\}$ and $\text{range}(T)= \{(3t,t) :t \in\mathbb{R}\}$. Can anyone explain how to get those answers?
How would this change if we use the following matrix instead of $A$?
$B=\left(\begin{matrix}5 & -1\\ 1 &1 \\ 1 & -1\end{matrix}\right)$;
($\text{range}(T) = \{(4s,4t,s-t) :s,t \in\mathbb{R}\}$)
Best Answer
Let's look at $A=\left(\begin{matrix} \frac{9}{10} & \frac{3}{10}\\ \frac{3}{10} & \frac{1}{10} \end{matrix}\right)$. If we envision multiplying a vector $(v_1, v_2)$ by $A$, then we see that we get $(\frac{9}{10} v_1+\frac{3}{10}v_2, \frac{3}{10}v_1+\frac{1}{10}v_2)$. Now, if $(v_1, v_2)$ is in the kernel of $A$ precisely when $A$ multiplied by $(v_1, v_2)$ is the zero vector. So this is true when $\frac{9}{10} v_1+\frac{3}{10}v_2=0$ and so $3v_1+v_2=0$. The vectors that satisfy this are precisely those of the form $\{(t, -3t): t \in \mathbb{R} \}$. Notice also that any vector $v$ of this form will satisfy $Av=(0,0)$ and so this set of vectors is the kernel.
The range of $A$ consists of the vectors $v$ so that there exists some vector $w$ with $Aw=v$. Try using a similar approach to figure out the form of any vector in the range.