Question

Show that the map : T : R4 -3 R2 given by

T(x1 , x2, x3, x4) = (2x1 + x3, 2x3 + x1) is a linear transformation. Find its image and the kernel.

Answer 1

Let us show that the map $T : \mathbb R^4\to\mathbb R^2$ given by $T(x_1, x_2, x_3, x_4) = (2x_1 + x_3, 2x_3 + x_1)$ is a linear transformation:

$T(a(x_1, x_2, x_3, x_4)+b(y_1, y_2, y_3, y_4)) =T(ax_1+by_1, ax_2+by_2, ax_3+by_3, ax_4+by_4) = (2(ax_1+by_1) + ax_3+by_3, 2(ax_3+by_3) + ax_1+by_1)=(2ax_1+2by_1 + ax_3+by_3, 2ax_3+2by_3 + ax_1+by_1)= (2ax_1+ ax_3, 2ax_3+ax_1)+(2by_1+by_3, +2by_3 +by_1)=a(2x_1+ x_3, 2x_3+x_1)+b(2y_1+y_3, +2y_3 +y_1)=aT(x_1, x_2, x_3, x_4)+bT(y_1, y_2, y_3, y_4).$

Let us find its kernel:

$\ker(T)=\{ (x_1, x_2, x_3, x_4)\in\mathbb R^4\ |\ T(x_1, x_2, x_3, x_4)=(0,0) \} =$

$=\{ (x_1, x_2, x_3, x_4)\in\mathbb R^4\ |\ (2x_1 + x_3, 2x_3 + x_1)=(0,0) \} =$

$=\{ (x_1, x_2, x_3, x_4)\in\mathbb R^4\ |\ 2x_1 + x_3=0\text{ and }2x_3 + x_1=0 \} =$

$=\{ (x_1, x_2, x_3, x_4)\in\mathbb R^4\ |\ x_3= -2x_1\text{ and }-3x_1 =0 \} =$

$=\{ (x_1, x_2, x_3, x_4)\in\mathbb R^4\ |\ x_1=0\text{ and }x_3 =0 \} =$

$=\{ (0, x_2, 0, x_4)\ |\ x_2,x_4\in\mathbb R\}$.

Let us find its image. For this it is sufficient to find all $(a,b)\in\mathbb R^2$ such that $T(x_1, x_2, x_3, x_4) =(a,b)$, that is the system

$\begin{cases}2x_1 + x_3=a\\ x_1 + 2x_3=b\end{cases}$ has a solution.

Since $\left|\begin{array}{cc}2 & 1\\1 &2\end{array}\right|=4-1=3\ne 0$, this system has a solution for all all $(a,b)\in\mathbb R^2$

Therefore, T is a surjection, that is $Im(T)=\mathbb R^2.$