Answer to Question #298594 in Linear Algebra for Hamis

Question

Show that T(x1, x2, x3, x4) = 3x1 −7x2 + 5x4 is a linear transformation by finding the

matrix for the transformation. Then find a basis for the null space of the transforma￾tion.

Answer 1

$\forall \vec{a}=(x_1,x_2,x_3,x_4)\\ T(\vec{a})=3x_1-7x_2+5x_4\\ 1. \forall \vec{a}=(x_1,x_2,x_3,x_4), \vec{b}=(y_1,y_2,y_3,y_4)\\ T(\vec{a}+\vec{b})=T(\vec{a})+T(\vec{b})\\ \vec{a}+\vec{b}=(x_1,x_2,x_3,x_4)+(y_1,y_2,y_3,y_4)=\\ =(x_1+y_1,x_2+y_2,x_3+y_3,x_4+y_4)\\ T(\vec{a}+\vec{b})=3(x_1+y_1)-7(x_2+y_2)+5(x_4+y_4)\\ T(\vec{a})=3x_1-7x_2+5x_4\\ T(\vec{b})=3y_1-7y_2+5y_4\\ T(\vec{a})+T(\vec{a})=3x_1-7x_2+5x_4+3y_1-7y_2+5y_4$

$2. \forall \alpha\in R, \forall \vec{a}=(x_1,x_2,x_3,x_4)\\ T(\alpha\vec{a})=\alpha T(\vec{a})\\ \alpha\vec{a}=(\alpha x_1,\alpha x_2,\alpha x_3,\alpha x_4)\\ T(\alpha\vec{a})=3\alpha x_1-7\alpha x_2+5\alpha x_4=\\ =\alpha(3x_1-7x_2+5x_4)=\alpha T(\vec{a})$

$T$ is a linear transformation.

Basis:

$\vec{e_1}=(1,0,0,0)\\ \vec{e_2}=(0,1,0,0)\\ \vec{e_3}=(0,0,1,0)\\ \vec{e_4}=(0,0,0,1)\\$

$T(\vec{e_1})=3x_1= 3\vec{e_1}+0\vec{e_2}+0\vec{e_3}+0\vec{e_4}\\ T(\vec{e_2})=-7x_2= 0\vec{e_1}-7\vec{e_2}+0\vec{e_3}+0\vec{e_4}\\ T(\vec{e_3})=0= 0\vec{e_1}+0\vec{e_2}+0\vec{e_3}+0\vec{e_4}\\ T(\vec{e_4})=5x_4= 0\vec{e_1}+0\vec{e_2}+0\vec{e_3}+5\vec{e_4}\\ T=\begin{pmatrix} 3 & 0&0&0 \\ 0 & -7&0&0 \\ 0 & 0&0&0 \\ 0 & 0&0&5 \end{pmatrix}$

$\forall \vec{a}=(x_1,x_2,x_3,x_4)\\ T(\vec{a})=3x_1-7x_2+5x_4=\vec{0}\\ 3x_1=0, x_1=0\\ -7x_2=0, x_2=0\\ 5x_4=0, x_4=0\\ \forall x_3\\ (0,0,x_3,0), x_3\in R$