Question

Define the linear function f:R³→R² by f(x,y,z)=(x-z,y-x,z-y).Find

i) the kernel of f

ii)the nulllity of f

iii)the rank of f and a basis for the range of f

Answer 1

Consider the linear function $f:\R^3→\R^3,\ f(x,y,z)=(x-z,y-x,z-y).$ .

i) Let us find the kernel of $f:$

$ker f=\{(x,y,z)\in R^3,\ f(x,y,z)=(0,0,0)\}\\ =\{(x,y,z)\in R^3,\ (x-z,y-x,z-y)=(0,0,0)\}\\ =\{(x,y,z)\in R^3,\ x-z=0,y-x=0,z-y=0\}\\ =\{(x,y,z)\in R^3,\ x=y=z\}\\ =\{(x,x,x),\ x\in \R\}\\ =\{x(1,1,1),\ x\in \R\}$

ii) It follows that the vector $(1,1,1)$ form a basis of the kernel, and hence the nulllity of $f$ is 1.

iii) Let the vector $(a,b,c)$ belongs to the range of $f.$ Then $f(x,y,z)=(x-z,y-x,z-y)=(a,b,c)$ for some $(x,y,z)\in\R^3.$ It follows that $x-z=a,y-x=b,z-y=c.$ Then $(x-z)+(y-x)=a+b,z-y=c,$ that is $-z+y=a+b,z-y=c.$ We conclude that $a+b=-c.$ Since $(a,b,c)=(a,b,-(a+b))=a(1,0,-1)+b(0,1,-1),$ and by the rank-nullity theorem $rank f=3-1=2,$ we conclude that $\{(1,0,-1),(0,1,-1)\}$ is a basis of for the range $f$.