Answer to Question #278547 in Linear Algebra for Tharun

Question

Verify Rank-nullity theorem for the linear transformation T:R³----->R³ defined by T(x,y,z)=(x-y, 2y+z, x+y+z).

Answer 1

Let us verify Rank-nullity theorem for the linear transformation $T:\R^3\to \R^3$ defined by $T(x,y,z)=(x-y, 2y+z, x+y+z).$

For this let us find the kernal and the range of $T.$

$Ker(T)=\{(x,y,z)\in\R:T(x,y,z)=(0,0,0)\} \\=\{(x,y,z)\in\R:(x-y, 2y+z, x+y+z)=(0,0,0)\} \\=\{(x,y,z)\in\R:x-y=0, 2y+z=0, x+y+z=0\} \\=\{(x,y,z)\in\R:x=y, z=-2y, x+y+z=0\} \\=\{(y,y,-2y)\in\R:y\in\R\} \\=\{y(1,1,-2)\in\R:y\in\R\}.$

It follows that $\dim(Ker(T))=1.$

Let $(a,b,c)\in Range(T).$ Then $T(x,y,z)=(a,b,c)$ for some $(x,y,z)\in\R.$

Therefore, $x-y=a, 2y+z=b, x+y+z=c.$

It follows that $x-y=(x+y+z)-(2y+z)=c-b,$ and hence $a=c-b.$

Then

$Range(T)=\{(a,b,c)\in\R: T(x,y,z)=(a,b,c)\} \\=\{(c-b,b,c):b,c\in\R\} \\=\{c(1,0,1)+b(-1,1,0):b,c\in\R\}.$

Consequently, $\dim(range(T))=2.$

Since $\dim(Ker(T))+\dim(range(T))=1+2=3=\dim(\R^3),$ we conclude that the Rank-nullity theorem is indeed true.