Answer to Question #160247 in Linear Algebra for Nikhil

Question

Let T: R^3→R^2 be given by:

T(x2,x2,x3)= (X1+x2+x3,x2+x3)

Prove that T is a linear transformation. Also find the rank and nullity of T.

Answer 1

T: R³→R² be given by:

T(x₁,x₂,x₃)=(x₁+x₂+x₃,x₂+x₃)

$\implies$ First, we need to check T is a linear transformation.

let (x₁,x₂,x₃), (y₁,y₂,y₃) $\in$ R³ and $\alpha$ $\isin$ R

T($\alpha$ (x₁,x₂,x₃) + (y₁,y₂,y₃)) = T(($\alpha$x₁,$\alpha$x₂,$\alpha$x₃) + (y₁,y₂,y₃))

= T( $\alpha$x₁+y₁, $\alpha$x₂+y₂, $\alpha$x₃+y₃)

= ($\alpha$x₁+y₁+$\alpha$x₂+y₂+$\alpha$x₃+y₃ , $\alpha$x₂+y₂+$\alpha$x₃+y₃)

= ($\alpha$(x₁+x₂+x₃) + (y₁+y₂+y₃), $\alpha$(x₂+x₃)+(y₂+y₃))

= $\alpha$(x₁+x₂+x₃,x₂+x₃) + (y₁+y₂+y₃,y₂+y₃)

= $\alpha$T(x₁,x₂,x₃) + T(y₁,y₂,y₃)

since,it is satisfying the definition of linear transformation.

Hence,

T is a linear transformation.

$\implies$ Now we need to find the rank and nullity of T.

as given;

T(x₁,x₂,x₃)=(x₁+x₂+x₃,x₂+x₃)

here first to find the matrix corresponding to this linear transformation,

let $\beta$ ={(1,0,0),(0,1,0),(0,0,1)}

T(1,0,0) = (1,0) = 1(1,0) + 0(0,1)

T(0,1,0) = (1,1) = 1(1,0) + 1(0,1)

T(0,0,1) = (1,1) = 1(1,0) + 1(0,1)

[T] =$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}$

since matrix have two non-zero rows. therefore rank of a matrix is 2.

$\implies$ rank (T)= 2

using rank nullity theorem which states that rank(T) + nullity(T) = n;

rank(T) + nulity(T) = 3 $\because$ n = 3

nulity(T) = 3 - rank(T) = 3 - 2 =1

$\implies$ nulity(T) = 1