Question

et T : R3 —> R2 be given by : T (x1, x2, ,X3 = (x1 +x2 + X3, X2 + X3 ). Prove that T is a linear transformation. Also find the rank and nullity of T. 

Answer 1

Solution;

To prove that T is a linear transformation,we have to show that it preserves vector addition and scalar multiplication.

Let $(x_1,x_2,x_3),(y_1,y_2,y_3)$ $\epsilon R^3$

and a,b are scalar quantities such that ;

$a(x_1,x_2,x_3)+b(y_2,y_2,y_3)=(ax_1+by_1,ax_2+by_2,ax_3+by_3)$

Hence;$[a(x_1,x_2,x_3)+b(y_1,y_2,y_3)]=T(ax_1+by_1,ax_2+by_2,ax_3+by_3)$=$((ax_1+by_1+ax_2+by_2+ax_3+by_3),(ax_2+by_2+ax_3+by_3))$

$=(a(x_1+x_2+x_3)+b(y_1+y_2+y_3)),a(x_2+x_3)+b(y_2+y_3))$

$=a(x_1+x_2+x_3,x_2+x_3)+b(y_1+y_2+y_3,y_2+y_3)$

=$aT(x_1,x_2,x_3)+bT(y_1,y_2,y_3)$

Therefore T is a linear transformation.

To find Rank and nullity,

Write the coefficient matrix of the transformation,

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

The reduced row achelon form is;

$\begin{bmatrix} 1 & 0&0\\ 0 & 1&1 \end{bmatrix}$

The rank of the transformation is equal to the number of non- zero rows;

Rank(T)=2

Nullity of T can be found using the Rank Nullity theorem;

Nullity(T)=Domain(T)-Rank(T)=3-2=1