Answer to Question #218715 in Linear Algebra for Priyu

Question #218715

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. 


1
Expert's answer
2021-07-20T17:54:38-0400

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}\n 1\\\\\n 1\\\\\n1\n\\end{bmatrix}" ="\\begin{bmatrix}\n 1&1& 1\\\\\n 0 & 1&1\n\\end{bmatrix}"

The reduced row achelon form is;

"\\begin{bmatrix}\n 1 & 0&0\\\\\n 0 & 1&1\n\\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



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