Answer to Question #274210 in Linear Algebra for Dhruv rawat

Question

Check that T = R^3 to R^3, defined by

T(x1,x2,X3)= (x1+X3, x2+2x3, x1-x2-x3) is a linear operator. Also find the kernel

Answer 1

properties of linear operator:

$T(x+y)=T(x)+T(y)$

$T(cx)=cT(x)$

we have:

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

$=(x_1+x_3+y_1+y_3, x_2+2x_3+y_2+2y_3, x_1-x_2-x_3+y_1-y_2-y_3)$

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

$=(x_1+x_3, x_2+2x_3, x_1-x_2-x_3) +(y_1+y_3, y_2+2y_3, y_1-y_2-y_3)=$

$=(x_1+x_3+y_1+y_3, x_2+2x_3+y_2+2y_3, x_1-x_2-x_3+y_1-y_2-y_3)$

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

$T(cx_1,cx_2,cx_3)=(c(x_1+x_3), c(x_2+2x_3),c( x_1-x_2-x_3))$

$cT(x_1,x_2,x_3)=c(x_1+x_3, x_2+2x_3, x_1-x_2-x_3)=$

$=(c(x_1+x_3), c(x_2+2x_3),c( x_1-x_2-x_3))$

$T(cx_1,cx_2,cx_3)=cT(x_1,x_2,x_3)$

so, $T(x_1,x_2,x_3)$ is linear operator

for kernel:

$T(x_1,x_2,x_3)=(x_1+x_3, x_2+2x_3, x_1-x_2-x_3) =0$

$x_1+x_3=0$

$x_2+2x_3=0$

$x_1-x_2-x_3=0$

$x_1=-x_3,x_2=-2x_3$

kernel:

$(x_1,x_2,x_3)=(-x_3,-2x_3,x_3)$