Answer to Question #111480 in Linear Algebra for Priya jaiswal

Question

Define T:R^3 -R^3 by
T(x1,x2,x3) = (x1+x2,x2+x3,x1-x3,2x1+x2-x3)
Check that T is a linear operator . Find the kernal and range of T. Find the dimension of the kernal.

Answer 1

To show, that T is a linear operator let's check the following property:

$T(\lambda_1 v_1 + \lambda_2 v_2) = \lambda_1T(v_1) + \lambda_2 T(v_2)$, where $v_1 = (x_1, y_1, z_1), v_2 = (x_2, y_2, z_2)$. Indeed:

$T(\lambda_1 v_1 + \lambda_2 v_2) = T((\lambda_1 x_1, \lambda_1 y_1, \lambda_1 z_1)+ (\lambda_2 x_2, \lambda_2 y_2, \lambda_2 z_2))=$

$= T(\lambda_1 x_1 + \lambda_2 x_2, \lambda_1 y_1 + \lambda_2 y_2, \lambda_1 z_1 + \lambda_2z_2)$ $= (\lambda_1(x_1 + y_1) + \lambda_2(x_2 + y_2), \lambda_1(y_1 + z_1) + \lambda_2(y_2 + z_2), \lambda_1(x_1 -z_1) + \lambda_2(x_2 - z_2), \lambda_1(2x_1 + y_1 - z_1) + \lambda_2(2x_2 + y_2 - z_2)) =$

$= \lambda_1(x_1 + y_1, y_1 + z_1, x_1 - z_1, 2x_1 + y_1 - z_1) + \lambda_2((x_2 + y_2, y_2 + z_2, x_2 - z_2, 2x_2 + y_2 - z_2) = \lambda_1 T(v_1) + \lambda_2T(v_2)$

Let's now find the kernal and range of T.

$Ker(T) = v \in R^3 | T(v) = 0$ , so:

$T(v) = T(x, y, z) = (x+y, y+z, x-z, 2x+y-z) = 0.$ This means, that:

$\begin{cases} x+y = 0 \\ y+z = 0 \\ x-z = 0\\ 2x+y-z = 0 \end{cases}$ Thus $\begin{cases} x+y=0\\ z= x \\ \end{cases}$ . Hence, $Ker (T) = Lin (1,-1,1), dim(Ker (T)) = 1.$

Now let's do the same for the range:

$T(v) = T(x, y, z) = (x+y, y+z, x-z, 2x+y-z) =$

$= x(1, 0, 1, 2) + y(1,1,0,1)+z(0,1,-1,-1).$ But, since $(0,1,-1,-1) = -(1,0,1,2) + (1,1,0,1),$ we obtain that: $T(x,y,z) = (x-z)(1,0,1,2) + (y+z)(1,1,0,1),$ and:

$Range(T) = Lin((1,0,1,2),(1,1,0,1)).$