Question

Let T:R^3--->R^4 be defined by T(x1,x2,x3)=(x1+x2, x2+x3, x1-x3, 2x1+x2-x3). Check T is operator. Find kernel and range of T. Find dimensions of kernel.

Answer 1

First we prove that T is a linear operator:

$T(\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}+ \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}) = T\begin{pmatrix} x_1 + y_1 \\ x_2 + y_2 \\ x_3 + y_3 \end{pmatrix} = \begin{pmatrix} x_1 + y_1 + x_2 + y_2\\ x_2 + y_2 + x_3 + y_3\\ x_1 + y_1 - x_3 - y_3 \\ 2x_1 + 2y_1 + x_2 + y_2 - x_3 - y_3 \end{pmatrix} = \\ = \begin{pmatrix} x_1 + x_2 \\ x_2 + x_3 \\ x_1 - x_3 \\ 2x_1 + x_2 - x_3 \end{pmatrix} + \begin{pmatrix} y_1 + y_2\\ y_2 + y_3\\ y_1 - y_3 \\ 2y_1 + y_2 - y_3 \end{pmatrix} = T\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}+ T\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} \\ T(\alpha\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}) = T(\begin{pmatrix} \alpha x_1 \\ \alpha x_2 \\ \alpha x_3 \end{pmatrix}) = \begin{pmatrix} \alpha x_1 + \alpha x_2 \\ \alpha x_2 + \alpha x_3 \\ \alpha x_1 - \alpha x_3 \\ 2\alpha x_1 + \alpha x_2 - \alpha x_3 \end{pmatrix} = \begin{pmatrix} \alpha(x_1 + x_2) \\ \alpha(x_2 + x_3) \\ \alpha(x_1 - x_3) \\ \alpha(2x_1 + x_2 - x_3) \end{pmatrix} = \\ = \alpha\begin{pmatrix} x_1 + x_2 \\ x_2 + x_3 \\ x_1 - x_3 \\ 2x_1 + x_2 - x_3 \end{pmatrix} = \alpha T\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \\$

Then we find the kernel:

$\begin{pmatrix} x_1 + x_2 \\ x_2 + x_3 \\ x_1 - x_3 \\ 2x_1 + x_2 - x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \\ \begin{pmatrix} x_1 + x_2 \\ x_2 + x_3 \\ x_1 - x_3 \\ 2x_1 + x_2 - x_3 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & -1 \\ 2 & 1 & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \\ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & -1 \\ 2 & 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 1 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \\ x_2 = -x_3 \\ x_1 = -x_2 = x_3$

So the kernel is all vectors given as$\begin{pmatrix} x_3\\ -x_3 \\ x_3 \end{pmatrix}$ . Its dimension is 1, its basis is $\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ .

Finally we find the range of T:

$\begin{pmatrix} x_1 + x_2 \\ x_2 + x_3 \\ x_1 - x_3 \\ 2x_1 + x_2 - x_3 \end{pmatrix} = \begin{pmatrix} x_1 \\ 0 \\ x_1 \\ 2x_1 \end{pmatrix} + \begin{pmatrix} x_2 \\ x_2 \\ 0 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0 \\ x_3 \\ -x_3 \\ -x_3 \end{pmatrix}$

Check if the vectors are linearly independent:

$\begin{pmatrix} 1 & 0 & 1 & 2 \\ 1 & 1 & 0 & 1 \\ 0 & 1 & -1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 1 & 2 \\ 0 & 1 & -1 & -1 \\ 0 & 1 & -1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 1 & 2 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

So its basis is $\begin{pmatrix} 1 \\ 0 \\ 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ -1 \\ -1 \end{pmatrix}$ .