Question

Find the range space and the kernel of the
linear transformation :
T: R4 --> R4, T(x1, x2, x3, x4) =
(x1+ x2+ x3+ x4, x1+ x2, x3+ x4, 0)

Answer 1

$T: \R^4\to \R^4,$

$T(x_1,x_2,x_3,x_4)=(x_1+x_2+x_3+x_4,x_1+x_2,x_3+x_4,0).$ (1)

For $T\in L(\R^4,\R^4),$ the range of $T$ is the subset of $\R^4$ consisting of those vectors that are of the form $T(x_1,x_2,x_3,x_4)$ for some $(x_1,x_2,x_3,x_4)\in \R^4:$

$range T=\{ T(x_1,x_2,x_3,x_4): (x_1,x_2,x_3,x_4)\in \R^4\}.$

So, $T\in L(\R^4,\R^4)$ is defined by (1), then

$range T=\{(x_1+x_2+x_3+x_4,x_1+x_2,x_3+x_4,0): x_1,x_2,x_3,x_4\in\R\}.$

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

is  the matrix satisfying $T(x)=Ax, x\in\R^4.$ The range of the linear transformation $T$ is the same as the range of the matrix $A.$

The range of $A$ is the columns space of $A$ . Thus it is spanned by columns

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

From the below (2) reduction of the augmented matrix, we see that these vectors are linearly independent, thus a basis for the range.

Therefore, we have

$range T= range A=span\{\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix},\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}\}$

and

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

is a basis for $range T.$

$\ker T=\{(x_1,x_2,x_3,x_4)\in\R^4: T(x_1,x_2,x_3,x_4)=0\}.$

System of LE associated to the implicit equations of the kernel, resulting from equalling to zero the components of the linear transformation formula.

$x_1+x_2+x_3+x_4=0$

$x_1+x_2=0$

$x_3+x_4=0$

Transform the augmented matrix $A$ to row echelon form.

$r_2\to r_2-r_1$

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

$r_3\to r_3+r_2$

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

$r_1\to r_1+r_2$

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

$r_2\to -1r_2$

$\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$ (2)

Translate the row echelon form matrix to the associated system of linear equations

$x_1+x_2=0$

$x_3+x_4=0$

The implicit equations of the kernel are the equations above

$\ker T=\{(x_1,x_2,x_3,x_4)\in\R^4 | x_1+x_2=0; x_3+x_4=0 \}$

$\ker T =\begin{bmatrix} s \\ -s \\ t \\ -t \end{bmatrix}: t,s\in\R,$

$\ker T=span\{\begin{bmatrix} 1 \\ -1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \\ -1 \end{bmatrix}\}$

and

$\{\begin{bmatrix} 1 \\ -1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \\ -1 \end{bmatrix}\}$ is a basis for $\ker T.$