Answer to Question #122109 in Linear Algebra for Javeria

Question

Qs: 04 Let T(x, y, z, w) = ( 3y -4z, x +2y+4z-w, 7z, -y-w)

Write the standard matrix for T. Check if T is one-to-one and onto.

Answer 1

$T(x, y, z, w) = ( 3y -4z, x +2y+4z-w, 7z, -y-w)$

Write the standard matrix for T.

$$A=\begin{bmatrix} 0 & 3 & -4 & 0\\ 1 & 2 & 4 & -1 \\ 0 & 0 & 7 & 0\\ 0 & -1 & 0 & -1 \end{bmatrix}$$

One-to-one is the same as onto for square matrices.

In general, a transformation T is both one-to-one and onto if and only if $T(\bf x )=\bf b$has exactly one solution for all $\bf b$ in $\Bbb{R}^m.$

$$\begin{bmatrix} 0 & 3 & -4 & 0 & &0\\ 1 & 2 & 4 & -1 & &0 \\ 0 & 0 & 7 & 0 & &0\\ 0 & -1 & 0 & -1 & &0 \end{bmatrix}$$

Swap rows 1 and 2:

$$\begin{bmatrix} 1 & 2 & 4 &-1& &0\\ 0 & 3 & -4 & 0 & &0 \\ 0 & 0 & 7 & 0 & &0\\ 0 & -1 & 0 & -1 & &0 \end{bmatrix}$$

$R_2={R_2/3}$

$$\begin{bmatrix} 1 & 2 & 4 &-1& &0\\ 0 & 1 & -4/3 & 0 & &0 \\ 0 & 0 & 7 & 0 & &0\\ 0 & -1 & 0 & -1 & &0 \end{bmatrix}$$

$R_1=R_1-(2)R_2$

$$\begin{bmatrix} 1 & 0 & 20/3 &-1& &0\\ 0 & 1 & -4/3 & 0 & &0 \\ 0 & 0 & 7 & 0 & &0\\ 0 & -1 & 0 & -1 & &0 \end{bmatrix}$$

$R_4=R_4+R_2$

$$\begin{bmatrix} 1 & 0 & 20/3 &-1& &0\\ 0 & 1 & -4/3 & 0 & &0 \\ 0 & 0 & 7 & 0 & &0\\ 0 & 0 & -4/3 & -1 & &0 \end{bmatrix}$$

$R_3=R_3/7$

$$\begin{bmatrix} 1 & 0 & 20/3 &-1& &0\\ 0 & 1 & -4/3 & 0 & &0 \\ 0 & 0 & 1 & 0 & &0\\ 0 & 0 & -4/3 & -1 & &0 \end{bmatrix}$$

$R_1=R_1-(20/3)R_3$

$$\begin{bmatrix} 1 & 0 & 0 &-1& &0\\ 0 & 1 & -4/3 & 0 & &0 \\ 0 & 0 & 1 & 0 & &0\\ 0 & 0 & -4/3 & -1 & &0 \end{bmatrix}$$

$R_2=R_2+(4/3)R_3$

$$\begin{bmatrix} 1 & 0 & 0 &-1& &0\\ 0 & 1 & 0 & 0 & &0 \\ 0 & 0 & 1 & 0 & &0\\ 0 & 0 & -4/3 & -1 & &0 \end{bmatrix}$$

$R_4=R_4+(4/3)R_3$

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

$R_1=R_1-R_4$

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

$R_4=R_4\cdot(-1)$

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

The columns of matrix are linearly independent, which happens precisely when the matrix has a pivot position in every column. 

Therefore $T$ is one-to-one.