Matrices and images

Question

Let T: $\mathbb{R}^4 \to \mathbb{R}^5$ defined by

$T (x_1, x_2, x_3, x_4) = (x_4, x_1, x_3, x_2, x_1 - x_3)$

A) Find the standard matrix for the linear transformation $T$

B) Find the image of $x = (1, 2, 1, 2)$ by $T$

Can anyone help me out with this question. I have been through my notes but can not understand how to do it. Hopefully, a detailed solution would be the best so I can get a better understanding.

Answer 1

First, it is important to see how matrices act on vectors. In our case, we are transforming vectors from $\mathbb{R}^4$ into vectors in $\mathbb{R}^5$ (i.e. T will transform a vector $x = (x_1,x_2,x_3,x_4)$ into $b =(b_1,b_2,b_3,b_4,b_5)$. If we write a general transformation T: \begin{equation} \begin{bmatrix} t_{11} & t_{12} & t_{13} & t_{14}\\ t_{21} & t_{22} & t_{23} & t_{24}\\ t_{31} & t_{32} & t_{33} & t_{34}\\ t_{41} & t_{42} & t_{43} & t_{44}\\ t_{51} & t_{52} & t_{53} & t_{54}\\ \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ \end{bmatrix}= \begin{bmatrix} b_{1} \\ b_{2} \\ b_{3} \\ b_{4} \\ b_{5} \\ \end{bmatrix} \end{equation}

where

\begin{cases} b_1 = t_{11} x_1 + t_{12} x_2 + t_{13} x_3 + t_{14} x_4\\ b_2 = t_{21} x_1 + t_{22} x_2 + t_{23} x_3 + t_{24} x_4\\ b_3 = t_{31} x_1 + t_{32} x_2 + t_{33} x_3 + t_{34} x_4\\ b_4 = t_{41} x_1 + t_{42} x_2 + t_{43} x_3 + t_{44} x_4\\ b_5 = t_{51} x_1 + t_{52} x_2 + t_{53} x_3 + t_{54} x_4 \end{cases}

Back to our exercise:

\begin{cases} b_1 & = x_4&= 0 \cdot x_1 &+ 0 \cdot x_2 &+ 0 \cdot x_3 &+ 1 \cdot x_4 \\ b_2 &=x_1&= 1 \cdot x_1 &+ 0\cdot x_2 &+ 0 \cdot x_3 &+ 0 \cdot x_4 \\ b_3 & = x_3&= 0 \cdot x_1 &+ 0 \cdot x_2 &+1 \cdot x_3 &+ 0 \cdot x_4\\ b_4 & = x_2&= 0 \cdot x_1 &+ 1 \cdot x_2 &+0 \cdot x_3 &+ 0 \cdot x_4\\ b_5 & = x_1-x_3&= 1 \cdot x_1 &+ 0 \cdot &+ (-1) \cdot x_3 &+ 0 \cdot x_4 \end{cases}

Then,

\begin{equation} T= \begin{bmatrix} 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & -1 & 0\\ \end{bmatrix} \end{equation}

For the image of $x = (1,2,1,2)$ just compute:

\begin{equation} T (1,2,1,2) = \begin{bmatrix} 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & -1 & 0\\ \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 1 \\ 2 \\ \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 1 \\ 2 \\ 0 \\ \end{bmatrix} \end{equation}

Answer 2

Each component of the right hand side is a dot product of $(x_1,x_2,x_3,x_4)$ with the first row of the matrix. So the first row $a$ is the solution to $(a_1,a_2,a_3,a_4) \cdot (x_1,x_2,x_3,x_4) = x_4$.