Definition

We got a glimpse of linear transformations when we looked at the correspondence between elementary row operations and elementary matrices. We now look at linear transformations more generally.

Let $V$ and $W$ be vector spaces with scalars coming from the same field $\mathbb{F}$. A mapping $T:V\to W$ is a linear transformation if for any two vectors $x$ and $y$ in $V$ and any scalar $\alpha \in \mathbb{F}$, the following are satisfied: $$\begin{array}{rcl}T(x+y)& =& T(x)+T(y)\\ T(\alpha x)& =& \alpha T(x)\end{array}$$

A couple of consequences of this definition are:

1. $T({\alpha }_1{x}_1+\cdots +{\alpha }_m{x}_m)={\alpha }_{1}T(x_1)+\cdots +{\alpha }_{m}T(x_m)$ for any vectors $x_1,\dots ,x_m\in V$ and scalars ${\alpha }_1,\dots ,{\alpha }_m\in \mathbb{F}$.

2. $T(0_V)=0_W$ where $0_V$ denotes the zero vector in $V$ and $0_W$ denotes the zero vector in $W$.

Examples

1. Let $A\in {\mathbb{F}}^{m\times n}$. Then $T(x)=Ax$ is a linear transformation from ${\mathbb{F}}^n$ to ${\mathbb{F}}^m$. To see this, we need to check the two conditions. Let $x,y\in {\mathbb{F}}^n$ and $\alpha \in \mathbb{F}$. Then $T(x+y)=A(x+y)=Ax+Ay=T(x)+T(y)$ and $T(\alpha x)=A(\alpha x)=\alpha (Ax)=\alpha T(x)$.

2. Let $T$ be a mapping from $P_2$, the vector space of polynomials in $x$ with real coefficients having degree at most $2$, to ${\mathbb{R}}^2$ given by $T(a{x}^2+bx+c)=\left[\begin{array}{c}a+3c\\ a\end{array}\right]$. For example, $T(2x^2+1)=\left[\begin{array}{c}2+3(1)\\ 2\end{array}\right]=\left[\begin{array}{c}5\\ 2\end{array}\right]$ since $2x^2+1$ can be written as $2x^2+0x+1$.

   To check that $T$ is a linear transformation, take two vectors $a_1{x}^2+b_{1}x+c_1$ and $a_2{x}^2+b_{2}x+c_2$ from $P_2$ and a real number $\gamma$. Then, $$\begin{array}{rcl}& & T((a_1{x}^2+b_{1}x+c_1)+(a_2{x}^2+b_2+c_2))\\ & =& T((a_1+a_2)x^2+(b_1+b_2)x+(c_1+c_2))\\ & =& \left[\begin{array}{c}(a_1+a_2)+3(c_1+c_2)\\ a_1+a_2\end{array}\right]\\ & =& \left[\begin{array}{c}(a_1+3c_1)+(a_2+3c_2)\\ a_1+a_2\end{array}\right]\\ & =& \left[\begin{array}{c}{a}_1+3c_1\\ a_1\end{array}\right]+\left[\begin{array}{c}{a}_2+3c_2\\ a_2\end{array}\right]\\ & =& T(a_1{x}^2+b_{1}x+c_1)+T(a_2{x}^2+b_{2}x+c_2),\end{array}$$ and $$\begin{array}{rcl}& & T(\gamma (a_1{x}^2+b_{1}x+c_1))\\ & =& T((\gamma a_1)x^2+(\gamma b_1)x+\gamma c_1)\\ & =& \left[\begin{array}{c}\gamma a_1+3(\gamma c_1)\\ \gamma a_1\end{array}\right]\\ & =& \gamma \left[\begin{array}{c}{a}_1+3(c_1)\\ a_1\end{array}\right]\\ & =& \gamma T(a_1{x}^2+b_{1}x+c_1).\end{array}$$

Quick Quiz

Exercises

1. Let $T:{\mathbb{R}}^5\to \mathbb{R}$ be such that $T(x)=0$ for all $x\in {\mathbb{R}}^5$. Is $T$ a linear transformation? Explain your answer.  

   Yes.

2. Let $T:{\mathbb{R}}^3\to {\mathbb{R}}^2$ be a linear transformation such that $T\left(\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]\right)=\left[\begin{array}{c}1\\ -4\end{array}\right]$ and $T\left(\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]\right)=\left[\begin{array}{c}0\\ 2\end{array}\right]$. What is $T\left(\left[\begin{array}{c}4\\ -1\\ 5\end{array}\right]\right)$?  

   Since $\left[\begin{array}{c}4\\ -1\\ 5\end{array}\right]=3\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right](-1)\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right],$ $T\left(\left[\begin{array}{c}4\\ -1\\ 5\end{array}\right]\right)=3T\left(\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]\right)-T\left(\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]\right)=\left[\begin{array}{c}3\\ -14\end{array}\right]$.