# How do you show whether a transformation is linear or not?

Consider the transformation $T$ from $\mathbb{R^2}$ to $\mathbb{R^3}$ given by:

$$T \left( \begin{array}{ccc} x_1 \\ x_2 \\ \end{array} \right) = x_1 \left( \begin{array}{ccc} 1 \\ 2 \\ 3 \end{array} \right) + x_2\left( \begin{array}{ccc} 4 \\ 5 \\ 6 \end{array} \right)$$

How do we determine whether the transformation is linear, and subsequently find its matrix?

I understand that for a transformation to be linear, we need to show $T(\alpha x+ \beta y) = \alpha T(x)+\beta T(y)$. Can someone provide an instance where this would not be the case, so I can understand how this "proves" linearity?

• If it is too difficult for you to do it, try to separate the proof into two parts, namely 1) $T(x+y)=T(x)+T(y)$ and 2) $T(\alpha x)=\alpha T(x)$, for all $\alpha \in \mathbb{R}$. – Nighty Oct 11 '14 at 6:08

By definition, a transformation is linear if it is closed under vector addition and scalar multiplication. This means that for any two vectors $x$ and $y$, and any scalar $\alpha$, $$T(x+\alpha y) = T(x) + \alpha T(y).$$ A simple example of a nonlinear transformation is $T(x) = x^2.$

In this problem, you want to look for a $3\times2$ matrix that, when left-multiplied with $\begin{pmatrix} x_1\\x_2\end{pmatrix}$, will give you $$x_1\begin{pmatrix} 1\\2\\3\end{pmatrix}+x_2\begin{pmatrix} 4\\5\\6\end{pmatrix}.$$

It may be easier to see if you combine the terms on the right-hand side, $\begin{pmatrix}x_1+4x_2\\2x_1+5x_2\\3x_1+6x_2\end{pmatrix}$.

It should be clear at this point that the transformation matrix $T = \begin{pmatrix}1&4\\2&5\\3&6\end{pmatrix}$.

Just check by definition, let $\alpha, \beta \in \mathbb{R}$, $x=(x_1,x_2),y=(y_1,y_2)\in \mathbb{R^2}$, then $\alpha x+ \beta y=(\alpha x_1+\beta y_1,\alpha x_2+\beta y_2),$

$$T \left( \begin{array}{ccc} \alpha x_1+\beta y_1 \\ \alpha x_2+\beta y_2 \\ \end{array} \right) = (\alpha x_1+\beta y_1) \left( \begin{array}{ccc} 1 \\ 2 \\ 3 \end{array} \right) + (\alpha x_2+\beta y_2)\left( \begin{array}{ccc} 4 \\ 5 \\ 6 \end{array} \right)$$ Then can you rearrange the terms so that the RHS=$\alpha T\left(\begin{array}{ccc} x_1\\x_2\end{array}\right)+\beta T\left(\begin{array}{ccc} y_1\\y_2\end{array}\right)$ Hence it proves the linearity by definition.

• Can you provide an example of a transformation that is not linear? At what point in our algebraic manipulation would we find that it was impossible to arrange the equation in that? – Joel B Oct 11 '14 at 6:25
• The transformation given in your form is always linear. For a simple nonlinear transformation, $T:\mathbb{R}\to \mathbb{R}$ given by $T(x)=x^2$. Then $T(-1)=T(1)=1$,but $T(1+(-1))=0 \neq T(-1)+T(1)$. – John Oct 11 '14 at 6:33
• @JohnBayley Please see above comment. – John Oct 11 '14 at 6:36