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 \(T:V\rightarrow W\) 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{eqnarray} T(x+y) & = & T(x) + T(y) \\ T(\alpha x) & = & \alpha T(x) \end{eqnarray}

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 \(x_1,\ldots, x_m \in V\) and scalars \(\alpha_1,\ldots,\alpha_m \in \mathbb{F}\).

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 \(T(ax^2 + bx + c) = \begin{bmatrix} a+3c \\ a\end{bmatrix}\). For example, \(T(2x^2 + 1) = \begin{bmatrix} 2 + 3(1) \\ 2\end{bmatrix} =\begin{bmatrix} 5 \\2 \end{bmatrix}\) since \(2x^2 + 1\) can be written as \(2 x^2 + 0x + 1\).

   To check that \(T\) is a linear transformation, take two vectors \(a_1x^2+b_1x+c_1\) and \(a_2x^2+b_2x+c_2\) from \(P_2\) and a real number \(\gamma\). Then, \begin{eqnarray} & & T( (a_1x^2+b_1x+c_1) + (a_2x^2+b_2+c_2) ) \\ & = & T( (a_1+a_2)x^2+(b_1+b_2)x+(c_1+c_2) ) \\ & = & \begin{bmatrix} (a_1+a_2)+3(c_1+c_2) \\ a_1+a_2 \end{bmatrix} \\ & = & \begin{bmatrix} (a_1+3c_1)+(a_2+3c_2) \\ a_1+a_2 \end{bmatrix} \\ & = & \begin{bmatrix} a_1+3c_1 \\ a_1 \end{bmatrix} + \begin{bmatrix} a_2+3c_2 \\ a_2 \end{bmatrix} \\ & = & T(a_1x^2+b_1x+c_1) + T(a_2x^2+b_2 x+c_2), \end{eqnarray} and \begin{eqnarray} & & T( \gamma(a_1x^2+b_1x+c_1) ) \\ & = & T( (\gamma a_1)x^2+(\gamma b_1)x+\gamma c_1 ) \\ & = & \begin{bmatrix} \gamma a_1+3(\gamma c_1) \\ \gamma a_1 \end{bmatrix} \\ & = & \gamma\begin{bmatrix} a_1+3(c_1) \\ a_1 \end{bmatrix} \\ & = & \gamma T(a_1x^2+b_1x+c_1). \end{eqnarray}

Quick Quiz

Exercises

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

   Yes.

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

   Since \( \begin{bmatrix} 4 \\ -1 \\ 5\end{bmatrix} = 3\begin{bmatrix} 1 \\ 0 \\ 2\end{bmatrix} (-1)\begin{bmatrix} -1 \\ 1 \\ 1\end{bmatrix},\) \(T\left(\begin{bmatrix} 4 \\ -1 \\ 5\end{bmatrix}\right) = 3T\left(\begin{bmatrix} 1 \\ 0 \\ 2\end{bmatrix}\right) -T\left(\begin{bmatrix} -1 \\ 1 \\ 1\end{bmatrix}\right)= \begin{bmatrix} 3 \\ -14\end{bmatrix}\).