
# 6.1: The Consequence of Linearity


Now that we have a sufficiently general notion of vector space it is time to talk about why linear operators are so special. Think about what is required to fully specify a real function of one variable. One output must be specified for each input. That is an infinite amount of information.

By contrast, even though a linear function can have infinitely many elements in its domain, it is specified by a very small amount of information.

Example 62

If you know that the function $$L$$ is linear and that

$$L\begin{pmatrix}1\\0\end{pmatrix} =\begin{pmatrix}5\\3\end{pmatrix}$$

$$L\begin{pmatrix}2\\0\end{pmatrix},~L\begin{pmatrix}3\\0\end{pmatrix}~,L\begin{pmatrix}4\\0\end{pmatrix},~L\begin{pmatrix}5\\0\end{pmatrix} ,\ldots,$$

because by homogeneity

$$L\begin{pmatrix}5\\0\end{pmatrix}=L\left[ 5\begin{pmatrix}1\\0\end{pmatrix} \right] = 5 L\begin{pmatrix}1\\0\end{pmatrix}=5\begin{pmatrix}5\\3\end{pmatrix}=\begin{pmatrix}25\\15\end{pmatrix}.$$

In this way an infinite number of outputs is specified by just one.

Likewise, if you know that $$L$$ is linear and that

$$L\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}5\\3\end{pmatrix} {\rm ~and~} L\begin{pmatrix}0\\1\end{pmatrix}= \begin{pmatrix}2\\2\end{pmatrix}$$

$$L\begin{pmatrix}1\\1\end{pmatrix}$$

$$L\begin{pmatrix}1\\1\end{pmatrix}= L \left[ \begin{pmatrix}1\\0\end{pmatrix} + \begin{pmatrix}0\\1\end{pmatrix} \right] =L\begin{pmatrix}1\\0\end{pmatrix} + L \begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}5\\3\end{pmatrix} +\begin{pmatrix}2\\2\end{pmatrix} =\begin{pmatrix}7\\5\end{pmatrix}.$$

In fact, since every vector in $$\Re^{2}$$ can be expressed as

$$\begin{pmatrix}x\\y\end{pmatrix}= x\begin{pmatrix}1\\0\end{pmatrix}+y\begin{pmatrix}0\\1\end{pmatrix}\, ,$$

we know how $$L$$ acts on every vector from $$\Re^{2}$$ by linearity based on just two pieces of information;

$$L\begin{pmatrix}x\\y\end{pmatrix} = L \left[ x\begin{pmatrix}1\\0\end{pmatrix}+y\begin{pmatrix}0\\1\end{pmatrix} \right] = x L\begin{pmatrix}1\\0\end{pmatrix}+y L\begin{pmatrix}0\\1\end{pmatrix} = x \begin{pmatrix}5\\3\end{pmatrix}+ y\begin{pmatrix}2\\2\end{pmatrix} =\begin{pmatrix}5x+2y\\3x+2y\end{pmatrix}.$$

Thus, the value of $$L$$ at infinitely many inputs is completely specified by its value at just two inputs. (We can see now that $$L$$ acts in exactly the way the matrix

$$\begin{pmatrix} 5&2\\ 3&2 \end{pmatrix}$$

acts on vectors from $$\Re^{2}$$.)

This is the reason that linear functions are so nice; they are secretly very simple functions by virtue of two characteristics:

1. They act on vector spaces.
2. They act additively and homogeneously.

A linear transformation with domain $$\Re^{3}$$ is completely specified by the way it acts on the three vectors

$$\begin{pmatrix}1\\0\\0\end{pmatrix}\, ,\:\begin{pmatrix}0\\1\\0\end{pmatrix}\, ,\:\begin{pmatrix}0\\0\\1\end{pmatrix}\, .$$

Similarly, a linear transformation with domain $$\Re^{n}$$ is completely specified by its action on the $$n$$ different $$n$$-vectors that have exactly one non-zero component, and its matrix form can be read off this information. However, not all linear functions have such nice domains.