# 6.1: The Consequence of Linearity

- Page ID
- 1892

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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

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

then you do not need any more information to figure out

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

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

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

then you don't need any more information to compute

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

because by additivity

$$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:

- They act on vector spaces.
- 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.

## Contributor

David Cherney, Tom Denton, and Andrew Waldron (UC Davis)