Basis for CPL$\left\{0,1,2,3 \right\}$

Question

I have the following question from a linear algebra textbook:

Let $V = CPL \left\{0,1,2,3 \right\}$, where this denotes the continuous piecewise linear functions on $[0,3]$, ie. the set of continuous functions on $[0,3]$ that are linear in each interval $I_i = [x_i, x_{i+1}]$, where $x_0 = 0, x_1=1, x_2=2$ and $x_3=3$. Show that $\left\{1,t,|t-1|,|t-2| \right\}$ is a basis for $V$.

Showing linear independence is easy enough, but I'm stuck on how to show these vectors span $V$. Since it isn't easy (for me at least) to represent any piecewise linear function on this domain in a single expression, this is where I'm having trouble. I also tried considering each vector restricted to the interval $I_i$, e.g. in the interval $[1,2]$ the vectors will be $\left\{ 1,t, t-1, 2-t\right\}$ but this doesn't appear to shed any light on the situation where we have a general piecewise linear function on $[0,3]$. If anyone can provide a general outline of how this proof is supposed to look, I should be able to fill in the details. Thanks very much.

Answer 1

Note that this essentially comes down to prove that $CPL\{0,1,2,3\}$ has dimension 4 (because 4 linearly independant vectors in a 4-dimensional space form a basis of this space, and you proved already they were linearly independant)

Consider the map $T : CPL\{0,1,2,3\} \to \mathbb{R}^4$ such that $T(f) = (f(0),f(1),f(2),f(3))$. It is clear that this map is linear. And it should not be too hard to prove that it is an isomorphism.

Essentially, what I did was choosing an easier basis to work with than the one you provided (I chose the basis $f_0,f_1,f_2,f_3$ such that $f_0(0) = 1$ and $f_0(i) = 0$ for $i = 1,2,3$, etc, for example $f_0$ is given by $1-t$ on $[0,1]$ and $0$ on $[1,2]$ and $[2,3]$). If you wish to do so, you can do the necessary matrix computations in order to get an explicit way of expressing any $f \in CPL\{0,1,2,3\}$ as a linear combination of $1,t,|t-1|,|t-2|$ :

$f(t) = a + bt + c|t-1| + d|t-2|$

and $a,b,c,d$ can be expressed as linear combinations of $f(0),f(1),f(2),f(3)$. If you find it not clear how to do it, I can add more details.

EDIT : the details.

So first, we apply $T$ to the 4 functions you provided

$T(1) = (1,1,1,1) \quad T(t) = (0,1,2,3) \quad T(|t-1|) = (1,0,1,2) \quad T(|t-2|) = (2,1,0,1)$.

The fact that $T$ is an isomorphism is equivalent to the fact that this is a basis of $\mathbb{R}^4$. The idea now is that we want to express a function $f \in CPL\{0,1,2,3\}$ as a linear combinations of the previous functions $1,t,|t-1|,|t-2|$. To do this we want to compute the inverse the matrix

$M = \begin{pmatrix}1&0&1&2 \\ 1&1&0&1 \\ 1&2&1&0 \\ 1&3&2&1 \end{pmatrix}$

made up of the images of $1,t,|t-1|,|t-2|$ by $T$. Now you can check that the inverse of this matrix is given by

$N = \frac{1}{2}\begin{pmatrix}1&0&3&-2 \\ -1&1&-1&1 \\ 1&-2&1&0 \\ 0&1&-2&1\end{pmatrix}$

This tells us that for any $f \in CPL\{0,1,2,3\}$, we have $f(t) = \frac{1}{2}\left((f(0) + 3f(2) - 2f(3))\times 1 + (-f(0) + f(1) - f(2) + f(3))\times t + (f(0) - 2f(1) + f(2))\times |t-1| + (f(1) - 2f(2) + f(3))\times |t-2|\right)$