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

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.

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 $$f(t) = \frac{1}{2}\left((f(0) + 3f(2) - 2f(3))\times 1 + (-f(0) + f(1) - f(2) + f(3))\times t + (f(0) - 2f(1) + f(2))\times |t-1| + (f(1) - 2f(2) + f(3))\times |t-2|\right)$$

• Thank you, yes can you add the details? – Elliot Herrington Jun 25 at 0:18
• Great, thanks so much for your time and effort – Elliot Herrington Jun 26 at 11:57