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I am stuck on a homework problem I am not sure if my solution makes any sense. We are given the basis {$1,t,t^2,t^3,t^4,t^5$}.

Here is the question verbatim: How does the operator $D^{2}$ act on a polynomial? What is its matrix?

Here is my attempt:

The matrix for $D$ is $\left[\begin{array}{cccccc} \frac{d^{2}}{dt^{2}} & 0 & 0 & 0 & 0 & 0\\ 0 & \frac{d^{2}}{dt^{2}} & 0 & 0 & 0 & 0\\ 0 & 0 & \frac{d^{2}}{dt^{2}} & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{d^{2}}{dt^{2}} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{d^{2}}{dt^{2}} & 0\\ 0 & 0 & 0 & 0 & 0 & \frac{d^{2}}{dt^{2}} \end{array}\right]$

Given any polynomial then $D^{2}$ will give the $4^{th}$ derivative of the polynomial in the span of $\mathbb{P}_{6}$. The matrix can be represented by:

$D^{2}=\left[\left[\begin{array}{cccccc} \frac{d^{2}}{dt^{2}} & 0 & 0 & 0 & 0 & 0\\ 0 & \frac{d^{2}}{dt^{2}} & 0 & 0 & 0 & 0\\ 0 & 0 & \frac{d^{2}}{dt^{2}} & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{d^{2}}{dt^{2}} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{d^{2}}{dt^{2}} & 0\\ 0 & 0 & 0 & 0 & 0 & \frac{d^{2}}{dt^{2}} \end{array}\right]\left[\begin{array}{cccccc} \frac{d^{2}}{dt^{2}} & 0 & 0 & 0 & 0 & 0\\ 0 & \frac{d^{2}}{dt^{2}} & 0 & 0 & 0 & 0\\ 0 & 0 & \frac{d^{2}}{dt^{2}} & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{d^{2}}{dt^{2}} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{d^{2}}{dt^{2}} & 0\\ 0 & 0 & 0 & 0 & 0 & \frac{d^{2}}{dt^{2}} \end{array}\right]\right] $

But this is just: $[\begin{array}{cccccc} \frac{d^{4}}{dt^{4}} & 0 & 0 & 0 & 0 & 0\\ 0 & \frac{d^{4}}{dt^{4}} & 0 & 0 & 0 & 0\\ 0 & 0 & \frac{d^{4}}{dt^{4}} & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{d^{4}}{dt^{4}} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{d^{4}}{dt^{4}} & 0\\ 0 & 0 & 0 & 0 & 0 & \frac{d^{4}}{dt^{4}} \end{array}]$

I am just worried though because when I think about the vector: $(6,t-1,(t-1)^2,(t-1)^3,(t-1)^4,(t-1)^5)$ when I represent this as a set of vectors I get: $\left[\begin{array}{c} 1\\ 0\\ 0\\ 0\\ 0\\ 0 \end{array}\right],\left[\begin{array}{c} -1\\ 1\\ 0\\ 0\\ 0\\ 0 \end{array}\right],\left[\begin{array}{c} 1\\ -2\\ 1\\ 0\\ 0\\ 0 \end{array}\right],\left[\begin{array}{c} -1\\ 3\\ -3\\ 1\\ 0\\ 0 \end{array}\right],\left[\begin{array}{c} 1\\ -4\\ 6\\ -4\\ 1\\ 0 \end{array}\right],\left[\begin{array}{c} 1\\ 5\\ -10\\ 10\\ -5\\ 1 \end{array}\right]$

But how do I apply my differentiation matrix to any of these vectors? It is always 0! I have a feeling my matrix for D is wrong but I am not sure how to fix it.

1 Answers 1

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The matrix representing $D$ should contain numbers, not operators! The $i$-th column of the matrix representing $D$ with respect to the basis $\mathcal{B} = (p_0,p_1,\dots,p_5)$ (where $p_i(t) = t^i$) should contain the column vector $[D(p_i)]_{\mathcal{B}}$ which is the coordinate representation of the result of applying $D$ to $p_i$ with respect to the basis $\mathcal{B}$.

For example,

$$ D(p_0) = 0, D(p_1) = (t)'' = 0, D(p_2) = (t^2)'' = (2t)' = 2, D(p_3) = (t^3)'' = 6t, \dots $$

so

$$ [D]_{\mathcal{B}} = \begin{pmatrix} 0 & 0 & 2 & 0 & ? & ? \\ 0 & 0 & 0 & 6 & ? & ? \\ 0 & 0 & 0 & 0 & ? & ? \\ 0 & 0 & 0 & 0 & ? & ? \\ 0 & 0 & 0 & 0 & ? & ? \\ 0 & 0 & 0 & 0 & ? & ? \\ \end{pmatrix}. $$

I'll leave the question marks for you to fill out.

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    Oooh yes I see now. Thank you. For $D^2$ then that will be $[D]_\mathcal{B} ^2$. Not the 4th derivative but the square of my matrix. What does it mean: "How does $D^2$ act on the polynomial"? – 2017-01-26
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    oh wait no I see, never mind. – 2017-01-26
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    Your notation is somewhat confusion. It is better to call the second derivative operator by $D$ and denote by $[D]_{\mathcal{B}}$ the matrix representing $D$. The operator $D^2$ is then the fourth derivative operator and the matrix representing $D^2$ with respect to $\mathcal{B}$ can be calculated either directly (which is better in this case) or by squaring the matrix of $D$ as in: $[D^2]_{\mathcal{B}} = ([D]_{\mathcal{B}})^2$. – 2017-01-26