Linear Algebra and Its Applications, Exercise 2.6.8

Exercise 2.6.8. If $P_3$ is the space of cubic polynomials in $t$ , what matrix would represent $d^2/dt^2$ ? What are the nullspace and column space of this matrix? What polynomials would they represent?

Answer: $P_3$ consists of all polynomials of the form $a_3t^3 + a_2t^2 + a_1t + a_0$ . The first derivative $d/dt$ of such a polynomial is $3a_3t^2 + 2a_2t + a_1$ and the second derivative $d^2/dt^2$ is $6a_3t + 2a_2$ .

If we choose as our basis the vectors $p_1 = 1$ (representing the constant term), $p_2 = t$ , $p_3 = t^2$ , and $p_4 = t^3$ then the polynomial $a_3t^3 + a_2t^2 + a_1t + a_0$ can represented as the vector $p = \begin{bmatrix} a_0&a_1&a_2&a_3 \end{bmatrix}^T$ .

For the matrix corresponding to the second derivative $d^2/dt^2$ we must have $Ap_1 = 0$ , $Ap_2= 0$ , $Ap_3 = 2p_1$ , and $Ap_4 = 6p_2$ . The resulting matrix is

$A = \begin{bmatrix} 0&0&2&0 \\ 0&0&0&6 \\ 0&0&0&0 \\ 0&0&0&0\end{bmatrix}$

so that

$d^2p/dt^2 = Ap = \begin{bmatrix} 0&0&2&0 \\ 0&0&0&6 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \end{bmatrix} = \begin{bmatrix} 2a_2 \\ 6a_3 \\ 0 \\ 0 \end{bmatrix}$

with the vector on the right corresponding to the polynomial $2a_2 + 6a_3t$ .

The matrix $A$ has two pivots (in the third and fourth columns) so its rank $r = 2$ .

The nullspace $\mathcal{N}(A)$ is the set of all vectors $x$ for which $Ax = 0$ or

$\begin{bmatrix} 0&0&2&0 \\ 0&0&0&6 \\ 0&0&0&0 \\ 0&0&0&0\end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = 0$

From the first row we have $2x_3 = 0$ and from the second row $6x_4 = 0$ so that $x_3 = x_4 = 0$ . In this case $x_1$ and $x_2$ are free variables that can take on any value. The null space is therefore the set of all vectors of the form

$a_0 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + a_1 \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$

which corresponds to polynomials of the form $a_1t + a_0$ . The nullspace has rank $n-r = 4-2 = 2$ .

The column space $\mathcal{R}(A)$ is the set of all linear combinations of the last two columns of $A$ :

$c_1 \begin{bmatrix} 2 \\ 0 \\ 0 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 6 \\ 0 \\ 0 \end{bmatrix}$

$= (c_{1}/2) \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + (c_{2}/6) \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$

This set is the same as the nullspace $\mathcal{N}(A)$ , namely the set of all vectors of the form

$a_0 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + a_1 \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$

corresponding to polynomials of the form $a_1t + a_0$ .

UPDATE: Corrected the paragraph discussing the effects of $A$ when applied to the basis vectors $p_1$ through $p_4$ . Thanks go to Abdalsalam Fadeel for pointing out a mistake with this.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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2 Responses to Linear Algebra and Its Applications, Exercise 2.6.8

Abdalsalam Fadeel says:

February 21, 2016 at 2:16 am

Before the first matrix Ap1=0 , Ap2=0 , Ap3=2a2 , and Ap4=6a3 Not as written I think can you make sure for this please? thank you

- hecker says:
  
  February 21, 2016 at 4:31 am
  
  You were correct that for the third formula I needed to reference Ap_3, not Ap_2 as I had it. However we have Ap_3 = 2p_1, not Ap_3 = 2a_2. That’s because p_1 through p_4 are unit vectors of the form p_1 = [1, 0, 0, 0]^T, p_2 = [0, 1, 0, 0]^T, p_3 = [0, 0, 1, 0]^T, and p_4 = [0, 0, 0, 1]^T. The vector involving a_0, a_1, a_2, and a_3 is the general vector p = [a_0, a_1, a_2, a_3]^T representing the corresponding polynomial.
  
  When A multiplies p we get a vector involving a_0, etc. But when A multiples p_1, p_2, p_3, or p_4 we get a vector involving the elements of A, which have nothing to with a_0, etc., and the 0 or 1 values from p_1, etc.