## 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

1. Abdalsalam Fadeel says:

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:

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.