## Linear Algebra and Its Applications, Exercise 2.6.9

Exercise 2.6.9. Considering taking a polynomial from $P_3$, the space of cubic polynomials in $t$, and multiplying it by the polynomial $2+3t$ to produce a polynomial in $P_4$, the space of polynomials in $t$ of degree four. Describe a matrix representing this multiplication by considering its effect on the basis vectors $x_1 = 1$ (representing the constant term), $x_2 = t$, $x_3 = t^2$, and $x_4 = t^3$.

Answer: The matrix $A$ representing multiplying a member of $P_3$ by $(2+3t)$ will have 4 columns because there are 4 basis vectors of $P_3$ and will have 5 rows because it must produce a member of $P_4$.

Multiplying the basis vectors of $P_3$ by $(2+3t)$ we have

$x_1 (2+3t) = 1 \cdot (2+3t) = 2 + 3t$

$x_2 (2+3t) = t (2+3t) = 2t + 3t^2$

$x_3 (2+3t) = t^2 (2+3t) = 2t^2 + 3t^3$

$x_4 (2+3t) = t^3 (2+3t) = 2t^3 + 3t^4$

The resulting matrix $A$ representing the multiplication is

$A = \begin{bmatrix} 2&0&0&0 \\ 3&2&0&0 \\ 0&3&2&0 \\ 0&0&3&2 \\ 0&0&0&3 \end{bmatrix}$

Consider the polynomial $p = 1+t+t^2+t^3$. Using matrix multiplication we have

$Ap = \begin{bmatrix} 2&0&0&0 \\ 3&2&0&0 \\ 0&3&2&0 \\ 0&0&3&2 \\ 0&0&0&3 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \\ 5 \\ 5 \\ 3 \end{bmatrix}$

Using normal multiplication we have

$(2+3t)(1+t+t^2+t^3)$

$= (2+3t) \cdot 1 + (2+3t)t + (2+3t)t^2 + (2+3t)t^3$

$= 2 + 3t + 2t + 3t^2 + 2t^2 + 3t^3 + 2t^3 + 3t^4$

$= 2 + 5t + 5t^2 + 5t^3 + 3t^4$

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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