## 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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