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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