# Math Help - p1(x) = x^2 - 9, p2(x) = x + 3....show that _____ for the given matrix

1. ## p1(x) = x^2 - 9, p2(x) = x + 3....show that _____ for the given matrix

p1(x) = x^2 - 9, p2(x) = x + 3, p3(x) = x- 3. Show that p1(A)=p2(A)p3(A).

A = 2 0
4 1
_________________________________________________

I inserted the equation times the matrix A and then noticed p1(A) = (p2*p3)(A) and then decided to cancel A's. Which left only the equations and then it's obvious that by foiling p2*p3 you get p1. Way too easy. What am I missing?

2. ## Re: p1(x) = x^2 - 9, p2(x) = x + 3....show that _____ for the given matrix

the matrix A =

2 0
4 1

Anybody?

4. ## Re: p1(x) = x^2 - 9, p2(x) = x + 3....show that _____ for the given matrix

they probably want you to actually compute A2 - 9I, and (A + 3I)(A - 3I):

$A^2 - 9I = \begin{bmatrix}2&0\\4&1 \end{bmatrix} \begin{bmatrix}2&0\\4&1 \end{bmatrix} - \begin{bmatrix}9&0\\0&9 \end{bmatrix}$

$= \begin{bmatrix}4&0\\12&1\end{bmatrix} - \begin{bmatrix}9&0\\0&9 \end{bmatrix} = \begin{bmatrix}-5&0\\12&-8\end{bmatrix}$.

$(A + 3I)(A - 3I) = \left(\begin{bmatrix}2&0\\4&1 \end{bmatrix} + \begin{bmatrix}3&0\\0&3 \end{bmatrix}\right)\left(\begin{bmatrix}2&0\\4&1 \end{bmatrix} - \begin{bmatrix}3&0\\0&3 \end{bmatrix}\right)$

$= \begin{bmatrix}5&0\\4&4 \end{bmatrix} \begin{bmatrix}-1&0\\4&-2 \end{bmatrix} = \begin{bmatrix}-5&0\\12&-8\end{bmatrix}$.

but, yes, working backwards, we have, for any nxn matrix A:

$p_2(A)p_3(A) = (A + 3I)(A - 3I) = A(A - 3I) + 3I(A - 3I)$

$= A^2 - 3A + 3A - 9I^2 = A^2 - 9I = p_1(A)$

it appears the exercise is designed to convince you that polynomials in a (square) matrix A work very similar to polynomials in x.

this lets us have polynomials "act" on vectors: given a vector v, and a polynomial p(x), and a matrix A, we define:

p(x).v = p(A)(v), so using a matrix A as a "go-between", we can multiply polynomial functions, and vectors, which is sort of weird, but it works.

polynomials and vector spaces have more in common than you might think.