Need help to understand a solution to a polynomial problem.

Question

Let $r$ and $s$ be roots of $x^2-(a+d)x+(ad-bc)=0$.

Prove that $r^3$ and $s^3$ are the roots of $y^2-(a^3+d^3+3abc+3bcd)y+(ad-bc)^3=0$.

The solution given : The solution didn't give full details. It said $r$ and $s$ are eigenvalues of the matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}.$$

The $y$-equation is the characteristic polynomial of the cube of this matrix.

I don't have much knowledge about matrices. So, I would really appreciate if someone explains this solution to me.

Answer 1

An escalar$,\lambda$, is said to be an eigenvalue of a linear operator $A$ when for some vector $v \neq 0: Av=\lambda v$. It is know that the eigenvalues can be obtained as the roots of the characteristic polynomial, $p(\lambda)=det(A-\lambda I)$. Consider $A$ the linear operator described by the matrix given.

The characteristic polynomial of $A$ is $p_{A}(\lambda)=(a-\lambda)(d-\lambda)-bc=\lambda^2-(a+d)\lambda+(ad-bc)$ which has as solutions $s$ and $r$. Therefore $s$ and $r$ are the eigenvalues of $A$,which means that

1. exists $v_{r}\neq0$ such that $Av_{r}=rv_{r}$
2. exists $v_{s}\neq0$ such that $Av_{s}=sv_{s}$

Then you must calculate $B=A^3$ and verify that $p_{B}(\lambda)=det(B-\lambda I)=\lambda^2−(a^3+d^3+3abc+3bcd)\lambda+(ad−bc)^3$ . To conclude that $s^3$ and $r^3$ are solutions of $p_{B}(\lambda)$ then they must be eigenvalues of $B$ and in fact:

1. $Bv_{r}=A^3v_{r}=A^2(Av_{r})=(A^2)(rv_{r})=r(A^2v_{r})=rA(Av_{r})=rA(rv_{r})=r^2Av_{r}=r^3v_{r}$
2. It is analogous to conclude that $s^3$ is also an eigenvalue of $B$.

Answer 2

Here are some facts you need:

1. $\lambda$ is an eigenvalue of the matrix $A$ if $Av = \lambda v$ for some vector $v$. (definition)

2. $\det(A - tI)$ is called the characteristic polynomial of $A$. (definition)

3. $\lambda$ is an eigenvalue iff $\det(A - \lambda I) = 0$.

See if you can understand the proof from there.

Answer 3

Call the matrix $A$, then $r$ and $s$ are eigenvalues of $A$. This means, $r$ and $s$ are roots of the characteristic polynomial

$$\chi_A(x) = x^2 - (a + d)x + (ad - bc).$$

Now consider the matrix $A^3$. It is a general theorem in Linear Algebra that $A^3$ has the eigenvalues $r^3$ and $s^3$, if $A$ has the eigenvalues $r$ and $s$ (see e.g. here). Hence, $r^3$ and $s^3$ are roots of the characteristic polynomial

$$\chi_{A^3}(y) = y^2 - (a^3 + d^3 + 3abc + 3bcd)y + (ad - bc)^3.$$