# Need help to understand a solution to a polynomial problem.

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.

• You can alternatively prove it by using only Viete's formulas. Aug 9, 2020 at 18:32
• Yes,I noticed that.But I wanted to understand this proof. Aug 9, 2020 at 18:35
• The characteristic polynomial for a 2 by 2 is $x^2 - (\mbox{trace})x + \mbox{determinant}.$ Write out the cube of the matrix Aug 9, 2020 at 18:39
• For a $2\times2$ matrix $A$, the eigenvalues are roots of $x^2-$trace$(A)x+\det(A)=0$ Aug 9, 2020 at 18:40

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 $$B$$.
• Is there a way to reasonably calculate $B=A^3$ has that specific characteristic polynomial? I tried multiplying out $A^3$ by hand but it's like four $3$-degree terms multiplying by four $3$-degree terms and it's literally impossible to simplify. Aug 9, 2020 at 18:58
• The best answer I can give you is that when $n=2$ the polynomial characteristic is given by $p(\lambda)=\lambda ^2 -tr(A)\lambda+det(A)$.
– user723846
Aug 9, 2020 at 19:19

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.

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

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

• Is there a fast way to see $A^3$ has characteristic polynomial equal to the $y$ equation? I tried to multiply out $A^3$ manually but found it almost undoable by hand. Aug 9, 2020 at 18:44
• At first glance, I don't see it. We want $\chi_{A^3} (y) = \det (A^3 - y I)$. Maybe you find a clever factorisation of $A^3 - yI$ in terms of $A$, then you could use the multiplicativity of the determinant. I think about something like $a^3 - b^3 = (a - b) (a^2 + ab + b^2)$ and so on.
– Jan
Aug 9, 2020 at 18:50