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.

  • $\begingroup$ You can alternatively prove it by using only Viete's formulas. $\endgroup$
    – Berci
    Aug 9, 2020 at 18:32
  • $\begingroup$ Yes,I noticed that.But I wanted to understand this proof. $\endgroup$ Aug 9, 2020 at 18:35
  • 1
    $\begingroup$ The characteristic polynomial for a 2 by 2 is $x^2 - (\mbox{trace})x + \mbox{determinant}.$ Write out the cube of the matrix $\endgroup$
    – Will Jagy
    Aug 9, 2020 at 18:39
  • 1
    $\begingroup$ For a $2\times2$ matrix $A$, the eigenvalues are roots of $x^2-$trace$(A)x+\det(A)=0$ $\endgroup$ Aug 9, 2020 at 18:40

3 Answers 3


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 $B$.
  • $\begingroup$ 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. $\endgroup$
    – cr001
    Aug 9, 2020 at 18:58
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    $\begingroup$ The best answer I can give you is that when $n=2$ the polynomial characteristic is given by $p(\lambda)=\lambda ^2 -tr(A)\lambda+det(A)$. $\endgroup$
    – 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

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

  • $\begingroup$ 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. $\endgroup$
    – cr001
    Aug 9, 2020 at 18:44
  • $\begingroup$ 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. $\endgroup$
    – Jan
    Aug 9, 2020 at 18:50

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