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We are studying for a qualifying exam and have come across the following problem in a previous exam.

Determine the solutions (if any) of the matrix equation $X^3=I-X$ in the $2 \times 2$-matrices over $\mathbb{R}$.

Any hints in the right direction would be much appreciate for this.

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Start from the diagonal case.Then you will identify necessary conditions for eigenvalues. After this it is easy to prove that if $X$ is solution then $VXV^{-1}$ for any non singular $V$ is also a solution.Actually you even don't have to diagonalize. If $\lambda$ is eigenvalue and $u$ is eigenvector than you have $\lambda^3=1-\lambda$ . Solve the equation. All matrices with such eigenvalues will be solutions. –  Alexander Vigodner Aug 11 at 22:34

5 Answers 5

up vote 10 down vote accepted

The polynomial $x^3+x-1$ has one real root $\alpha$ and two conjugate complex non real roots $\beta$ and $\overline \beta$. Since this polynomial annihilates the matrix $X$ then their eigenvalues belong to the set of roots of the polynomial and since $X$ is a real matrix then there's two cases:

  • $\alpha$ is the only eigenvalue of $X$ and if $X$ isn't diagonalizable then the minimal polynomial is $(x-\alpha)^2$ divides the given polynomial which isn't true so $X$ is diagonalizable and then $X$ is similar to $\alpha I_2$.
  • $\beta$ and $\overline\beta$ are the eigenvalues of $X$ so the minimal polynomial of $X$ is $(x-\beta)(x-\overline\beta)=x^2-2\operatorname{Re}(\beta) x+|\beta|^2$ so let $$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ and since $$\chi_X(x)=\det(X-xI_2)=(a-x)(d-x)-cb=x^2-(a+d)x+ad-cb$$ hence we choose $a,b,c$ and $d$ such that $$a+d=2\operatorname{Re}(\beta)\quad;\quad ad-cb=|\beta|^2$$
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Being similar to $\alpha I$ means being equal to $\alpha I$. –  Thomas Andrews Aug 11 at 23:27
Absolutely right:-) –  Sami Ben Romdhane Aug 11 at 23:28
Also, possibly worth noting that $|\beta|^2 = \beta\bar\beta = \alpha^{-1}$, and $\beta+\bar\beta=2\mathrm{Re}(\beta)=-\alpha$. –  Thomas Andrews Aug 11 at 23:35
Why would $\beta$ be related to $\alpha$? –  bartgol Aug 12 at 15:23

Hint: The matrix satisfies $X^3+X-I=0$. What can you say about the minimal polynomial?

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The equation is $x^3+x-1=0$ now a two by two matrix has a minimal polynomial of degree $\leq 2$. So which quadratic or linear polynomials divide tbe above cubic ?

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Notice that, if there is a quadratic polynomial that divides $x^3+x-1$ then there must be also a linear polynomial that divides it (the quotient). Hence, it suffices to check whether or not there is a linear polynomial that divides the matrix. –  bartgol Aug 11 at 23:17
@bartgol Well it has just one real root. So see the accepted answer, it is not really clear what the best from to express the matrices with quadratic minimal poly. I would like to know what the intended answer was, since it is for a qualifying exam. –  Rene Schipperus Aug 11 at 23:26

Note that this polynomial has one real root and two complex conjugates. For a real matrix, if one of the complex solutions is an eigenvalue then so is the other. A real $2 \times 2$ matrix with eigenvalues $a \pm bi$ is $$ \pmatrix{ s & t \cr -((a-s)^2 + b^2)/t & 2a - s }$$ for arbitrary $s$ and $t \ne 0$.

On the other hand, if the real root $r$ is an eigenvalue, it is the only one, and has geometric multiplicity $2$, so the matrix is $r I$.

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You can start looking at the eigenvalues of $X$. What kind of relationship there is? Can you find the eigenvalues of $X$?

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