I just analysed this equation for real matrices $$ A^2=\begin{pmatrix}a&b\\c&d\end{pmatrix}^2=I $$ From the main diagonal of $A^2$ we must have $a^2+bc=bc+d^2=1$ showing that $d=\pm a$.

CASE 1: If $b=0$ we have $a^2=d^2=1$ so then $c=0$. If $c=0$ the same argument renders $b=0$.

CASE 2: Otherwise assume $b,c\neq 0$. Then if $d=a$ the opposite diagonal of $A^2$ shows that $2ab=2ac=0$. Thus $a=d=0$ and the equation $a^2+bc=1$ yields $bc=1$.

CASE 3: Now, if $d=-a$ we have $ab-ab=ac-ac=0$ which is always true, so then only the first equation $a^2+bc=1$ matters.

So we have three sets of solutions:

  • CASE 1: gives the four solutions $b=c=0$ and $a,d\in\{-1,1\}$.
  • CASE 2: gives an infinite family of solutions $a=d=0$ and $b=x,c=x^{-1}$ with $x\in\mathbb R\setminus\{0\}$.
  • CASE 3: gives a last family of solutions: choose $(x,y)\in\mathbb R\times\left(\mathbb R\setminus\{0\}\right)$ and define $a=x,d=-x$ and $b=y,c=(1-x^2)y^{-1}$.

I find it interesting how the solutions divide nicely into advancing stages of first discrete, then one-dimensional, and finally two-dimensional parametrisations in this analysis. I wonder if that is purely a coincidence ...

QUESTION: What would a similar analysis of $A^2=I$ where we are dealing with $3\times 3$ matrices look like? The case of $3\times 3$ looks rather complicated already, so I would hardly dream of considering the general case of $n\times n$ matrices.

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    $\begingroup$ You can tackle the general case using reduction (diagonalization). $\endgroup$ – Beni Bogosel Aug 15 '14 at 11:29

Since $A^2=I$, the minimal polynomial of $A$ has simple roots, and therefore $A$ is diagonalisable. $A=P^{-1}DP$ with $P$ invertible and $D$ diagonal. The equation $A^2 = I$ implies $D^2=I$ so $D$ is a diagonal matrix with $1$ or $-1$ on the diagonal.

Conversely, if $D$ is a diagonal matrix with $\pm 1$ on the diagonal, then $A=P^{-1}DP$ satisfies the required equation.

Therefore the solutions of the equations $A^2=I$ are precisely the matrices of the form $A=P^{-1}DP$ with $P$ invertible and arbitrary and $D$ a diagonal matrix with $1$ or $-1$ on the diagonal.

  • $\begingroup$ why $A$ is diagonalisable? $\endgroup$ – user437713 Dec 22 '17 at 23:47
  • $\begingroup$ Because the minimal polynomial of $A$, which divides $X^2-1$, has simple roots. $\endgroup$ – Beni Bogosel Dec 24 '17 at 11:25
  • $\begingroup$ Thanks, that makes sense. $\endgroup$ – user437713 Dec 24 '17 at 21:08

This is an alternative method to find all possible $2\times 2$ matrix $A$ such that $A^2=I$ so the polynomial $x^2-1$ with simple roots $\pm1$ annihilates $A$ and then $A$ is diagonalisable over $\Bbb R$ and $$\operatorname{sp} A\in\{-1,1\}$$ hence

  • if $\operatorname{sp} A=\{1\}$ then $A=I$
  • if $\operatorname{sp} A=\{-1\}$ then $A=-I$
  • and if $\operatorname{sp} A=\{-1,1\}$ then $A=P\operatorname{diag}(-1,1)P^{-1}$ where $P$ is an invertible matrix.

The case of $3\times 3$ matrix isn't very different and the diagonal matrix similar to $A$ can be $$I_3;\;-I_3;\;\operatorname{diag}(-1,-1,1);\;\operatorname{diag}(-1,1,1)$$


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