I tried a basic approach and wrote x as a matrix of four unknown elements $\begin{pmatrix} a && b \\ c && d \end{pmatrix}$ and squared it when I obtained $\begin{pmatrix} a^2 + bc && ab + bd \\ ca + dc && cd + d^2\end{pmatrix}$ and by making it equal with $I_2$ I got the following system

$a^2 + bc = 1$

$ab +bd = 0$

$ca + dc = 0$

$cd + d^2 = 1$

I don't know how to proceed. (Also, if anyone knows of a better or simpler way of solving this matrix equation I'd be more than happy to know).

  • 1
    $\begingroup$ familiar with Jordan canonical forms??? $\endgroup$ – user87543 Oct 21 '13 at 14:51
  • $\begingroup$ how can i write one of those here? $\endgroup$ – andreas.vitikan Oct 21 '13 at 14:51
  • 2
    $\begingroup$ Your last equation should be $bc + d^2 = 1$. $\endgroup$ – Robert Israel Oct 21 '13 at 15:02
  • $\begingroup$ Are you allowed to work with complex matrices, or are you sticking with reals? $\endgroup$ – alex.jordan Oct 26 '13 at 16:33
  • $\begingroup$ I'm allowed to work with complex matrices $\endgroup$ – andreas.vitikan Oct 27 '13 at 5:23

The second and third equalities are

$$b(a+d)=0$$ $$c(a+d)=0$$

Split the problem in two cases:

Case 1: $a+d \neq 0$. Then $b=0, c=0$, and from the first and last equation you get $a,d$.

Case 2: $a+d=0$. Then $d=-a$ and $bc=1-a^2$. Show that any matrix satisfying this works.


What you need to know about JCF is that for any $2\times 2$ matrix $x$, there is an invertible matrix $P$ such that $P^{-1}xP=\left(\begin{smallmatrix}t&0\\0&t\end{smallmatrix}\right)$ or $P^{-1}xP=\left(\begin{smallmatrix}t&1\\0&t\end{smallmatrix}\right)$ or $P^{-1}xP=\left(\begin{smallmatrix}t&0\\0&r\end{smallmatrix}\right)$. Those three possibilities are the various Jordan forms you might get.

We have $x^2=I$. Multiply on the left by $P^{-1}$ and on the right by $P$ and you get $P^{-1}x^2P=P^{-1}P=I$. But then $P^{-1}x^2P=P^{-1}xIxP=P^{-1}x(PP^{-1})xP=(P^{-1}xP)(P^{-1}xP)=I$.

Hence if $x^2=I$, then the JCF also squares to $I$. This means that the second case is out, and in the first or third case, $t^2=r^2=1$. Hence the only possible JCF are the four matrices $\left(\begin{smallmatrix}\pm1 &0\\0&\pm1\end{smallmatrix}\right)$.


$\begin{pmatrix} a^2 + bc & ab + bd \\ ca + dc & bc + d^2\end{pmatrix}=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$

You have $b(a+d)=0$ which implies $b=0$ or $a+d=0$

suppose $b=0,a+d\neq0$, you have $a^2=1$ concluding $a=\pm 1$ you can find $d$ and then $c$

suppose $a+d=0, b\neq 0$, you have some other possibility....

You have $c(a+d)=0$ which implies $c=0$ or $a+d=0$ repeating same condition you can find other entities...

  • $\begingroup$ sorry, my course in linear algebra doesn't cover those, I was looking for a more simple explanation or solution $\endgroup$ – andreas.vitikan Oct 21 '13 at 14:56
  • $\begingroup$ That is why i have asked are you familiar with Jordan canonical forms... $\endgroup$ – user87543 Oct 21 '13 at 14:57

You are looking for matrices $A$ satisfying the polynomial equation $A^2-I=0$. Note that the corresponding polynomial splits $X^2-1=(X+1)(X-1)$, and since you are presumably not working over a field of characterisitic$~2$ but rather over the rational, real or complex numbers (although the question is not entirely clear about this), these two linear factors are distinct. It is a general fact that matrices satisfying a polynomial equation that splits into distinct linear factors are diagonalisable, and its eigenvalues must be among the roots of those linear factors (since the polynomial in $A$ is supposed to kill eigenvectors in particular). So no Jordan forms are needed. For the case $A^2=I$ you can also argue "by hand" that any vector $v$ is the sum of $\frac12(v+Av)$ in the eigenspace for$~1$ and $\frac12(v-Av)$ in the eigenspace for$~{-}1$, so those eigenspaces span the whole space and $A$ is diagonalisable. All this does not depend in the size of the square matrix$~A$.

Now to find all solutions for $A$, you just need to determine what the eigenspaces for $1$ and $-1$ can be. They can be any pair of complementary subspaces. If one of the eigenspaces is reduced to $\{0\}$ (so it is not really an eigenspace) one gets $A=I$ or $A=-I$, and these are isolated solutions. In the $2\times 2$ case the only other possibility is having two complementary eigenspaces of dimension$~1$. If $\binom pq$ and $\binom rs$ are linearly independent vectors, then you get a matrix with eigenspace for$~1$ spanned by the first and the eignespace for$~{-}1$ spanned by the second as $$ A=\begin{pmatrix}p&r\\q&s\end{pmatrix} \begin{pmatrix}1&0\\0&-1\end{pmatrix} \begin{pmatrix}p&r\\q&s\end{pmatrix}^{-1} =\frac1{ps-qr}\begin{pmatrix}ps+qr&-2pr\\2qs&-ps-qr\end{pmatrix}, $$ and that is the general form for this case. There is some redundancy in this expression since scaling either $\binom pq$ or $\binom rs$ has no effect; the set of such involutions only has dimension$~2$, and it can also be described as the set of matrices $$ \begin{pmatrix}a&b\\c&-a\end{pmatrix} \qquad\text{with $a^2+bc=1$,} $$ in other words as those with characteristic polynomial $X^2-1$.


$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ Let's use the $\large\mbox{Pauli matrices}$:

Given $\quad I_{2} = \alpha + \vec{\beta}\cdot\vec{\sigma},\quad$ let $\quad x \equiv a + \vec{b}\cdot\vec{\sigma}.\quad$ Then $$ x^{2} = \pars{a + \vec{b}\cdot\vec{\sigma}}\pars{a + \vec{b}\cdot\vec{\sigma}} = a^{2} + \vec{b}\cdot\vec{b} + 2a\vec{b}\cdot\vec{\sigma} = \alpha + \vec{\beta}\cdot\vec{\sigma} = I_{2} $$ We have two conditions \begin{align} a^{2} + \vec{b}\cdot\vec{b} &= \alpha\tag{1} \\ 2a\vec{b} &= \vec{\beta}\tag{2} \end{align} such that $\vec{b} = \vec{\beta}/2a$ with $a \not= 0$. Then $$ a^{2} + {\vec{\beta}\cdot\vec{\beta} \over 4a^{2}} = \alpha \quad\imp\quad a^{4} -\alpha\,a^{2} + {\vec{\beta}\cdot\vec{\beta} \over 4} = 0 $$ which yields the solutions $$ a_{\pm}^{2} = {\alpha \pm \root{\alpha^{2} - \vec{\beta}\cdot\vec{\beta}} \over 2} \quad\mbox{and}\quad \pars{a_{\pm}}_{\pm} = \pm\root{{\alpha \pm \root{\alpha^{2} - \vec{\beta}\cdot\vec{\beta}} \over 2}} $$ $$ \mbox{For each value of}\ a\ \mbox{we have a value of}\ \vec{b} = {\vec{\beta} \over 2a}\,,\qquad a \not= 0 $$ Notice that $x$ can be rewritten as $$ x = a + {1 \over 2a}\pars{\beta\cdot\vec{\sigma}} = a + {I_{2} - \alpha \over 2a} = \pars{a - {\alpha \over 2a}} + {I_{2} \over 2a} $$ such that we just need to evaluate $a$.

Whenever we get a value of $a = 0$, we return to conditions $\pars{1}$ and $\pars{2}$. In that case, those conditions are reduced to $\vec{b}\cdot\vec{b} = \alpha$ and $\vec{0} = \vec{\beta}$. So, we see that it occurs when $I_{2}$ is proportional to the identity matrix.


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