3
$\begingroup$

This question is triggered by another one:

Suppose we seek a $2\times2$ matrix equivalent of the imaginary unit, that is a matrix $\,i\,$ such that $\,i^2 = -1\,$ , or, with $\;a,b,c,d \in \mathbb{R}$ : $$ i^2 = \begin{bmatrix} a&b\\c&d \end{bmatrix} \begin{bmatrix} a&b\\c&d \end{bmatrix} = \begin{bmatrix} a^2+bc&ab+bd\\ac+cd&bc+d^2 \end{bmatrix} = - \begin{bmatrix} 1&0\\0&1 \end{bmatrix} $$ Leading to the following equations: $$ a^2+bc=-1 \quad ; \quad b(a+d)=0 \quad ; \quad c(a+d)=0 \quad ; \quad bc+d^2=-1 $$ Subtracting the first from the last one gives two possible solutions: $$ a^2-d^2=0 \quad \Longrightarrow \quad a = \pm\, d $$ The solution $\,a=d\ne0\,$ leads to $\,b=0\,$ and $\,c=0\,$ , giving $\,a^2 = -1$ , which is impossible in the reals. The other possibility is $\,d = -a$ : $$ i = \begin{bmatrix} a&b\\c&-a \end{bmatrix} \quad \mbox{with} \quad a^2+bc = -1 \quad \mbox{or} \quad \begin{vmatrix} a&b\\c&-a \end{vmatrix} = 1 $$ But here I'm stuck. IMO it does not follow that $\,i\,$ is a special case of the above matrix, namely the one with $\,a=0\,$ and $\,b=-1$ , $c=1$ : $$ i = \begin{bmatrix} 0&-1\\1&0 \end{bmatrix} $$ Am I missing something obvious?

Note. In a comment by Meelo it is remarked that <quote>One can notice that this representation is not unique, though one must always treat $1$ as the matrix $$ \pmatrix{1&&0\\0&&1} $$ any matrix with characteristic polynomial $x^2+1$ suffices to represent $i$ . For instance, you could use $$\pmatrix{1&&-2\\1&&-1}$$ for $i$.</quote> That's right, because $$ \begin{vmatrix}a-\lambda&b\\c&-a-\lambda\end{vmatrix}=(a-\lambda)(-a-\lambda)-bc=\lambda^2+1 $$ I don't see, however, how this can be matched with my answer at the same place: $$ e^{\begin{bmatrix} 1 & -2 \\ 1 & -1 \end{bmatrix}\theta} = \mbox{?} = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} $$

$\endgroup$
  • $\begingroup$ What is wrong ? You do have $i^2=-I$. $\endgroup$ – Yves Daoust Feb 9 '15 at 15:23
  • 1
    $\begingroup$ Notice that the matrix I posted is a conjugate of the typical matrix; that is, $$\pmatrix{1&&-2\\1&&-1}=\pmatrix{1&&1\\0&&1}\cdot \pmatrix{0&&-1\\1&&0}\cdot\pmatrix{1&&1\\0&&1}^{-1}.$$ Conjugation preserves addition and multiplication of matrices, ergo we can conjugate the ordinary matrix for $e^{i\theta}$ to get: $$e^{\pmatrix{ 1&&-2\\1&&-1}\theta}=\pmatrix{1&&1\\0&&1}\cdot \pmatrix{\cos(\theta)&&-\sin(\theta)\\ \sin(\theta)&&\cos(\theta)}\cdot\pmatrix{1&&1\\0&&1}^{-1}$$ which can be computed to some which this margin is too narrow to contain. $\endgroup$ – Milo Brandt Feb 10 '15 at 23:14
  • 1
    $\begingroup$ @Meelo: Exactly what I've been thinking about last night. That explains it all. Thanks! $\endgroup$ – Han de Bruijn Feb 11 '15 at 14:08
  • 2
    $\begingroup$ @Meelo: $$\begin{pmatrix}\cos(t)+\sin(t) & -2\sin(t) \\ \sin(t) & \cos(t)-\sin(t)\end{pmatrix}$$ $\endgroup$ – robjohn Feb 14 '15 at 11:05
4
$\begingroup$

I would suggest looking at the problem from another angle. Suppose you wanted to find the matrix representation of the linear map $$ L: \mathbb{C} \rightarrow \mathbb{C},\; z \mapsto i\cdot z $$.

If we write $z$ in the form $z = a + bi$, we see

$$L(z) = i\cdot (a+bi) = -b + ai$$

So L maps the number $(1 + 0i)$ to $(0 + 1i)$ and the number $(0+1i)$ to $(-1+0i)$. Identifying $\mathbb{C}$ with $\mathbb{R}^2$ we see that the matrix representation of this map is

$$ L = \begin{bmatrix} 0&-1\\1&0 \end{bmatrix} $$

$\endgroup$
2
$\begingroup$

From:

$$ \left[ \begin{array}{cccc} a^2+bc&ab+bd\\ ac+cd&bc+d^2 \end {array} \right] = \left[ \begin{array}{cccc} -1&0\\ 0&-1 \end {array} \right] $$

we have $$ \begin{cases} b(a+d)=0\\ c(a+d)=0\\ a^2+bc=-1\\ d^2+bc=-1 \end{cases} $$ If $a+d\ne0$ then $b=0$ and $c=0$ and we have $a^2=-1$ that is impossibile. So we must have $$ \begin{cases} a+d=0\\ a^2+bc=-1 \end{cases} $$ From the second equation we find $ a^2=-1-bc$ that has real solutions iff $bc \le -1$ and in such a case $a=\pm\sqrt{-1-bc}$.

The simplest case is when $bc=-1$ so that we have only one value for $a$ i.e. $a=0$ and the matrix we are searching for has the form:

$$ \left[ \begin{array}{cccc} 0&b\\ -1/b&0 \end {array} \right] $$ Note that the $i$ matrix cited in OP is of this form. But, as we see there are many other matrices that represent $i$ and for every such matrix the subring generated by the matrix and the identity is isomorphic to the complex numbers.

$\endgroup$
2
$\begingroup$

Allow me to repeat a piece of theory which shall be well known to many readers of Mathematics Stack Exchange, but it is rather unfamiliar to myself and probably to some others as well; some keywords are Matrix similarity and Equivalence relation.
In the sequel, all matrices are $2\times 2$, real valued and non-singular. Two matrices $A$ and $B$ are called similar, written as $A \sim B$ if there exists a matrix $P$ such that: $$ B = P^{-1}\, A\, P \qquad \mbox{with:} \quad P = \begin{bmatrix}p&q\\r&s\end{bmatrix} $$ Example: $$ \begin{bmatrix}0&-1\\1&0\end{bmatrix} \sim \begin{bmatrix}0&1\\-1&0\end{bmatrix} \quad \mbox{because} \quad \begin{bmatrix}0&1\\1&0\end{bmatrix} \begin{bmatrix}0&-1\\1&0\end{bmatrix} \begin{bmatrix}0&1\\1&0\end{bmatrix}= \begin{bmatrix}0&1\\-1&0\end{bmatrix} $$ With other words: $\;i \sim -i$ , which is somehow relevant to what follows.
Similarity is sort of an equality, because it's easy to prove that: $$ A \sim A \quad ; \quad (A \sim B) \; \Longleftrightarrow \; (B \sim A) \quad ; \quad (A \sim B) \; \wedge \; (B \sim C) \; \Longrightarrow \; (A \sim C) $$ Now let $\left[i\right]$ denote the "standard" matrix representation of the imaginary unit. Then we have for an arbitrary $2\times 2$ matrix $P$ : $$ P^{-1}\,\left[i\right]\,P = \begin{bmatrix}s&-q\\-r&p\end{bmatrix}/D \begin{bmatrix}0&-1\\1&0\end{bmatrix} \begin{bmatrix}p&q\\r&s\end{bmatrix} \qquad\mbox{with:}\quad D = \begin{vmatrix}p&q\\r&s\end{vmatrix} = (ps-qr) $$ $$ \Longrightarrow \qquad P^{-1}\,\left[i\right]\,P = \begin{bmatrix}(pr+qs)&-(p^2+q^2)\\ (r^2+s^2)&-(pr+qs)\end{bmatrix}/(ps-qr) $$ The determinant is (there is a shortcut for this, though): $$ \begin{vmatrix}-(pq+rs)/(ps-qr)&-(q^2+s^2)/(ps-qr)\\ (p^2+r^2)/(ps-qr)&(pq+rs)/(ps-qr)\end{vmatrix} = \frac{-(pq+rs)^2+(q^2+s^2)(p^2+r^2)}{(ps-qr)^2} = 1 $$ Hence $\;P^{-1}\,\left[i\right]\,P\;$ has the form as required: see my question. $$ \begin{bmatrix}a&b\\c&-a\end{bmatrix} \qquad\mbox{with:}\quad a^2+bc=-1 $$ It is concluded that all matrix representations of the imaginary unit are similar (but not equal).
If we do the above for powers of $(P^{-1}\left[ i \right]P)$ , then: $$ \left(P^{-1}\left[ i \right]P\right)^2 = P^{-1}\left[ i \right]^2 P \\ \left(P^{-1}\left[ i \right]P\right)^3 = P^{-1}\left[ i \right]^3 P \\ \cdots \\ \left(P^{-1}\left[ i \right]P\right)^n = P^{-1}\left[ i \right]^n P $$ In this way, it's easy to see how the result in the comment by Meelo can be generalized: $$ e^{P^{-1}\,\left[i\right]\theta\,P} = P^{-1} e^{\left[i\right]\theta} P $$ Or, whatever matrix equivalent of $\,i\,$ may be preferred: $$ e^{\left[i\right]\,\theta} = P^{-1}\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} P \sim \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} $$ This answers all my questions. But it feels like walking on thin ice. So please correct me if I have contributed to more confusion instead of clarification somewhere.

$\endgroup$
  • $\begingroup$ Is what you are showing above simply the isomorphism $\bar z:\mathbb{C}\mapsto\mathbb{C}$? $\endgroup$ – robjohn Feb 14 '15 at 11:02
  • $\begingroup$ @robjohn: "simply", uhm yes, I think that must be so. $\endgroup$ – Han de Bruijn Feb 14 '15 at 11:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.