# Ambiguous matrix representation of imaginary unit?

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: $$e^{\begin{bmatrix} 1 & -2 \\ 1 & -1 \end{bmatrix}\theta} = \mbox{?} = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$$

• What is wrong ? You do have $i^2=-I$. – Yves Daoust Feb 9 '15 at 15:23
• 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. – Milo Brandt Feb 10 '15 at 23:14
• @Meelo: Exactly what I've been thinking about last night. That explains it all. Thanks! – Han de Bruijn Feb 11 '15 at 14:08
• @Meelo: $$\begin{pmatrix}\cos(t)+\sin(t) & -2\sin(t) \\ \sin(t) & \cos(t)-\sin(t)\end{pmatrix}$$ – robjohn Feb 14 '15 at 11:05

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

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

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.

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.

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