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I'm practising for my linear algebra exam that is tomorrow by doing old exams. I stumbled in this one question, and got something out of it but I'm not really sure if I did it right.

The problem:

What links multiplication of complex numbers and multiplication of matrix $\begin{bmatrix} a&-b\\ b&a \end{bmatrix}\begin{bmatrix} c&-d\\ d&c \end{bmatrix}=...$ , where $a{,}\ b{,}\ c{,}\ d\in\mathbb{R}$?

My solution:

$\begin{bmatrix} a&-b\\ b&a \end{bmatrix}\begin{bmatrix} c&-d\\ d&c \end{bmatrix}=\begin{bmatrix} ac-bd&-ad-bc\\ ad+bc&ac-bd \end{bmatrix}$

We get $v_1=\begin{bmatrix} ac-bd\\ ad+bc \end{bmatrix}$ and $v_2=\begin{bmatrix} -ad-bc\\ ac-bd \end{bmatrix}$.

Then multiplication with complex numbers:

$\left(a{,}\ b\right)\cdot\left(c{,}\ d\right)$

$=\left(a+bi\right)\cdot\left(c+di\right)$

$=ac+adi+bci+bdi^2\ \parallel i^2=-1$

$=\left(ac-bd\right)+\left(ad+bc\right)i$

$=\left(ac-bd{,}\ ad+bc\right)$

$=v_1$

and

$\left(-b{,}\ a\right)\cdot\left(-d{,}\ c\right)$

$=\left(-b+ai\right)\cdot\left(-d+ci\right)$

$=bd-bci-adi+aci^2\ \parallel i^2=-1$

$=\left(-ac+bd\right)+\left(-ad-bc\right)i $

$=\left(-ac+bd{,}\ -ad-bc\right)$
This should equal to $v_2$ but it doesn't. Why is that, and why is it "flipped" (shouldn't $-ad-bc$ be first and then $-ac+bd$)?

I really appreciate any help :)

linearalgebrayay
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    Compare with this post. $(a,b)\cdot (c,d)=(ac-bd,ad+bc).$ Now replace $a$ by $-b$ etc. – Dietrich Burde Feb 24 '22 at 12:18
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    Factor the matrix as $aI +bJ$ where $I$ is the identity matrix and $J$ has a $1$ and a $-1$ on the antidiagonal and zeroes elsewhere. $I$ will act like the number $1$ and $J$ will act like the number $i$. It can simplify this problem. – CyclotomicField Feb 24 '22 at 12:21
  • See https://math.stackexchange.com/questions/849403/embedding-fields-in-matrix-rings – lhf Feb 24 '22 at 13:02
  • @Ihf So is the idea just to say that because $z=a+ib$ corresponds to a matrix $\begin{bmatrix} a&-b\ b&a \end{bmatrix}$ and $w=c+id$ corresponds to a matrix $\begin{bmatrix} c&-d\ d&c \end{bmatrix}$ so then $(a+ib)(c+id)$ is basically same thing as $\begin{bmatrix} a&-b\ b&a \end{bmatrix}\begin{bmatrix} c&-d\ d&c \end{bmatrix}$? – linearalgebrayay Feb 24 '22 at 13:13

1 Answers1

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Let $I_2$ denote the $2\times2$ identity matrix (which we get with $a=1,\,b=0$), and let $J$ denote the matrix for $a=0,\,b=1$ so $J^2=-I_2$, and in the analogy $a+bi$ is asociated with $aI_2+bJ$. You first verified$$(a+bi)(c+di)=e+fi\iff(aI_2+bJ)(cI_2+dJ)=eI_2+fJ.$$Your manipulation of the rightmost column in each matrix shows we can't similarly associate $aI_2+bJ$ with$$-b+ai=i(a+bi).$$Well, we can with a small modification: if $z^\prime:=iz,\,w^\prime:=iw^\prime,\,(zw)^\prime:=izw$ then$$(zw)^\prime=\underbrace{i^{-1}}_{-i}z^\prime w^\prime,$$whereas if $Z^\prime:=JZ,\,W^\prime:=JW,\,(ZW)^\prime:=JWZ$ then (since all these matrices commute)$$(ZW)^\prime=\underbrace{J^{-1}}_{-J}Z^\prime W^\prime.$$

J.G.
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