Question

Is it possible for A^2=0 and A is not equal to 0?

Answer 1

I'll admit, I'm being ambiguous on purpose here haha.

Answer 2

\[A=\begin{bmatrix}0&\alpha \\0&0\end{bmatrix}\]

Answer 3

Alright you win, I can't pull a fast one on you for a nilpotent matrix.

Answer 4

https://en.wikipedia.org/wiki/Dual_number

Answer 5

interesting, is that the only matrix form with degree 2 ?

Answer 6

Yeah, including the transpose I think so. It's interesting because dual numbers are like the complex number representation of a matrix, only instead of giving us rotations they give us shearing transformations instead when we multiply them.

Answer 7

so they do stuff like transforming a square into rhombus etc is it

Answer 8

So for contrast to see more vividly: \[a+\epsilon b= a\left[ \begin{array}c 1 & 0\\0 & 1\\\end{array} \right] + b\left[ \begin{array}c 0 & 1\\0 & 0\\\end{array} \right] = \left[ \begin{array}c a & b\\0 & a\\\end{array} \right]\]

\[a+i b= a\left[ \begin{array}c 1 & 0\\0 & 1\\\end{array} \right] + b\left[ \begin{array}c 0 & 1\\-1 & 0\\\end{array} \right] = \left[ \begin{array}c a & b\\-b & a\\\end{array} \right]\]

Answer 9

I just found this now so I don't really know anything much but you can tell by simply multiplying out the rules:

\[(a+b \epsilon)(x+y \epsilon) = ax+(bx+ay)\epsilon\] and the part \(-by\) we would normally have is thrown away which is the difference between the rotation and shear! Interesting!

Answer 10

Okay I have a little experience with transformation matrices, so im trying to relate that with these nilpoint matrices

Consider a 2 dimensional point on plane \(z=1\) : \((x,y,1)\) \[\begin{pmatrix}x'\\y'\\1\end{pmatrix}=\begin{bmatrix} a&\color{red}{c}&e\\\color{red}{b}&d&f\\0&0&1\end{bmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}\]

if i remember correctly those red numbers represent the shear effect

Answer 11

Ahhh perhaps what I've written is confusing for the same reason that if I write out:

\[(a+b i)(x+y i) = (ax-by)+(bx+ay)i\] This represents not only a rotation but a stretching. By the same reasoning the equation below is not only strictly shearing, but also extension at the same time.

\[(a+b \epsilon)(x+y \epsilon) = ax+(bx+ay)\epsilon\]

I'm essentially using their line her to come to this conclusion from the wikipedia article:

http://prntscr.com/7o8mzb

Specifically in my mind the only difference is the real part (x axis) of the complex numbers is shifted to the left by \(yb\) while in the dual numbers we aren't shifting left by \(yb\) and are instead solely moving upwards (and stretching in this case) but in that article they do a pure shear I believe.

Answer 12

Specifically I'm interested in grasping the "geometry" section of the article here: https://en.wikipedia.org/wiki/Dual_number#Geometry which should answer our questions about shearing.

Answer 13

Just playing around right now, I noticed there's an interesting form for powers of dual numbers:

\[d=a+b \epsilon\]
\[d^n = (a+ b \epsilon)^n = a^n + n a^{n-1}b \epsilon\]
Which is just the first two terms of the binomial theorem.

I'm sort of just playing around and trying to figure it out since I don't really understand this but it's interesting and new and might have some interesting uses for solving problems since it's so simple to use.

Answer 14

don't have an A matrix with 0 entires ...then A^2 won't be equal to 0

Answer 15

A^2 = A(A) with matrix multiplication

Answer 16

I'm talking about this matrix specifically: \[\left[ \begin{array}c 0 & 1\\0 & 0\\\end{array} \right] \left[ \begin{array}c 0 & 1\\0 & 0\\\end{array} \right] = \left[ \begin{array}c 0 & 0\\0 & 0\\\end{array} \right]\]

Answer 17

ohhhhhhh

Answer 18

I thought it's not possible. I wasn't thinking about matrices.

Answer 19

nice thats because \(\epsilon^k=0\) for \(k\gt 1\)

Answer 20

In effect it seems as though I've really calculated more impressive looking haha:

\[\left[ \begin{array}c a & b\\0 & a\\\end{array} \right]^n = a^n\left[ \begin{array}c 1 & \frac{nb}{a}\\0 & 1\\\end{array} \right]\]

Answer 21

What about: \[
\delta = \begin{bmatrix}
0&0 \\ 1&0
\end{bmatrix} \implies \delta^2=
\begin{bmatrix}
0&0 \\ 1&0
\end{bmatrix}^2 =\begin{bmatrix}
0&0 \\ 0&0
\end{bmatrix} = 0
\]And we might think: \[
i = \epsilon - \delta
\]

Answer 22

Is there any way to compare dual numbers?

Answer 23

Actually it seems \(\delta = \epsilon^T\), so we would say: \[
i = \epsilon - \epsilon^T
\]

Answer 24

Interesting one application of dual numbers is that they allow you to do automatic differentiation! Which makes sense I think when you look at them.

http://prntscr.com/7o8sqt

So specifically I'm guessing because we have \[(a+ b \epsilon)^n = a^n+na^{n-1} b \epsilon\] we subtract a^n and divide by epsilon (well whatever that means). Since these are not invertible.

I don't know, that's a good question @wio . One thing I'm noticing is we can possibly make a "dual conjugate" which makes the epsilon term negative but what about if we transpose the matrix, that almost seems to correspond to another operation altogether too.

Answer 25

Let's suppose we define \[\epsilon^* = -\epsilon \] \[\epsilon ^T\]and allow us to just put a transpose on it to represent the transposed matrix, there's no other thing we will do.

Then if we take the conjugate transpose of a dual number and add it to itself we get a complex number.

Answer 26

Example:
\[(a+b \epsilon)^* = a-b \epsilon \]\[(a+b \epsilon)^T = a+b \epsilon^T\] Combine these operations to make a "Hermitian" operator:\[(a+ \epsilon)^H = a-b \epsilon^T\]

Now we add:
\[(a+b \epsilon) + (a+ b \epsilon)^H = 2a+bi\]

Hmm... good idea @wio!!!

Answer 27

Interesting thing I've just read is by considering the power series of a function it naturally falls out that:

\[f(a+b \epsilon) = f(a)+b f'(a) \epsilon\]

Answer 28

I wonder if there's a quaternion analog for dual numbers as well

Answer 29

\[
\epsilon^T\epsilon = \begin{bmatrix}
0&0 \\ 1&0
\end{bmatrix}\begin{bmatrix}
0&1 \\ 0&0
\end{bmatrix} = \begin{bmatrix}
0&0 \\ 0&1
\end{bmatrix}
\]Therefore \[
\epsilon \epsilon^T +\epsilon^T\epsilon = 1
\]And we can say: \[
a\epsilon\epsilon^T+b\epsilon+c\epsilon^T+d\epsilon^T\epsilon = \begin{bmatrix}
a&b \\ c&d
\end{bmatrix}
\]From \(\epsilon\) and its transpose, you can get a full \(2\times2\) matrix.

Answer 30

Interesting, we can get to all 4 parts by algebra and kind of throwing away the matrix in a sense. Hmmm this seems more fundamental than complex numbers in a sense.

Answer 31

I don't think there is a single means of generalizing dual numbers. A while back I came up with a way to generalize imaginary numbers that didn't lead to quaternions.

Answer 32

Hmmm well in this light it almost feels as though we have a companion to the "minus sign" and we are using the "transpose" as sort of another minus sign in a different sense here.

Specifically when I said the thing about quaternions I meant the Cayley-wingspanson construction https://en.wikipedia.org/wiki/Cayley%E2%80%93wingspanson_construction

Answer 33

What happens when you do \(\epsilon^0\)? Are we saying it is \(1\) or indeterminate?

Answer 34

When I say more fundamental I also mean that we can see that all rotations are really a specific form of combining two shears. Now that we've decoupled the shears into two parts, it seems like maybe we'd be able to rotate along an ellipse or hyperbola if we desire.

According to the wikipedia article \(\epsilon^0 = 1\) from calculating it here: http://prntscr.com/7o8yrw

Answer 35

I am fine with the idea of interpreting it as the identity matrix I think.

Answer 36

I feel as though instead of previously we were just throwing complex numbers into a matrix because it happened to work, now it seems to be crucial since we are now giving each entry of the matrix a specific role to play in our algebra. One shears in one axis, one shears in the other axis, one stretches in one axis and the other stretches in the other axis. At least I believe this is what we can now understand a 2x2 matrix to represent.

Answer 37

Complex numbers are just the case when the shearing is identical and the stretching on the axes is identical which causes circular motion and uniform stretching that commutes with it. I could be wrong on this but this is my hunch, I don't know, this is the direction I'm sorta being dragged along I guess haha weird stuff.

Answer 38

Makes me wonder what happens if you do something like: \[
a+b\alpha = \begin{bmatrix}
a+b& b \\ 0&a
\end{bmatrix}
\]It looks like \[
(b\alpha)^2= \begin{bmatrix}
b&b \\ 0&0
\end{bmatrix}^2= \begin{bmatrix}
b^2&b ^2\\ 0&0
\end{bmatrix}= b^2\alpha
\]

Answer 39

But might not be closed under multiplication:

Answer 40

Interesting yeah.

\[(aI+b \alpha)^2 = a^2I+2ab \alpha + b^2 \alpha^2\]
We can easily and quickly show \[\alpha^2=\alpha\] so

\[(aI+b \alpha)^2 = a^2I+(2ab + b^2) \alpha\]

Answer 41

I had just computed: \[
(a+b\alpha)(c+d\alpha) = ac+(ad+bc+bd)\alpha
\]

Answer 42

Another thing that I see them saying is that no dual number has an inverse but you can still do division. I think this is essentially the same idea as clifford algebra division of vectors kind of thing but it doesn't exactly translate to matrices as we know them so I'd like to get a better foundation on this operation:

\[\frac{a+b \epsilon}{x+ y \epsilon} = \frac{a+b \epsilon}{x+ y \epsilon} \frac{x- y \epsilon}{x- y \epsilon} = \frac{a}{x} + \frac{ bx-ay}{x^2} \epsilon\]

Answer 43

\[
(a+b\alpha)^n = \sum_{k=0}^n{n\choose k}a^{n-k}(b\alpha)^k = a^n\alpha^0+\sum_{k=1}^n{n\choose k}a^{n-k}b^k\alpha
\]Not sure what \(\alpha^0\) would look like.

Answer 44

I think in general we can say that for matrices \(A^0 = I\)

Answer 45

As long as A isn't the zero matrix. The reasoning for why that would be a suitable definition is just by how we define the exponent rules:

\(A^{n+m} = A^nA^m\)

Answer 46

Actually, that does make sense here: \[
(a+b\alpha)^n = a^n + \left(\sum_{k=1}^n{n\choose k}a^{n-k}b^k\right)\alpha
\]Seems similar to what you got for \(n=2\).

Answer 47

How would you define modulus or magnitude for dual numbers?

Answer 48

I guess it's really asking what your inner product would be.

Answer 49

I think that imaginary numbers would use the determinate for the inner product.

Answer 50

To find out what an inverse dual matrix representation would be, I calculated out multiplying a matrix with values and plugged it in to get:

\[\left[ \begin{array}c \frac{1}{x} & -y\\0 & \frac{1}{x}\\\end{array} \right]\] is the inverse of \[\left[ \begin{array}c x & y\\0 & x\\\end{array} \right]\]

Which supposedly doesn't exist but now I'm curious if multiplying these gives us 1.

Inner product is a good question, I don't know. Since we can turn a dual into a complex number we might be able to just hijack that

Answer 51

But for dual numbers, the determinate seems to give \(\langle a+b\epsilon, a+b\epsilon\rangle = a^2\).

Answer 52

For the alpha concept, it would seem determinate inner product would give \[
\langle a+b\alpha, a+b\alpha\rangle = a^2+ab
\]

Answer 53

I guess I don't know if I even understand these things geometrically yet in the "dual plane". While I can visualize complex number multiplication I can't quite visualize dual multiplication yet, so I guess that's what's to look into now for me.

Answer 54

Hmmm \[(a+b\alpha)^{-1}=\begin{bmatrix}
a+b&b\\0&a
\end{bmatrix}^{-1} =\frac{1}{a^2+ab}\begin{bmatrix}
a&-b\\0&a+b
\end{bmatrix}^{-1} =\frac{(a+b)+-b\alpha}{a^2+ab} \]

Answer 55

So while \(e^{i t} = \cos t + i \sin t\) it appears that we have \(e^{\epsilon t} = 1+ \epsilon t\) which is supposedly some kind of parabolic rotation. I'm not sure, I'm just about to download and skim this paper: http://arxiv.org/abs/0707.4024

Answer 56

See @wio I said earlier I was studying probability and now I'm onto this already haha

Answer 57

Interesting, another alternative is that we can use a different matrix other than i or epsilon to create a hyperbola.

Answer 58

\[
e^{\epsilon t} = \sum_{n=0}^\infty \frac{(\epsilon t)^n}{n!} = 1 + \epsilon t + \sum_{n=2}^\infty \frac{(\epsilon t)^n}{n!}
\]Where \[
\sum_{n=2}^\infty \frac{(\epsilon t)^n}{n!} = \sum_{k=0}^\infty \frac{(\epsilon t)^{k+2}}{(k+2)!} = \sum_{k=0}^\infty \frac{0\cdot (\epsilon t)^k}{(k+2)!} = 0
\]

Answer 59

All of the shapes traced out by the "rotations" are defined by this single function:

\[x^2-i^2y^2=1\]

So if i is the traditional imaginary number we have

\[x^2+y^2=1\]

if i is our new dual epsilon we have:

\[x^2=1\]

and if we have this other alternative hyperbolic i we have

\[x^2-y^2=1\]

I wonder what matrix they're using for this hyperbolic i that squares to 1 but isn't 1.

Answer 60

Probably \[\left[ \begin{array}c 0 & 1\\1 & 0\\\end{array} \right]\]

Answer 61

This is illuminating: http://prntscr.com/7o9ctw

Answer 62

One way to generalize dual numbers: \[
\epsilon_3 = \begin{bmatrix}
0&1&0 \\ 0&0&1 \\ 0&0&0
\end{bmatrix}, (\epsilon_3)^2 = \begin{bmatrix}
0&0&1 \\ 0&0&0 \\ 0&0&0
\end{bmatrix}
\]We have: \[
(\epsilon_3)(\epsilon_3)= (\epsilon_3)^2 \\
(\epsilon_3)^3 = (\epsilon_3)^2(\epsilon_3) = 0
\]This is similar to my previous generalization of imaginary numbers, where tried to start with an identity: \[
(i_3)^3 = -1
\]It turned out that \(i_3\) was just a 6th of a rotation around the unit circle, if you projected onto the imaginary plane.

Answer 63

complex plane

Answer 64

However, the 'quaternion' approach to it might be more along the lines of: \[
\epsilon = \begin{bmatrix}
0&0&1&1 \\ 0&0&1&1 \\ 0&0&0&0 \\ 0&0&0&0
\end{bmatrix}
\]Or \[
\epsilon = \begin{bmatrix}
0&0&0&1 \\ 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0
\end{bmatrix}
\]

Answer 65

The amazing parabolic sine and cosine functions lol, and they satisfy the hilarious
\[e^{\epsilon t} = cosp(t) + \epsilon sinp(t) = 1+\epsilon t\]
\[sinp(x) = x=\frac{e^{\epsilon x}-e^{-\epsilon x}}{2 \epsilon} \]\[cosp(x)=1=\frac{e^{\epsilon x}+e^{-\epsilon x}}{2 }\]

\[\frac{d}{dx} sinp(x) = cosp(x)\]\[\frac{d}{dx} cosp(x) = 0\]

So at least it feels natural in a stupid way.

Answer 66

Ah interesting I see what you're saying. I think I am sort of seeing how pauli spin matrices might just be how we write quaternions using matrices and complex numbers or something, I'm not sure.

Answer 67

Out of curiosity can we extend Fermat's Last theorem to matrices?
\[A^n+B^n=C^n\]

Answer 68

\[
e^{\alpha t} = \sum_{n=0}^\infty \frac{(\alpha t)^n}{n!} = \sum_{n=0}^\infty \frac{\alpha^n t^n}{n!} = 1+ \sum_{n=1}^\infty \frac{\alpha t^n}{n!}
\]We can can start that \(n\) back at zero by adding and subtracting \(\alpha\):\[
1+ \sum_{n=1}^\infty \frac{\alpha t^n}{n!} + \alpha t^0 - \alpha t^0 = 1 - \alpha + \alpha\sum_{n=0}^\infty \frac{t^n}{n!} = 1-\alpha + \alpha e^{t}
\]In summary:\[
e^{\alpha t} =1-\alpha + \alpha e^{t}
\]

Answer 69

This is very illuminating to me since the difference between these three matrices is simply the bottom left value being -1, 0 or 1. http://prntscr.com/7o9h0g It almost seems like we should consider all of these simultaneously with a value in the bottom left that we can change spaces with.

Answer 70

Which could also be said as: \[
e^{\alpha t} = 1 + (e^t-1)\alpha
\]So we could say: \[
\cos\alpha(t) = 1 \\
\sin\alpha(t) = e^t-1 \\
\]Very weird result.

Answer 71

Well, not completely sure about that last one but

Answer 72

\[\left[ \begin{array}c 0 & 1\\s & 0\\\end{array} \right]^2 = s\left[ \begin{array}c 1& 0\\0 & 1\\\end{array} \right]\]

So we have this matrix where the bottom left entry is the value of the square. 0, 1, or -1.

How are you coming to that result wio?

Answer 73

I am just using the weird identity: \[
e^{\alpha t} = \cos_\alpha(t) + \alpha \sin_\alpha(t) = 1+(e^t-1)\alpha
\]

Answer 74

Though I'm not sure what the underlying assumptions of that identity are.

Answer 75

It's more of a definition though, so...

Answer 76

well as long as you define sine and cosine by this structure:
\[sinp(x) = x=\frac{e^{\epsilon x}-e^{-\epsilon x}}{2 \epsilon} \]\[cosp(x)=1=\frac{e^{\epsilon x}+e^{-\epsilon x}}{2 }\]

I think you will in a good position.

Answer 77

I'll try to play around with the alpha values a bit to see if I get to it as well

Answer 78

That structure? Why?

Answer 79

It produces an even and an odd function that add to the exponential function because:

\[f(x) = \frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}=g(x)+h(x)\] and we have the identities that g(x)=g(-x) and h(x)=-h(-x) like we expect from a sine or cosine.

Answer 80

Also I'm noticing we might be able to generalize your matrix a bit:

\[\left[ \begin{array}c s & 1\\0 & 0\\\end{array} \right]^2=s\left[ \begin{array}c s & 1\\0 & 0\\\end{array} \right]\] or in other words \[\alpha_s^2=s \alpha_s\] So you have some control over what the square is, you can make s=0, 1, or -1 for different interesting but also easier to track properties. Something to play with.

Answer 81

Let's see \[
e^{-\alpha t} = 1 + (e^{-t}-1)\alpha
\]So \[
e^{\alpha t}+e^{-\alpha t} = 2 + \alpha\bigg(e^t-1 + e^{-t}-1\bigg) = 2+\alpha (e^t+e^{-t}-2)
\]So we have: \[
\frac{e^{\alpha t}+e^{-\alpha t} }{2} = 1+\cosh(t)
\]For the other one: \[
e^{\alpha t}- e^{-\alpha t} =\alpha\bigg(e^t - e^{-t}\bigg) =
\]So \[
\frac{e^{\alpha t}- e^{-\alpha t}}{2\alpha} = \sinh(t)
\]

Answer 82

Hmmm, the only issue thought is that I know division by \(\alpha\) is problematic, just as it is for \(\epsilon\).

Answer 83

It is weird I'll agree with you hmm.\[\frac{a+b \alpha}{x+ y \alpha} = \frac{a+b \alpha}{x+ y \alpha} \frac{a-b \alpha}{x- y \alpha} = \frac{ax}{x^2-y^2} + \alpha \frac{bx-ay-by}{x^2-y^2}\]

Answer 84

For \(a+b\epsilon\) we have division problems at \(a=0\) (regardless of \(b\)), and for \(a+b\alpha\) we have division problems at \(a=-b\). You could say that for complex we have division problems at \(a^2+b^2=0\) for obvious reasons.

Answer 85

Yeah, I am not sure how I feel about "dividing by a matrix" but somehow it seems to just sorta work even though it definitely has problems going on.

Answer 86

I think we have to say that at these places, the magnitude ought to be defined as \(0\).

Answer 87

Everything we have done so far can be generalized as "Using a \(2\times 2\) matrix to define multiplication for a \(2\) component vector."

Answer 88

Hahaha yeah basically.

Answer 89

But I think we have some interesting relationship that the matrices satisfy a relation like this:

\[A^2=sA\]
or
\[A^2=sI\]

So there's some sort of structure here. Hmmm. It's sorta all nonsense in a way though unless we can get something out of it I think.

Answer 90

We could even do Hadamard matrix for this: \[
a+bH = \begin{bmatrix}
a+b & a+b \\ a+b & a-b
\end{bmatrix}
\]It appears that \(H^2 = 2I\), so: \[
(a+bH )(c+dH ) = (ac+2bd)+(ad+bc)H
\]

Answer 91

That one quantum computing tutorial said something about Unitary matrices or something being special.

Answer 92

Oh interesting. I am sort of getting a new idea from this.

Yeah a unitary matrix is basically just the complex version of an orthogonal matrix. An orthogonal matrix is the inverse of its transpose while a unitary matrix is the inverse of its hermitian, which is the conjugate transpose. The DFT uses this for instance.

Answer 93

\[M=\left[ \begin{array}c a & b\\c & d\\\end{array} \right]=a\left[ \begin{array}c 1 & 0\\0 & 0\\\end{array} \right]+b\left[ \begin{array}c 0 & 1\\0 & 0\\\end{array} \right]+c\left[ \begin{array}c 0 & 0\\1 & 0\\\end{array} \right]+d\left[ \begin{array}c 0 & 0\\0 & 1\\\end{array} \right]\]

\[M=aA+bB+cC+dD\]

So now if I square this I'm going to get a more in depth understanding of how each part influences the square rather than looking at our specific cases like we've been doing.

Answer 94

\[M= \left[ \begin{array}c A & B & C & D\\\end{array} \right]\left[ \begin{array}c a\\ b\\ c\\ d\\\end{array} \right]\] So now I've got a matrix of matrices.... Maybe this isn't the right path to take on this haha. Nevermind.

Answer 95

\[
H^n = \begin{cases}
2^{k}I & n=2k\\
2^{k}H & n=2k+1
\end{cases}
\]I think the power of \(n\) is important, since we can use it in the MacLaurin series of \(e^x\).

Answer 96

Yeah I think you're right about that. Splitting it up like you've done is also going to make it nice for thinking about the sine and cosine functions

Answer 97

\[
e^{tH} = \sum_{n=0}^\infty \frac{(tH)^n}{n!} = \sum_{k=0}^\infty \frac{t^{2k}}{(2k)!} + H\sum_{k=0}^\infty \frac{t^{2k+1}}{(2k+1)!}
\]Hmmm, not quite trignometric, since we don't have the \(-1\) coefficient.