Why does A squared not equal A times A when k = Z2?

[Karl Karlsson]

TL;DR

Let $$A=\begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}$$ with $k=\mathbb{Z}_2$ I think k is the set of scalars for a vector that can be multiplied with the matrix A (I could definitely be wrong). Then for some reason $A^2= \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$ this is not the same as $A\cdot A$. Why? What has happened?

In my book no explanation for this concept is given and i can't find anything about it when I am searching. One example that was given was:
Let $$A=\begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}$$ with $k=\mathbb{Z}_2$ I think k is the set of scalars for a vector that can be multiplied with the matrix A (I could definitely be wrong). Then for some reason $A^2= \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$ this is not the same as $A\cdot A$. Why? What has happened?

 

[fresh_42]

$A^2=A\cdot A$. You only have to apply the multiplication given by the scalar field $\mathbb{Z}_2$ where $1\cdot 1+1\cdot 1=0.$

 

[Karl Karlsson]

$A^2=A\cdot A$. You only have to apply the multiplication given by the scalar field $\mathbb{Z}_2$ where $1\cdot 1+1\cdot 1=0.$

Hi! What do you mean multiplication given by the scalar field $\mathbb{Z}_2$ where $1\cdot 1+1\cdot 1=0$? Why is not $1\cdot 1+1\cdot 1=2$ ? I have never seen this before

 

[PeroK]

Look up $\mathbb{Z}_2$. This is the two-element field consisting of $\{0, 1 \}$ only, with arithmetic operations defined modulo 2.

 

[fresh_42]

Hi! What do you mean multiplication given by the scalar field $\mathbb{Z}_2$ where $1\cdot 1+1\cdot 1=0$? Why is not $1\cdot 1+1\cdot 1=2$ ? I have never seen this before

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

$\mathbb{Z}_p=\mathbb{Z}/p\cdot\mathbb{Z}$ which means that all numbers which differ by multiples of $p$ are considered equivalent. Hence we only have the representants of the equivalence classes as elements which are the remainders of the division by $p\, : \,0,1,2,\ldots,p-1$.

I told you in another thread to consider $\mathbb{Z}_{12}=\mathbb{Z}/12\cdot\mathbb{Z}$ in order to practice quotient building. $\mathbb{Z}_{12}$ isn't a field as $\mathbb{Z}_{2}$ is, but this doesn't matter, since we do not divide anywhere. The reason why I suggested $\mathbb{Z}_{12}$ is, that you have it on the clock!

 

[martinbn]

Sorry for the off topic comment, but I couldn't resist. I don't like the notation $\mathbb Z_p$ for the field of $p$ elements, because this is the standard notation for the ring of $p$-adic integers, and for the field of $p$ elements there are already other good notations.

 

[fresh_42]

Sorry for the off topic comment, but I couldn't resist. I don't like the notation $\mathbb Z_p$ for the field of $p$ elements, because this is the standard notation for the ring of $p$-adic integers, and for the field of $p$ elements there are already other good notations.

How do you notate $\mathbb{Z}_{12}$? As quotient, in which case I demand you to write $\mathbb{C}$ as $\mathbb{R}[x]/(x^2+1)$, or do you only oppose the prime field notation $\mathbb{Z}_p$, in which case you have two notations for the same class of rings: $\operatorname{GF}(2)$ and $\mathbb{Z}_{12}$? This is way more confusing than p-adic numbers and finite prime fields are, since they are rarely used side by side.

 

[martinbn]

How do you notate $\mathbb{Z}_{12}$? As quotient, in which case I demand you to write $\mathbb{C}$ as $\mathbb{R}[x]/(x^2+1)$, or do you only oppose the prime field notation $\mathbb{Z}_p$, in which case you have two notations for the same class of rings: $\operatorname{GF}(2)$ and $\mathbb{Z}_{12}$? This is way more confusing than p-adic numbers and finite prime fields are, since they are rarely used side by side.

No need to get upset, I only said what I don't like. Everyone else can like or use whatever he wants.

 

[Mark44]

Hi! What do you mean multiplication given by the scalar field $\mathbb{Z}_2$ where $1\cdot 1+1\cdot 1=0$? Why is not $1\cdot 1+1\cdot 1=2$ ? I have never seen this before

Look up $\mathbb{Z}_2$. This is the two-element field consisting of $\{0, 1 \}$ only, with arithmetic operations defined modulo 2.

To elaborate, here are the arithmetic operations in $\mathbb Z_2$.
Addition
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0
Multiplication
0 * 0 = 0
0 * 1 = 0
1 * 0 = 0
1 * 1 = 1

Therefore, 1 * 1 + 1 + 1 = 1 + 1 = 0

 

[Karl Karlsson]

Thanks for the answers! I have since found more info about $Z_2$ even though there were no info about it in my book

 

[fresh_42]

Thanks for the answers! I have since found more info about $Z_2$ even though there were no info about it in my book

$\mathbb{Z}_2$ is a light switch: on and off. If you switch it twice it's off again. The same principle every computer is based on. $\mathbb{Z}_{12}$ is the clock.

 

[Mark44]

$\mathbb{Z}_{12}$ is the clock.

But not quite. In $\mathbb{Z}_{12}$ there are the equivalence classes {0}, {1}, {2}, {3}, ..., {11}. So 11 + 1 = 0 (mod 12), but on the clock, 11:00 + 1 hour = 12:00.

 

[fresh_42]

$12:00 = 0:00.$ We say null o'clock (= midnight). But even if you insist on the usage of twelve o'clock, then simply change the equivalence classes to $\{1\},\{2\},\{3\},\{4\},\{5\},\{6\},\{7\},\{8\},\{9\},\{10\},\{11\},\{12\}$ with $\{12\}$ as neutral element. $\mathbb{Z}_{12}$ is exactly the clock. O.k., the hours on the clock.