Not likely. Hungarians are top Math people. He would probably blacklist him if he asked a simple question.No one said it won't be a boring solution. At least it is one which doesn't depend on assumptions made by others.
Not likely. Hungarians are top Math people. He would probably blacklist him if he asked a simple question.No one said it won't be a boring solution. At least it is one which doesn't depend on assumptions made by others.
As he spoke about absorbing elements, I couldn't but think of the world of algebras, since in groups, there are no such elements. But as soon as we talk about algebras, we are in the additive world. That doesn't imply, that ##\{\,M\,|\,[G,M]=0\,\}## hasn't an additional multiplicative structure, however, not a multiplicative group, an additive one (subspace) with additional multiplicative properties (e.g. ideal). So all in all, the additive component doesn't make it trivial, rather does it fit better to the rest of what has been given, than automatically make hidden assumptions. (Sorry, I'm really no friend of reading more out of a post than what is actually given. The question about the operation was the first we have asked, which was basically answered by: any.)Not likely. Hungarians are top Math people. He would probably blacklist him if he asked a simple question.
He already said it:Is this a class homework problem? It sounds like one. If so, you need to show some work and follow the template.
In any case, rather than getting caught up in a lot of discussions, you should go one-by-one through the required properties of a group and see if you can prove them or not. If you have problems, then you should ask more specific questions.
Since we cannot talk to his professor, we can only take what we have, and there is a solution with a "yes" as answer.I've been learning linear algebra for one month only, so I might not see your point immidiately. My teacher mentioned this problem and I found it interesting.
Yes, I assumed so but did not feel like looking to see if he eventually said it. But that should not excuse him from not not following the template and showing his work. The fact remains that he should go one-by-one through the required properties and ask a specific question if he has a problem. @phyzguy said that in post #2, but that appears to have been ignored.He already said it:
There's something not right here. How can we need ##\delta = 0##? The identity matrix, for which ##\delta = 1##, commutes with everything. And, how can we have "##M## cannot be equal to ##G##"? By your own analysis ##G## always commutes with itself.I started with simple 2x2 matrices. Let's say we have ## G = \begin{pmatrix} a & b \\c & d \end{pmatrix}##, where ##a,b,c,d \in \mathbb{R}##.
We're looking for all possible ## M = \begin{pmatrix} \alpha & \beta \\\gamma & \delta \end{pmatrix}##, ##\alpha,\beta,\gamma,\delta \in \mathbb{R}## such that ##GM = MG##
I did both multiplications, set up a system of equations and solved it.
##\alpha = \frac{\gamma (a-d)}{c}##
##\beta = \frac{\gamma b}{c}##
##\gamma \in \mathbb{R}##
##\delta = 0##
These ##M## matrices form a group with addition. Now what I don't know is how to generalize this to ##n \times n## matrices and I'm not sure if they form a group with other operations too. They don't with multiplication.
Also ##M## cannot be equal to ##G## but ##GG=GG## is necessarily true.
##(M-N)\cdot G - G\cdot (M-N) = GN-NG##. But what is ##N##? Element of ##X## or an arbitrary matrix?Given any matrix ##G## and a set ##X:=\{\,M\,|\,MG=GM\,\}##. Now what is ##(M-N)\cdot G - G\cdot (M-N)## and what does this mean for the properties of ##X\,?##
You're right. I was too tired last night so I used a linear equation solver and it seems it has some bugs when it comes to dealing with parameters. I solved it on my own and it seems I got it right this time.There's something not right here. How can we need ##\delta = 0##? The identity matrix, for which ##\delta = 1##, commutes with everything. And, how can we have "##M## cannot be equal to ##G##"? By your own analysis ##G## always commutes with itself.
I'd still question what happens if ##b = 0## in those equations.##(M-N)\cdot G - G\cdot (M-N) = GN-NG##. But what is ##N##? Element of ##X## or an arbitrary matrix?
You're right. I was too tired last night so I used a linear equation solver and it seems it has some bugs when it comes to dealing with parameters. I solved it on my own and it seems I got it right this time.
##\alpha \in \mathbb{R}##
##\beta \in \mathbb{R} ##
##\gamma = \beta \frac{c}{b}##
##\delta = \beta\frac{d-a}{b}+\alpha##
These matrices form a group too, and ##I,G## are also in it.
How to generalize this to nxn matrices?
Both have been meant to commutate with ##G##, so ##M,N\in X##.##(M-N)\cdot G - G\cdot (M-N) = GN-NG##. But what is ##N##? Element of ##X## or an arbitrary matrix?
Forget about the particulars of matrices. What you want to prove holds for any mathematical system that obeys the associative law (posts #12, #32)How to generalize this to nxn matrices?
But this too seems problematic. If ##\beta ## is in ##\mathbb R## then ##\gamma ## can be anything.##(M-N)\cdot G - G\cdot (M-N) = GN-NG##. But what is ##N##? Element of ##X## or an arbitrary matrix?
You're right. I was too tired last night so I used a linear equation solver and it seems it has some bugs when it comes to dealing with parameters. I solved it on my own and it seems I got it right this time.
##\alpha \in \mathbb{R}##
##\beta \in \mathbb{R} ##
##\gamma = \beta \frac{c}{b}##
##\delta = \beta\frac{d-a}{b}+\alpha##
These matrices form a group too, and ##I,G## are also in it.
How to generalize this to nxn matrices?