Matrices and linear transformations.

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Every m x n matrix defines a linear transformation from k^n to k^m through left multiplication. While matrices can be applied in contexts outside of linear maps, they inherently correspond to linear transformations when a basis is established. The discussion highlights the distinction between matrices operating within vector spaces and those applied to non-vector space contexts, where linearity may not hold. The conversation also touches on the nature of fields and the implications for defining linear transformations, particularly when elements do not belong to a field. Ultimately, the consensus is that matrices can represent linear transformations, but this is contingent upon the context and structure imposed on the elements involved.
  • #31
Studiot said:
\left[ {\begin{array}{*{20}{c}}<br /> 0 &amp; 1 &amp; 0 \\<br /> \end{array}} \right] \, \left[ {\begin{array}{*{20}{c}}<br /> \textbf{{blue}} \\<br /> \textbf{{red}} \\<br /> \textbf{{green}} \\<br /> \end{array}} \right] = red

I'm sorry, what is this supposed to mean?! Specifically, how do you define:

<br /> x \times \left( \mathrm{blue}/\mathrm{red}/\mathrm{green} \right) = ?<br />
where x \in \mathbb{R}.
 
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  • #32
Dickfore said:
I'm sorry, what is this supposed to mean?! Specifically, how do you define:

<br /> x \times \left( \mathrm{blue}/\mathrm{red}/\mathrm{green} \right) = ?<br />
where x \in \mathbb{R}.

Well, x can only be 0 or 1. So he defines 1\cdot \mathrm{color}=\mathrm{color} and 0\cdot \mathrm{color}=0. And further, he sets 0+\mathrm{color}=\mathrm{color}.

This is a completely consistent algebraic structure, but it of course has nothing to do with usual linear algebra.
 
  • #33
micromass said:
Well, x can only be 0 or 1. So he defines 1\cdot \mathrm{color}=\mathrm{color} and 0\cdot \mathrm{color}=0.
I guess you/he are/is trying to define multiplication of a vector (in the Hilbert space \left\lbrace \mathrm{red}, \mathrm{green}, \mathrm{blue} \right\rbrace) by a scalar. However, the set since the set \left\lbrace 0, 1 \right\rbrace does not form a field, since 1 does not have an additive inverse in this set. You need to extend it with at least -1. Then, we need to define -1 \times \mathrm{red}/\mathrm{green}/\mathrm{blue}.

micromass said:
And further, he sets 0+\mathrm{color}=\mathrm{color}.
I'm afraid I don't understand this part! You are adding a scalar with a vector! This is not defined in a vector space.

You need to define:
<br /> \begin{array}{lcccl}<br /> \mathrm{red} &amp; + &amp;\mathrm{red} &amp; = &amp; ? \\<br /> <br /> \mathrm{red} &amp; + &amp;\mathrm{green} &amp; = &amp; ? \\<br /> <br /> \mathrm{red} &amp; + &amp;\mathrm{blue} &amp; = &amp; ? \\<br /> <br /> \ldots<br /> \end{array}<br />

micromass said:
This is a completely consistent algebraic structure, but it of course has nothing to do with usual linear algebra.
So, it has been shown that this algebraic structure is not a linear space. Therefore, it does not support linear algebra.
 
  • #34
Dickfore said:
I guess you/he are/is trying to define multiplication of a vector (in the Hilbert space \left\lbrace \mathrm{red}, \mathrm{green}, \mathrm{blue} \right\rbrace) by a scalar.

What makes you think that \{\mathrm{red},\mathrm{green},\mathrm{blue}\} is a Hilbert space :confused:

However, the set since the set \left\lbrace 0, 1 \right\rbrace does not form a field, since 1 does not have an additive inverse in this set. You need to extend it with at least -1. Then, we need to define -1 \times \mathrm{red}/\mathrm{green}/\mathrm{blue}.

Well, we could set 1=-1 and obtain the field \mathbb{Z}_2. But that is not the point.

I'm afraid I don't understand this part! You are adding a scalar with a vector! This is not defined in a vector space.

The point is that Studiot wants to do something completely different from vector spaces. There are some relations with vector spaces as matrices multiply the same way, but the rest is completely different.

You need to define:
<br /> \begin{array}{lcccl}<br /> \mathrm{red} &amp; + &amp;\mathrm{red} &amp; = &amp; ? \\<br /> <br /> \mathrm{red} &amp; + &amp;\mathrm{green} &amp; = &amp; ? \\<br /> <br /> \mathrm{red} &amp; + &amp;\mathrm{blue} &amp; = &amp; ? \\<br /> <br /> \ldots<br /> \end{array}<br />

He does not need to define this as those additions will never show up in practice. Again, nobody claims that \{\mathrm{red},\mathrm{green},\mathrm{red}\} forms a vector space.
 
  • #35
micromass said:
What makes you think that \{\mathrm{red},\mathrm{green},\mathrm{blue}\} is a Hilbert space :confused:
Then, what kind of structure do they form. I am certainly not aware of such a structure.

micromass said:
Well, we could set 1=-1 and obtain the field \mathbb{Z}_2. But that is not the point.
Ok, that makes sense.


micromass said:
The point is that Studiot wants to do something completely different from vector spaces. There are some relations with vector spaces as matrices multiply the same way, but the rest is completely different.

What does he want to do, exactly?!
 
  • #36
Thank you all.
I can see, from the confusion expressed by several, that my initial statement may have been taken to read that I propose either that all matrices represent linear transformations or that I wanted to create a vectors tructure.

Neither were the case.

I apologise if I inadvertantly created a false impression, but I thought I has specified matters pretty clearly.

@Dickfore

Thank you for your thoughts.

So, it has been shown that this algebraic structure is not a linear space. Therefore, it does not support linear algebra.

Yes indeed.

However it is representable by matrices.

I guess you/he are/is trying to define multiplication of a vector (in the Hilbert space {red,green,blue}) by a scalar.

Why guess?

Why not just read the definition I gave?

I accept that if I wanted to create a vector space I would have to impose different criteria/structure. But I don't.
Further my scheme is capable of further development.

@Robert

Thank you for your further thoughts.
You still have not indicated what colours are possible for a monocoloured ball.

@TrickyDicky

Thank you for your support.
This thread was created because an apparently very competent mathematician was adament, several times, that all matrices represent linear transformations, full stop. I did not want to disturb the other thread (referenced) with the discussion.
 
  • #37
Dickfore said:
Then, what kind of structure do they form. I am certainly not aware of such a structure.


Ok, that makes sense.




What does he want to do, exactly?!

He gave an example of a certain type of matrix multiplication of specific types of matrices, such that this matrix multiplication was not just a matrix representation of a linear transformation. I think his initial intent was to provide a counter example to the claim "all matrices are linear transformations" - which he did. However, as micromass pointed out, this has nothing to do with ordinary linear algebra (which is probably the context in which the claim was first made.)
 
  • #38
I believe the point of the thread has already been addressed.
But I find it interesting to ramble on a bit.
Just like the others here apparently. ;)

I've been trying to deduce a algebraic structure here that is consistent and complete.
I'm concluding that the actual dataset of colors is ##\{nocolor, red, green, blue\}##.
Let's call it ##C##.

##C## supports a partial operation ##+##, which only defines addition of ##nocolor## with any of the other colors.
Note that in algebra and model theory, the definition of an operation includes that all combinations should be defined, which is not the case here.

Furthermore, ##1 \cdot color=color## and ##0 \cdot color=nocolor## (as mm said).

Then the matrix multiplication would be a map ##F_2^{1 \times 3} \times C^3 \to C##.
The only reason that this is not a linear map, is because ##+## is not defined for all combinations of elements in ##C##.
This restricts the allowed matrices to the ones that have at most one ##1## in them.
So there are 4 allowed matrices.
 
  • #39
Dickfore said:
Then, what kind of structure do they form. I am certainly not aware of such a structure.

They form just a set.

What he wants is just to define an algebraic structure where you can do something like

\left(\begin{array}{cc} 0 &amp; 1\end{array}\right)\left(\begin{array}{c} red\\ green\end{array}\right)=green

So we want a structure that is strong enough such that things like

0\cdot red+1\cdot green=green

make sense. There are multiple ways to formalize such a structure, but it should be intuitively clear what it does. Clear is that it has little to do with vector spaces or usual linear algebra.

Developing such a system might be interesting in programming and it would not be hard to implement it. However, I don't see much mathematical uses for it.
 
  • #40
Well, I guess you can never be wrong if you are intentionally too vague.
 
  • #41
Dickfore said:
Well, I guess you can never be wrong if you are intentionally too vague.

Was that directed to me?? I agree that I might be too vague, but why do you think it was intentional?
 
  • #42
micromass said:
Was that directed to me??

No, it was directed to the OP. I only saw there were two new posts after I had posted.
 
  • #43
Dickfore said:
Then, what kind of structure do they form. I am certainly not aware of such a structure.

micromass said:
Developing such a system might be interesting in programming and it would not be hard to implement it. However, I don't see much mathematical uses for it.

Look at differential geometry and what chiro said in post #5.
 
  • #44
What is this
<br /> \left[\begin{array}{c}<br /> \mathrm{blue} \\<br /> \mathrm{red} \\<br /> \mathrm{green}<br /> \end{array}\right] + \left[\begin{array}{c}<br /> \mathrm{green} \\<br /> \mathrm{blue} \\<br /> \mathrm{red}<br /> \end{array}\right] = ?<br />
equal to?
 
  • #45
Dickfore said:
What is this
<br /> \left[\begin{array}{c}<br /> \mathrm{blue} \\<br /> \mathrm{red} \\<br /> \mathrm{green}<br /> \end{array}\right] + \left[\begin{array}{c}<br /> \mathrm{green} \\<br /> \mathrm{blue} \\<br /> \mathrm{red}<br /> \end{array}\right] = ?<br />
equal to?

I don't think studiot wants to define addition of matrices.
 
  • #46
Dickfore said:
What is this
<br /> \left[\begin{array}{c}<br /> \mathrm{blue} \\<br /> \mathrm{red} \\<br /> \mathrm{green}<br /> \end{array}\right] + \left[\begin{array}{c}<br /> \mathrm{green} \\<br /> \mathrm{blue} \\<br /> \mathrm{red}<br /> \end{array}\right] = ?<br />
equal to?

It is undefined, since + on real colors is undefined.

However, making things consistent and complete, we might say:
<br /> \left[\begin{array}{c}<br /> \mathrm{nocolor} \\<br /> \mathrm{nocolor} \\<br /> \mathrm{green}<br /> \end{array}\right] + \left[\begin{array}{c}<br /> \mathrm{nocolor} \\<br /> \mathrm{blue} \\<br /> \mathrm{nocolor}<br /> \end{array}\right] = \left[\begin{array}{c}<br /> \mathrm{nocolor} \\<br /> \mathrm{blue} \\<br /> \mathrm{green}<br /> \end{array}\right]<br />
 
  • #47
Dickfore said:
What is this
<br /> \left[\begin{array}{c}<br /> \mathrm{blue} \\<br /> \mathrm{red} \\<br /> \mathrm{green}<br /> \end{array}\right] + \left[\begin{array}{c}<br /> \mathrm{green} \\<br /> \mathrm{blue} \\<br /> \mathrm{red}<br /> \end{array}\right] = ?<br />
equal to?
If studiot said explicitly he wasn't defining a vector space , why would you think that addition should be defined?
 
  • #48
TrickyDicky said:
If studiot said explicitly he wasn't defining a vector space , why would you think that addition should be defined?

Indeed Studiot never defined addition.
It's what we are inferring to fit the matrix multiplication in the framework we are used to.
And indeed, it is possible.
 
  • #49
TrickyDicky said:
If studiot said explicitly he wasn't defining a vector space , why would you think that addition should be defined?

http://en.wikipedia.org/wiki/Matrix_(mathematics)

Matrices of the same size can be added or subtracted element by element.
 
  • #50
Also, what is the meaning of:
<br /> \left[\begin{array}{ccc}<br /> 0 &amp; 1 &amp; 1<br /> \end{array}\right] \, \left[\begin{array}{c}<br /> \mathrm{blue} \\<br /> \mathrm{red} \\<br /> \mathrm{green}<br /> \end{array}\right] = ?<br />
 
  • #51
Dickfore said:
Also, what is the meaning of:
<br /> \left[\begin{array}{ccc}<br /> 0 &amp; 1 &amp; 1<br /> \end{array}\right] \, \left[\begin{array}{c}<br /> \mathrm{blue} \\<br /> \mathrm{red} \\<br /> \mathrm{green}<br /> \end{array}\right] = ?<br />

Such a multiplication would not be defined. The only multiplication that would be defined is if the left matrix has exactly one 1 on each row.
 
  • #52
Dickfore said:
http://en.wikipedia.org/wiki/Matrix_(mathematics)

That article does not say that addition has to be defined.
A matrix is just a rectangular set of symbols.
Furthermore, it allows to impose more mathematical structure to it, but that is not part of the definition of a matrix.

Dickfore said:
Also, what is the meaning of:
<br /> \left[\begin{array}{ccc}<br /> 0 &amp; 1 &amp; 1<br /> \end{array}\right] \, \left[\begin{array}{c}<br /> \mathrm{blue} \\<br /> \mathrm{red} \\<br /> \mathrm{green}<br /> \end{array}\right] = ?<br />

Explicitly undefined.
 
  • #53
So, nothing is defined. What we have are "red", "green", and "blue" by themselves. No need of matrices here.
 
  • #54
Dickfore said:
http://en.wikipedia.org/wiki/Matrix_(mathematics)

I'm not sure if that requirement is already referring to linear algebra matrices, but in any case it would be up to Studiot to define the result of that addition. I don't think he is going to develope a whole consistent non-linear system just to make his point which is already clear.
 
  • #55
Dickfore said:
So, nothing is defined. What we have are "red", "green", and "blue" by themselves. No need of matrices here.

Well, I certainly agree that there is no need of matrices here. But apparently studiot wanted them for some reason.
 
  • #56
But, they are not matrices, since they don't fit the usual definition of matrices. They are some weird notation of a row containing only a single one (not a row vector1) that picks out a "red", "green", and "blue" from another column array (not a column vector!) by some weird "rule" that 'Studiot' thought fitting to write.
 
  • #57
The matrix is designed as a notation to "pick" a color from an ordered set.

Studiot's point/question was that not every matrix defines a linear transformation.
And that is true, when you take it out of the context of linear algebra.
 
Last edited:
  • #58
Dickfore said:
But, they are not matrices, since they don't fit the usual definition of matrices. They are some weird notation of a row containing only a single one (not a row vector1) that picks out a "red", "green", and "blue" from another column array (not a column vector!) by some weird "rule" that 'Studiot' thought fitting to write.

Well, I guess this depends on your definition of matrix. In my point of view, a matrix is just a rectangular array with certain entries. In that respect, he certainly did define matrices. But they are kind of useless since almost no operation is defined.
 
  • #59
Dickfore said:
But, they are not matrices, since they don't fit the usual definition of matrices. They are some weird notation of a row containing only a single one (not a row vector1) that picks out a "red", "green", and "blue" from another column array (not a column vector!) by some weird "rule" that 'Studiot' thought fitting to write.

I already gave an example of a nonlinear matrix in #25, the 2X2 invertible complex matrix, to which you can associate a Mobius transformation, you can make all kinds of computations with it, and they fit the definition of matrix.
 
  • #60
micromass said:
Well, I guess this depends on your definition of matrix. In my point of view, a matrix is just a rectangular array with certain entries. In that respect, he certainly did define matrices. But they are kind of useless since almost no operation is defined.
Well, your definition of a matrix is certainly not the one commonly accepted by a math community.
 

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