# Matrices and linear transformations.

1. Oct 13, 2012

### Studiot

This thread is posted to examine the proposition that all matrices define linear transformations.

But what of the matrix equation?

$$\left[ {\begin{array}{*{20}{c}} 0 & 1 & 0 \\ \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {blue} \\ {red} \\ {green} \\ \end{array}} \right] = red$$

The left hand row matrix is not over a field since it is restricted to integers {0,1}

The right hand column matrix is not a vector since you cannot form the a linear combination (αblue+βred+γgreen) since this makes no sense.

Yet the equation makes perfect sense if I perform the experiment of withdrawing a coloured ball from a bag of balls and wish to input the result into a computer for processing.

2. Oct 13, 2012

### chiro

Is there a basis for your colors? Can you map these colors to vectors in R^n (preferrably orthonormal ones)?

Your choice of saying the colors don't make sense is a bit odd because these are just labels just like x,y,z and everything else. A label is a label, but something like <1,0,0,0> in R^n is something that is more specific and more constrained.

3. Oct 13, 2012

### Studiot

I can't seen any relation to Rn.

If you can display one I would be interested.

I'm not even sure that my statement about {1,0} being integers is correct. They are really Boolean truth values.

4. Oct 13, 2012

### Vaedoris

This is informal but you'll get the idea...

\begin{align*} \begin{pmatrix} 0 & 1 & 0 \end{pmatrix}\left(k_1\begin{pmatrix} a\\ b\\ c \end{pmatrix}+k_2\begin{pmatrix} e\\ f\\ g \end{pmatrix} \right )&=\begin{pmatrix} 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} k_1a+k_2e\\ k_1b+k_2f\\ k_1c+k_2g \end{pmatrix} \\&= k_1b+k_2f\\ &=k_1\begin{pmatrix} 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} a\\ b\\ c \end{pmatrix} +k_2\begin{pmatrix} 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} e\\ f\\ g \end{pmatrix} \end{align*}

Hence, it is a homomorphism and therefore a linear transformation.

5. Oct 13, 2012

### chiro

If you have independent quantities then these are just orthogonal basis vectors. As an example, consider <red,green,blue> Then we define <red,blue> = <red,green> = <blue,green> = 0.

Set red = <1,0,0>, blue = <0,1,0> and green = <0,0,1> and then apply linear algebra and vector algebra and everything behaves mathematically is if they were really independent attributes.

If you want them dependent, then just curve the geometry and relate things together that are dependent: this is all curved geometry really is.

Curved geometry (i.e not R^n) is just a way of describing spaces that have dependencies between the elements. It doesn't have to be some high level concept like space-time: it can be anything you want: it could be any system where you change one thing and something else also changes in some way.

Normal linear algebra under the R^n geometry assumes that everything is independent but the differential geometry extends this to the cases where it isn't, and this is the reason why you need differential geometry to study relativity.

6. Oct 13, 2012

### I like Serena

Hi Studiot,

Actually {0,1} is a field.
It's also denoted as $F_2$ or $\mathbb Z/2 \mathbb Z$. See: http://en.wikipedia.org/wiki/Field_(mathematics)
In particular $1+1=0$ and $1 \cdot 1=1$.

Furthermore, a vector space is defined as the combation of a set V with a field F that satisfy eight specific axioms.
See: http://en.wikipedia.org/wiki/Vector_space

In your case the vector space V is the set $\{r \cdot \mathbf{red}+ g \cdot \mathbf{green}+ b \cdot \mathbf{blue}| r,g,b \in F_2\}$ combined with the field F2, which defines the operations $+$ and $\cdot$.

$1 \cdot \mathbf{red} + 1 \cdot \mathbf{green}$ is one of the elements of V.
Mathematically it does not have to have a physical meaning, but of course it does.

Note that $1 \cdot \mathbf{red} + 1 \cdot \mathbf{red} = 0 \cdot \mathbf{red}$, which I guess has a more problemetic physical meaning. ;)

It particular it means that your matrix (0 1 0) is well defined and defines a linear transformation.

7. Oct 13, 2012

### Vaedoris

which means {red, green, blue} is a basis if mathematically (disregarding physical sense) we assume each color is independent

8. Oct 13, 2012

### Studiot

Thank you all for your replies so far, this is proving a most enlightening discussion.

@Vaedoris

Your equation is not the same as mine. I require a black box that accepts the left hand side as inputs and the RHS as the output. Specifically the output has to be simply red. (No pun intended)

@Chiro

Thank you for a most interesting viewpoint about curved geometry.

@ILS

Wikipedia also allows that the integers do not form a field, so perhaps I should not have restricted this to F2. Thank you for the insight. It is interesting that parts of a non field can form a field. I will have to think about that.

'1red+1green is one of the elements of V.' is specifically excluded from the dataset and is not defined.

This is rather like continuing the the curve of some variable plotted against temperature backwards below absolute zero. You can draw the curve but such a zone is specifically excluded from the domain.

9. Oct 13, 2012

### I like Serena

Agreed.

Which wikipedia article do you mean?

Your input is $\begin{pmatrix}blue \\ red \\ green\end{pmatrix}$.
I believe an alternative way to write this is 1blue + 1red + 1green.
This would be part of your input dataset?

Perhaps we need to make a distinction between your input dataset and your output dataset.
Your output dataset appears to be {0,red}.
This one indeed does not contain '1red+1green', but I think the input dataset does.

It's indeed usual to extend a definition beyond what is physically possible, then do calculations, and then restrict it again to what is physically possible.

10. Oct 13, 2012

### Studiot

This is as I was taught, and there plenty of further references.

11. Oct 13, 2012

### I like Serena

Yes, the integers do not form a field, but the integers modulo a prime number do form a field.

Furthermore, a linear transformation is only defined in the context of a vector space.
And a vector space requires a field, which is needed for the scalar multiplication.

12. Oct 13, 2012

### Studiot

Which has been my contention! (That some matrices are not linear transformations).

I can define my equation to avoid vector spaces; the fact that my set coincides with part of some vector space is irrelevant.

13. Oct 13, 2012

### I like Serena

Ah, but the proposition only claims that a matrix determines a linear transformation on a vector space, and even more specifically a vector space of the form k^n.

It does not say anything about what you get when you apply a matrix to something that is not a vector space.
If you apply a matrix to something that is not a vector space, then indeed that is not a linear transformation.

14. Oct 13, 2012

### Studiot

Maybe I was wrong, but I took the original comment (which was in another thread) to mean that a matrix can only be a linear transformation and nothing else.

To try to be fair and unbiased I did reproduce part of it in my initial post here.

It is like saying that the letter a is only the side of a triangle, because it can be applied as such, regardless of the fact that it can be a coefficient or a variable in an algebraic expression or many other things.

Thanks for the discussion.

15. Oct 13, 2012

### Robert1986

Yes, you can talk about a matrix being defined as a box that contains blanks that you write stuff in. But so what? It seems tha as soon as you define any sort of matrix multiplication, and if it is well defined (if it isn't well defined you haven't done anything really) then you must impose enough structure to get a vector space or at least a module but im not sure about that.

16. Oct 13, 2012

### Studiot

Please explain why you say that.

17. Oct 13, 2012

### AlephZero

I think you are confusing "concepts" with "notation". You can use notation that looks like a rectangular arrray or table of "stuff" in useful ways that have nothing to do with vector spaces or linear transformations. It might even make sense to apply some of the rules of matrix algebra to them (but not necessarily all the rules).

In the your OP it wasn't clear (to me at least) what your notation of color names meant. You seem to be usinig it in at least two dufferent ways. First taking linear combinations (αblue+βred+γgreen) which makes perfect sense if it represents the intensity of the components of a color display system for example. Then you switched to the "names" of three colored balls where linear combinations don't make much sense, except for integer coefficients, and even then the concept involved looks more like "set theory" than "matrix algebra" to me. If you want to use the notation [ i j k ] to represent a set with i red balls etc, that's fine, but just writing the symbols "[ i j k ]" doesn't make a matrix (in the mathematical sense) appear from nowhere.

18. Oct 13, 2012

### Robert1986

Because to define what matrix multiplication means, you have to know what it means to multiply something. And if you want everything to be well defined and work properly, then you have to impose some structure. You can't just say 2blue+3red unless you define what this means. Otherwise, you're just writing stuff down (and as I said before, this is ok, but it's just kind of pointless.)

19. Oct 13, 2012

### Studiot

You appear to be agreeing with me and disagreeing with me at the same time.

The thread is about a discussion of the definition of a matrix.

My maths dictionary and I hold that a matrix is

'A rectangular array of elements, usually themselves members of a field.....' (my bold)

It was proposed that a matrix always is (represents) a linear transformation.

I hold that the cases where the elements are not members of a field may well not be (represent) linear transformations.

It is true that two matrices picked at random may not be able to participate in all the rules of matrix algebra.
So what?
Some pairs whose members are all real numbers may still be non conformable.

But thank you for your thoughts.

20. Oct 13, 2012

### Studiot

But Robert, what isn't well defined or working properly?

The example I gave works perfectly, and my rules specifically excluded (2blue+3red) etc.