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Matrix Algebra question about identity matricies and scalars

Posted: Fri Feb 12, 2010 6:13 am UTC
So I was looking at some matrix algebra the other day when reading up on some AI algorithms, and I realized something.

Lets say you have an n*n matrix, A

A =
[2, 3, 5]
[4, 1, 7]
[4, 9, 6]

Now, you can multiply A by a scalar x, such as 3, and get B. Or, you can multiply that scalar 3 by an identity matrix, and then multiply it by A to get B.

So,
A * 3x = 3B
A * 3I * x = 3B (I = identity matrix)

therefore:

A * 3x = A * 3I * x
multiply both sides by A inverse
3x = 3I * x
multiply both sides by 1/3
x = l*x

3 =
[3, 0, 0]
[0, 3, 0]
[0, 0, 3]

This doesn't seem to make sense. I'm no mathematician and took linear algebra a year ago, so can someone tell me if I'm doing something wrong, this is totally normal, or if I have unlocked some great mathematical paradox that will get me a fields medal despite the fact that I am a psychology/neuroscience grad student and not a mathematician?

Re: Matrix Algebra question about identity matricies and scalars

Posted: Fri Feb 12, 2010 6:19 am UTC
It doesn't make sense because multiplication by a scalar and multiplication on matrices are two different things.

You can actually think of the scalar itself as a function from matrices to matrices.

So you can't just "cancel out" the A's; you'd have to cancel out the scalar function which takes as input the matrix A, which would leave you with Id = Id

Re: Matrix Algebra question about identity matricies and scalars

Posted: Fri Feb 12, 2010 7:06 am UTC
Syrin wrote:So you can't just "cancel out" the A's

This is actually exactly your error.
You started with [imath]A(3x) = A(3Ix)[/imath] and multiplied by [imath]A^{-1}[/imath]. But [imath]AA^{-1} = I[/imath], so when you multiply both sides by [imath]A^{-1}[/imath], what you really have is [imath]I(3x) = I(3Ix)[/imath], which is true and makes perfect sense.

Re: Matrix Algebra question about identity matricies and scalars

Posted: Fri Feb 12, 2010 4:07 pm UTC
Ok, thinking of scalar multiplication as a function on a matrix makes total sense and explains where I went wrong.

Re: Matrix Algebra question about identity matricies and scalars

Posted: Fri Feb 12, 2010 4:51 pm UTC
On the other hand, abusing notation and making the identification 3 = 3I often makes perfect sense.

Re: Matrix Algebra question about identity matricies and scalars

Posted: Fri Feb 12, 2010 11:11 pm UTC
antonfire wrote:On the other hand, abusing notation and making the identification 3 = 3I often makes perfect sense.

Especially since their is an obvious isomorphism between the scalar field and matrices of the form. In that sense, the OP has at least stumbled upon the idea that the underlying structure is the same.

Re: Matrix Algebra question about identity matricies and scalars

Posted: Sat Feb 13, 2010 7:00 am UTC
Cleverbeans wrote:
antonfire wrote:On the other hand, abusing notation and making the identification 3 = 3I often makes perfect sense.

Especially since their is an obvious isomorphism between the scalar field and matrices of the form. In that sense, the OP has at least stumbled upon the idea that the underlying structure is the same.

Hmm...thinking of the scalar X and the matrix X*I as being isomorphic is an interesting way of thinking about it.

Re: Matrix Algebra question about identity matricies and scalars

Posted: Wed Feb 17, 2010 3:00 am UTC
RockoTDF wrote:
Cleverbeans wrote:
antonfire wrote:On the other hand, abusing notation and making the identification 3 = 3I often makes perfect sense.

Especially since their is an obvious isomorphism between the scalar field and matrices of the form. In that sense, the OP has at least stumbled upon the idea that the underlying structure is the same.

Hmm...thinking of the scalar X and the matrix X*I as being isomorphic is an interesting way of thinking about it.

It's not so much an individual scalar being isomorphic to an individual matrix: it's an isomorphism between the set of scalars & the set of matrices of the form X*I, and the operations of addition & multiplication on these elements.

Re: Matrix Algebra question about identity matricies and scalars

Posted: Wed Feb 17, 2010 11:25 pm UTC
So what you mean is an isomorphism between the field of scalars and the field of matrices of the form λI.

Re: Matrix Algebra question about identity matricies and scalars

Posted: Thu Feb 18, 2010 12:06 am UTC
Note that there is also such an isomorphism between the field of scalars and the matrices with zeros everywhere but the upper left corner. However, you don't want to identify the scalars with those, because they don't interact the way you want with the other matrices.

Re: Matrix Algebra question about identity matricies and scalars

Posted: Thu Feb 18, 2010 1:08 am UTC
antonfire wrote:Note that there is also such an isomorphism between the field of scalars and the matrices with zeros everywhere but the upper left corner.

Meh, while we're doing this, there's also such an isomorphism between the field of scalars and the matrices of the form cP, where c is a scalar and P is any arbitrary (fixed) projection.

Re: Matrix Algebra question about identity matricies and scalars

Posted: Thu Feb 18, 2010 7:52 am UTC
Yes, but the relevant part of the [imath]f: \lambda \mapsto \lambda I[/imath] isomorphism is that [imath]\lambda \times_S A = f(\lambda) \times_M A[/imath] for any scalar [imath]\lambda[/imath] and matrix [imath]A[/imath].

Re: Matrix Algebra question about identity matricies and scalars

Posted: Thu Feb 18, 2010 4:47 pm UTC
Note that in Algebra, this kind of structure is called an Algebra. Therefore, you know that they have to be important.

The precise definition: Say we have a ring R (that is, a set which we can do multiplication and addition on. Think matrix multiplication and addition. Or polynomials with real coefficients and polynomial multiplication and addition), and that this ring is also a vector space* over some field K (that is, we can take elements of K and multiply them by ring elements r so that the usual linear relations hold. Think multiplying a matrix by a scalar, or multiplying a polynomial through by a constant).

Then R is a K-algebra if we can actually find the scalar field inside R. That is, there is a map [imath]f : K \to R[/imath] such that for any k, r, we have
• [imath]k \cdot r = f(k) \times r[/imath] (where the first multiplication is scalar multiplication, and the second is multiplication in the ring. From the above discussion: it doesn't matter if we view a scalar as [imath]\lambda[/imath] or [imath]\lambda I[/imath])
• And also [imath]f(k) \times r = r \times f(k)[/imath] (That is, we can multiply by (the representation of) a scalar on either the left or the right, and get the same answer. This is useful if we want things like [imath]A(\lambda B) = \lambda A B[/imath] to work out.)

* really, you just need a general S-module for some ring S, but if you knew that, why are you still reading this? (For everybody else, modules are like vector spaces, but instead of a scalar field, you have a scalar ring. Except we don't call them scalars anymore. And everything gets a different name, just to confuse what would otherwise be a perfectly simple structure (Though please don't ask about Simple modules. They are not simple at all, it turns out.))