Transforming between square matrices of different order

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Transforming between square matrices of different orders, such as a 2x2 matrix A and a 3x3 matrix B, is not straightforward due to dimensional incompatibility. A transformation matrix C cannot exist if A and B do not share the same intrinsic dimension, as this would require redundancy in the representation of B. The discussion suggests that a potential approach could involve using two transformation matrices, C1 and C2, but this would not yield a unique solution due to the presence of multiple unknowns. Additionally, the operation of multiplying by the inverse of A to derive C1 and C2 is not valid under matrix multiplication rules, as it does not account for the non-commutative nature of matrix operations. Ultimately, a clear understanding of the dimensions and properties of the matrices involved is essential for any transformation attempt.
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I have two known square matrices A and B of different order. Is there any way of constructing a transformation - e.g. a transformation matrix C - that transforms A to B? And, in that case, how do I determine C? Would it be something like this?

AC = B
Or maybe more general, how to determine the function f that transforms A to B? Like in:

f(A_{2x2}) = B_{3x3}
Would appreciate if someone could give me some guiding on this.
 
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Hey Mårten.

In terms of the equation AC = B, if A = 2x2 and B = 3x3 then C can't exist because A would have to be 3xn and C would have to be nx3 in order to get the right dimension.

The only way this could work with your general function is if the real dimension of A and B were the same which means that B would have some redundancy: An example redundancy is y = 2x where knowing x gives you y. This is a one-dimensional line over some region. An example of no redundancy in two dimensions is a general <x,y> vector where changing either x or y will never have any impact on the other co-ordinate.

Basically if you wanted to use matrices to do this, you would need something like f(A) = XAC where X is 3x2, A is your normal 2x2 matric and X is a 2x3 matrix.

If you have a true linear mapping, then X and C would be constant matrices (i.e. just matrices with numbers). If not, you would to use a general variable transformation (like the f(x) = x case) and that means you'll get a system of equations to generate each element of B.

The important though is to determine whether both A and B have the same intrinsic dimension because if they don't, you can't do it.
 
Hi Chiro,

Thanks a lot for your reply. Okey, I think I understand.

I was figuring maybe it could be possible to achieve something if I did (which you basically was suggesting as well)

C_1AC_2 = B
and then try to find C1 and C2. But having A and B known and C1 and C2 unknown, i.e. two unknowns, there is no unique solution for C1 and C2, isn't that so? Maybe that was what you said also.

P.s. Btw, is it allowed to do C_1AC_2 = B and then multiply by the inverse of A, i.e. A^{-1}C_1AC_2 = A^{-1}B to achieve C_1C_2 = A^{-1}B ? I mean, is that allowed according to matrix operation rules?
 
Mårten said:
I was figuring maybe it could be possible to achieve something if I did (which you basically was suggesting as well)

C_1AC_2 = B
and then try to find C1 and C2. But having A and B known and C1 and C2 unknown, i.e. two unknowns, there is no unique solution for C1 and C2, isn't that so?

In general there willl not be a unique solution for one pair of matrices A and B.

But if you think of the matrices are representing transformations or functions on vector spaces, you can find C1 and C2 that transform ANY matrix A into the corresponding B, which is a more useful idea in physics or math.

P.s. Btw, is it allowed to do C_1AC_2 = B and then multiply by the inverse of A, i.e. A^{-1}C_1AC_2 = A^{-1}B to achieve C_1C_2 = A^{-1}B ? I mean, is that allowed according to matrix operation rules?

No, first because in general matris multiplication is not commutative so ##A^{-1}C_1AC_2 \ne A^{-1}AC_1C_2##, and second because the product ##A^{-1}B## is not even defined if A and B don't have compatible numbers of rows and columns.
 
Okey, I think I understand, thanks for your help.
 
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