- #1

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[tex]AC = B[/tex]

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

[tex]f(A_{2x2}) = B_{3x3}[/tex]

Would appreciate if someone could give me some guiding on this.

- Thread starter Mårten
- Start date

- #1

- 126

- 1

[tex]AC = B[/tex]

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

[tex]f(A_{2x2}) = B_{3x3}[/tex]

Would appreciate if someone could give me some guiding on this.

- #2

chiro

Science Advisor

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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.

- #3

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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)

[tex]C_1AC_2 = B[/tex]

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 [itex]C_1AC_2 = B[/itex] and then multiply by the inverse of A, i.e. [itex]A^{-1}C_1AC_2 = A^{-1}B[/itex] to achieve [itex]C_1C_2 = A^{-1}B[/itex] ? I mean, is that allowed according to matrix operation rules?

- #4

AlephZero

Science Advisor

Homework Helper

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In general there willl not be a unique solution for one pair of matrices A and B.I was figuring maybe it could be possible to achieve something if I did (which you basically was suggesting as well)

[tex]C_1AC_2 = B[/tex]

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?

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.

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

- #5

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Okey, I think I understand, thanks for your help.

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