Why is b considered an element of R^m in the Ax=b theorem?

[bonfire09]

In this theorem it states " Let A be a m x n matrix. That is For each vector b in R^m, the column Ax=b has a solution..." Why do they say that bεR^m? Is that because b is a mx1 column matrix where it has m rows making it belong to R^m?

 

[HallsofIvy]

If A is an "m by n matrix" then it has n columns and m rows. That means that to multiply it by vector v, written as a column matrix, v must have n components, and then the product with be a vector, again written as a column matrix, will have m components. That is, A is from Rⁿ to R^m.

 

[bonfire09]

so what your saying is that (mxn)(mx1)=(mx1).
and that means that bεR^m?

 

[HallsofIvy]

so what your saying is that (mxn)(mx1)=(mx1).
and that means that bεR^m?

I'm not sure I know what you mean by "(mxn)(mx1)= (mx1)". I thought we were talking about matrices, not numbers. What I said before was that a matrix with n columns and m rows, multiplied by a matrix with 1 column and n rows, gives a matrix with 1 column and n rows.

If I understand you notation, that would be "(mxn)(nx1)= (mx1)".