Proving Non-Trivial Solutions of Ax = b

[HaLAA]

As I study today, I read through my textbook that says if Ax=b has non-trival solution, then b span in A.

I want to know how can I prove it.

 

[Fredrik]

I don't understand what you mean by "b span in A". Can you post the exact claim you came across? You should also tell us where you found it. What book? What page number?

 

[homeomorphic]

Sounds like it should be b is in THE span OF (the column vectors of) A. The proof would just be knowing the definition of span, plus multiplying A times x and writing it out in terms of the components of x.

 

[HallsofIvy]

The "range" of A, the set of all vector of the form Av for some v in the domain of A, is sometimes called the "span" of A. Since we can take, in succession, v= <1, 0, 0, ..., 0>, v= <0, 1, 0, ..., 0>, v= <0, 0, 1, ..., 0>, to v= <0, 0, 0, ..., 1> and applying A to each of those gives a column of A, it can be shown that the columns of A span the range of A. And, if A is invertible, form a basis for it.