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.

 

[blue_raver22]

To prove that if Ax=b has a non-trivial solution, then b spans in A, we can use the definition of spanning set. A set of vectors spans a space if every vector in that space can be written as a linear combination of the vectors in the set. In this case, A is a set of vectors and b is a vector in the space.

Now, if Ax=b has a non-trivial solution, it means that there exists a non-zero vector x such that Ax=b. This can also be written as x being a linear combination of the columns of A, since x is a vector and A is a matrix. Therefore, b can be expressed as a linear combination of the columns of A, meaning that b is in the span of the columns of A.

In other words, b can be written as a linear combination of the vectors in A, which is the definition of a spanning set. Therefore, we can conclude that if Ax=b has a non-trivial solution, then b spans in A.