Question Ax = 0 has only trivial solution and Ax = b has 1+ solution for every b

[Nynjal]

if A is a square matrix
Ax = 0 has exactly one solution if and only if Ax = b has at least one solution for every vector b.

why is this true?
I am new to this if you can tell...

 

[HallsofIvy]

What you are saying is not true. Ax= 0 has exactly one solution if and only if Ax= b has exactly one solution for every b, not "at least one".

First, since A0= 0, for any square matrix, A, if Ax= 0 has only one solution, that solution is x= 0. Now suppose that, for some b, Ax= b has two solutions x1 and x2. Since x1 and x2 are different, x1- x2 is NOT 0. But what is A(x1- x2)?

And, for the other way, of course, if Ax= b has one solution for every b, take b= 0.

 

[bpet]

And a counterexample to the original statement is A=[B,I] with solution x=[0;b] to Ax=b, but also x=[y;-Ay] is a nontrivial solution to Ax=0.