Linear algebra existence and uniqueness?
I have a homework question I am unsure about. Any help would be appreciated.
Let A be an m x n matrix. Prove or give a counterexample. If Ax=0 has only the trivial solution x=0, then Ax=b always has a unique solution.
My thinking...
Suppose there are vectors u and v such that Au=b and Av=b. Then, Au-Av=b-b=0. So A(u-v)=0. Because Ax=0 has only the trivial solution, this means u-v=0, so u=v. Hence, Ax=b has a unique solution.
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Verified answer
Your proof of uniqueness would work if we already knew that there is at least one solution to Ax = b. In other words, your proof shows that there cannot be more than one solution. However, the proof fails to show that there is a unique solution, since there might be no solutions at all.
Consider the following counterexample: choose A to be the matrix below
[1 0]
[0 1]
[1 0]
and choose b to be the column vector
[1]
[3]
[2].
Let x be the column vector
[x1]
[x2].
Then Ax = 0 represents the system
x1 = 0
x2 = 0
x1 = 0
and so has only the trivial solution x = 0.
However, Ax = b represents the system
x1 = 1
x2 = 3
x1 = 2
which clearly has no solution for x, since x1 = 1, x1 = 2 cannot both be true.
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