When Does a Square Matrix Ensure Unique Solutions?

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A square matrix A ensures that the equation Ax = 0 has exactly one solution if and only if the equation Ax = b has exactly one solution for every vector b. The unique solution to Ax = 0 is x = 0. If Ax = b has two solutions, it implies that A(x1 - x2) = 0, leading to a contradiction since x1 and x2 are distinct. The discussion clarifies that the original statement about "at least one solution" is incorrect, emphasizing the necessity of a unique solution for all b. Understanding these conditions is crucial for solving linear equations involving square matrices.
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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...
 
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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.
 
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.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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