When Does a Square Matrix Ensure Unique Solutions?

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SUMMARY

A square matrix A guarantees unique solutions for the equation Ax = 0 if and only if the equation Ax = b has exactly one solution for every vector b. This conclusion is derived from the properties of linear transformations and the implications of having multiple solutions. Specifically, if Ax = 0 has only the trivial solution x = 0, then Ax = b must also yield a unique solution for any b. Counterexamples, such as A = [B, I], demonstrate that having at least one solution does not suffice for uniqueness.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly matrix theory.
  • Familiarity with the properties of square matrices and their transformations.
  • Knowledge of the implications of unique versus multiple solutions in linear equations.
  • Basic proficiency in solving systems of linear equations.
NEXT STEPS
  • Study the properties of linear transformations in depth.
  • Learn about the rank and nullity of matrices and their relationship to solutions.
  • Explore counterexamples in linear algebra to solidify understanding of unique solutions.
  • Investigate the implications of the Invertible Matrix Theorem on solution uniqueness.
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Students of linear algebra, mathematicians, and educators seeking to deepen their understanding of matrix theory and solution uniqueness in linear equations.

Nynjal
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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.
 

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