Is there more than one way to prove this?

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SUMMARY

The discussion centers on proving that if an n × n matrix A satisfies Au = Av for vectors u and v in R^n, and u ≠ v, then A is not invertible. The established proof demonstrates that A(u - v) = 0 leads to a contradiction if A is assumed to be invertible, as it would imply u = v. An alternative proof using the existence of A^-1 also confirms that if A is invertible, then u must equal v, reinforcing the conclusion that A cannot be invertible under the given conditions.

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  • Familiarity with vector spaces and properties of vectors in R^n.
  • Knowledge of matrix invertibility and the implications of an invertible matrix.
  • Basic proof techniques in mathematics, including contradiction proofs.
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This discussion is beneficial for students and educators in mathematics, particularly those studying linear algebra, as well as researchers and professionals who require a solid understanding of matrix theory and its applications.

pyroknife
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Suppose that A is an n × n matrix and u and v
are vectors in R^n. Show that if Au = Av and
u ≠ v, then A is not invertible.


This is the book's proof:
From the fact that Au = Av, we have A(u − v) = 0. If
A is invertible, then u − v = 0, that is, u = v, which
contradicts the statement that u = v.

Mine proof is, if A is invertible thus A^-1 exists
(A^-1)(Au)=(A^-1)(Av)
A^-1*A=I
→u=v
if u≠v then then the above does not hold, implying that A^-1 does not exist.
 
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Your proof looks fine to me :approve:
 

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