Proving Simple Linear Algebra Statement: A^2 = A implies A is either 0 or I

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SUMMARY

The discussion centers on proving that if a matrix A satisfies the equation A^2 = A, then A must be either the zero matrix or the identity matrix. Participants explored the implications of this equation by manipulating it algebraically, leading to the conclusion that A(A - I) = 0. However, they noted the importance of recognizing that the product of two matrices equaling zero does not imply that either matrix must be zero, as demonstrated by the counterexample A = [1,0;0,0]. This highlights the necessity of careful reasoning in linear algebra proofs.

PREREQUISITES
  • Understanding of matrix multiplication and properties
  • Familiarity with the concepts of identity matrix and zero matrix
  • Basic knowledge of linear algebra and matrix equations
  • Awareness of counterexamples in mathematical proofs
NEXT STEPS
  • Study the properties of idempotent matrices in linear algebra
  • Learn about determinants and their role in matrix equations
  • Explore the implications of the rank-nullity theorem
  • Investigate other types of matrix decompositions and their applications
USEFUL FOR

Students of linear algebra, educators teaching matrix theory, and anyone interested in formal mathematical proofs related to matrix properties.

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Homework Statement


Prove that given a matrix A, and A^2 = A, then A must be either the zero matrix or the identity matrix.

The Attempt at a Solution


By multiplying both sides by A, you can deduce that A = A^2 = A^3 = A^4 ...
From there I think it's obvious that A must be either 0 or I, but I don't know how to start proving it formally.

Thanks very much for your help
-Patrick
 
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Try taking the determinant of A
 
Is there a way to prove it without using the determinant. This exercise is given before determinants are introduced.

Thanks
-Patrick
 
If A^2= A, then A^2- A= A(A- I)= 0.

However, you have to be careful here. With matrices it is NOT true that "if AB= 0 then either A= 0 or B= 0".
 
Is that true? What about A = [1,0;0,0]? This is not zero or I but A^2 = A
 
Ah. I didn't spot that buzzmath. Thank you. The question actually does say either prove or find a counterexample. I was just too sure that it was true.

Thanks
-Patrick
 

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