Matrices Proof> C=A-B, if Ax=Bx where x is nonzero, show C is singular

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SUMMARY

The discussion centers on proving that the matrix C, defined as C = A - B, is singular under the condition that Ax = Bx for a nonzero vector x. The proof follows from the equation Ax - Bx = 0, leading to x(C) = 0, which indicates that the matrix C has a nontrivial kernel. Consequently, since there exists a nonzero vector x such that Cx = 0, it is established that C is indeed singular.

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Homework Statement


Let A and B be n x n matrices and let C= A - B.
Show that if Ax=Bx, and x does not equal zero, then C must be singular.


Homework Equations





The Attempt at a Solution


Ax-Bx=0
x(A-B)=0
x(C)=0
So, Cx=0

Does that mean C is singular?
 
Physics news on Phys.org
If Cx=0 and x is not the zero vector, then what would C^(-1)(0) be? C0=0 as well. Would it be x or 0? Sure, it means C is singular.
 

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