Asked by David
Suppose A is a square matrix satisfying the equation A^3 - 2I = 0. Prove that A and (A - I) respectively are invertible. (the hint is to find an explicit equation for A^-1.
To prove A is invertible, this is what I did although I don't that it's right:
A^3 - 2I = 0
A(A^2 - 2A^-1) = 0 because A * A^-1 = I
so (A^2 - 2A^-1) must = A^-1
(A^2 - 2A^-1) = A^-1
so (1/3)A^2 = A^-1 thus proving A is invertible. However I don't know that this is correct, and I don't know how to prove that (A - I) is also invertible. I know I need to factor the equation, but the 2I makes it difficult.
To prove A is invertible, this is what I did although I don't that it's right:
A^3 - 2I = 0
A(A^2 - 2A^-1) = 0 because A * A^-1 = I
so (A^2 - 2A^-1) must = A^-1
(A^2 - 2A^-1) = A^-1
so (1/3)A^2 = A^-1 thus proving A is invertible. However I don't know that this is correct, and I don't know how to prove that (A - I) is also invertible. I know I need to factor the equation, but the 2I makes it difficult.
Answers
Answered by
David
Well I guess no one can help. :/
Oh well, I'll figure it out
Oh well, I'll figure it out
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