Recent content by Yoonique

A is invertible?

I assume |A^2| means determinant? Then determinant of A is 1 or -1. If it means length then it is 1.

(A+I)v=(A+I)(A-I)? I don't know how can it help me to conclude A+I=0

Oh wait. I think I have an idea. Does it mean that rank(A+I)=0?

I have no clue for this :/

Another equation? (A+I)v=0?

Eigenvector cannot be the zero vector?

If I know that the rank(I-A) + rank(I+A) = n, then does it mean that the dimension of eigenspace of A is n, therefore there are n eigenvectors so A is diagonalizable.

The only properties I know are that they are linearly independent and they form a basis for R^n.

I'm not sure whether it is right. But I compared them like coefficients.

I multiply both sides by identity matrix so it IAv is still Av.

I didn't learn anything about that in my course. Can you explain that more explicitly? I know the eigenvalue of A are -1 and 1.

Av=λv Av=-v Av=-Iv A=-I Can I deduce A=-I from that?

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >[/color] Let A be a square matrix of order n such that A^2 = I a) Prove that if -1 is the only eigenvalue of A, then A= -I b) Prove that if 1 is the only eigenvalue of A, then A= I c) Prove that A is...

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