Question

a) Let A be an n x n matrix all of whose eigenvalues equal +/- 1. Show that if A is diagonalizable, then A^2=I(n).

b) Let A be an n x n matrix all of whose eigenvalues equal 0 and 1. Show that if A is diagonalizable, then A^2 = A.

Answer 1

The key to this problem is the fact that if you compute:\[A^n\]and A is diagonalizable, then:\[A^n=(PDP^{-1})^n=(PDP^{-1})(PDP^{-1})\cdots(PDP^{-1})(PDP^{-1})\]\[=PD(P^{-1}P)D(P^{-1}P)\cdots (P^{-1}P)D(P^{-1}P)DP^{-1}\]\[=PDD\cdots DDP^{-1}\]\[=PD^nP^{-1}\]and calculating power of a diagonal matrix is easy, just take powers of the entries on the diagonal.

Answer 2

So for a, since A is diagonalizable, we know:\[A=PDP^{-1}\]Hence:\[A^2=PD^2P^{-1}\]

Answer 3

Since the entries of D were +/- 1, when you square them, you will always get 1. Therefore:\[D^2=I_n\]

Answer 4

Hence:\[A^2=PD^2P^{-1}=PI_nP^{-1}=PP^{-1}=I_n\]

Answer 5

on b) how do you know how many 0 and 1 eigenvalues there are... lets say you have 3 eigenvalues, and 2 of them =0 and 1 is equal to 1, will that still equal A, even if 2 of them =1 and only 1 is 0?

Answer 6

its saying that in either case, A^2 = A

Answer 7

It doesnt really matter how many 0s or 1s you have. The idea is that if you square a 0 or a 1, it stays a 0 or a 1. So the matrix doesnt change.\[D^2 = D\]

Answer 8

Well if it doesn't matter how many 1's I have then why is it not equal to I(n) like the first problem?

Answer 9

I thank you again for your help, I did answer my own question though... I am now official done with linear algebra homework for the entry year, we have a final project, and just studying for the final... so I hope to see you around... thanks for all your help.