Let A be a diagonalizable matrix whose eigenvalues are all e

Let A be a diagonalizable matrix whose eigenvalues are all either 1or -1. SHOW that A^2=In

Solution

Let A be a n x n matrix. Since A is diagonalizable, there exist an invertible matrix P and a diagonal matrix D such that A = PDP^-1. Further, all the entries on the leading diagonal of D are either 1 or -1 so that D² = I_n. Then A² = PD²P^-1 = PI_nP^-1 = PP^-1 = I_n

Alternatively, let A be a n x n matrix whose eigenvalues are either 1 0r -1. Then for any arbitrary vector v, we have either Av = v or Av = -v. If Av = v, then A²v = AAv = Av = v .

Also, if Av = -v, then A²v = AAv = A(-v) = -Av = -(-v) = v.

Thus A²v = v = I_nv in all cases so that (A² –I_n)v = 0, regardless of the choice of the vector. Hence A² –I_n = 0 or, A = I_n.