1 Let A be a symmetric nn matrix Show that rankA2 rankA 2

1 / Let A be a symmetric n×n matrix. Show that rank(A^2) = rank(A).

2/ Let Q be an n×n orthogonal matrix. Show that Q2I is nonsingular.

Solution

1)By the definiton of an eigenvalue we know from the given that there is a nonzero
vector x such that Ax = x. Multiplying A on both sides of this equation yields
A2x = A(x). Applying the definition again we obtain A2x = 2x. Since A is
idempotent we know that A2 = A. This means that A2x = Ax which implies
that x = 2x. Since x is nonzero we must have = 2. This then clearly
implies that = 0, 1. it is clear   than rank(A^2) = rank(A).

From part (a), we know that the eigenvalues of Q all have absolute
value 1. Then, |12 · · · n| = 1. We have defined that for a matrix A,
det(A)= 12 · · · n. So we have that |det(Q)| = 1.
(ii) Since Q is an orthogonal matrix, QTQ = I. Let us take the determinant of QTQ

det(QTQ) = det(I)
det(QTQ) = 1
det(QT )det(Q) = 1
So either det(QT ) = det(Q) = 1 or det(QT ) = det(Q) = 1. Therefore, we
have that |det(Q)| = 1. it means Q-2I is non singular