Suppose that A and Bare n times n matrices such that rankA

Suppose that A and Bare n times n matrices such that rank(A) = rank(B) = n. For each of the following statements, indicate whether that statement is true or false. 0 is not an eigenvalue of AB. rank (AB) = n rank(AB) = 2n det(BAB^-1)# 0 The matrix AB is a singular matrix. null (BA) = {0} The matrix B^-1A^-1 is an invertible matrix.

Solution

a.

True.

A is invertible and B is invertible hence, AB is invertible

So, det(AB) is non zero

HEnce, 0 is not eigenvalue of AB

b.

True.

Because, AB is invertible and of size nxn so rank(AB)=n

c.

False

FOr a matrix of size nxn its rank cannot be larger than n

d.

True

B is invertible, A is invertible and B^{-1} is invertible so all these three matrices have non zero determinant

Hence,

det(BAB^{-1})=det(B)det(A)det(B^{-1}) is non zero

e.

False.

AB is invertible and hence non singular

f.

True.

Invertible matrices have null={0}

g.

True.

B^{-1} and A^{-1} are both invertible

Product of invertible matrices is invertible