Suppose that A and B are two square matrices such that B is

Suppose that A and B are two square matrices such that B is similar to A. Prove that: (a) det (A) = det (B) (b) tr(A) = tr(B) (C) nullity (A) = nullity (B)

Solution

A and B are similar matrices :

a) Prove detA = detB

If A and B are similar then there is an invertible matrix P such that A = P^1BP. det A = det (P^ 1BP)

= (det P^1)(det B)(det P) = det(B) , as (det P^1)( det P) = 1.

Hence detA = detB

b) tr(A) = tr(B)

A and B are similar then A = P^1BP.

tr(A) = tr(P^1BP)

= tr[P^-1[BP])

=tr( [BP]P^-1)

=tr(BPP^-1) = tr(B)

Hence proved

c) nullity(A) = nullity(B)

A and B are similar matrices so, rank(A) = rank(B)

n = rank(A) + nullity(A) ;

n = rank(B) + nullity(B), and since we have rank(A) = rank(B).

So, nullity(A) = nullity(B)

Hence proved

 Suppose that A and B are two square matrices such that B is similar to A. Prove that: (a) det (A) = det (B) (b) tr(A) = tr(B) (C) nullity (A) = nullity (B) Sol

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