Page 243 - Linear Algebra Done Right

Chapter 10. Trace and Determinant
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all terms in which m j = m k for some j = k can be ignored because the
determinant of a matrix with two equal columns is 0 (by 10.29). Thus
instead of summing over all m 1 ,...,m n with each m j taking on values
1,...,n, we can sum just over the permutations, where the m j ’s have
distinct values. In other words,

det(AB) = b m 1 ,1 ...b m n ,n det[ Ae m 1 ... Ae m n ]
(m 1 ,...,m n )∈perm n

= b m 1 ,1 ...b m n ,n sign(m 1 ,...,m n ) det A
(m 1 ,...,m n )∈perm n

= (det A) sign(m 1 ,...,m n ) b m 1 ,1 ...b m n ,n
(m 1 ,...,m n )∈perm n
= (det A)(det B),

where the second equality comes from 10.30.
In the paragraph above, we proved that det(AB) = (det A)(det B).
Interchanging the roles of A and B, we have det(BA) = (det B)(det A).
The last equation can be rewritten as det(BA) = (det A)(det B), com-
pleting the proof.

Now we can prove that the determinant of the matrix of an oper-
ator is independent of the basis with respect to which the matrix is
computed.

10.32 Corollary: Suppose T ∈L(V).If (u 1 ,...,u n ) and (v 1 ,...,v n )
are bases of V, then

det M T, (u 1 ,...,u n ) = det M T, (v 1 ,...,v n ) .

Note the similarity of Proof: Suppose (u 1 ,...,u n ) and (v 1 ,...,v n ) are bases of V. Let

this proof to the proof A =M I, (u 1 ,...,u n ), (v 1 ,...,v n ) . Then
of the analogous result
−1
about the trace det M T, (u 1 ,...,u n ) = det A M T, (v 1 ,...,v n ) A

(see 10.10). −1
= det M T, (v 1 ,...,v n ) A A

= det M T, (v 1 ,...,v n ) ,
where the first equality follows from 10.3 and the second equality fol-
lows from 10.31. The third equality completes the proof.