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I =M I, (v 1 ,...,v n ), (u 1 ,...,u n ) M I, (u 1 ,...,u n ), (v 1 ,...,v n ) .
Now interchange the roles of the u’s and v’s, getting

Chapter 10. Trace and Determinant

I =M I, (u 1 ,...,u n ), (v 1 ,...,v n ) M I, (v 1 ,...,v n ), (u 1 ,...,u n ) .
These two equations give the desired result.
Now we can see how the matrix of T changes when we change
bases.
10.3 Theorem: Suppose T ∈L(V). Let (u 1 ,...,u n ) and (v 1 ,...,v n )

be bases of V. Let A =M I, (u 1 ,...,u n ), (v 1 ,...,v n ) . Then
−1
10.4 M T, (u 1 ,...,u n ) = A M T, (v 1 ,...,v n ) A.
Proof: In 10.1, replace U and W with V, replace w j with v j , replace
T with I, and replace S with T, getting

10.5 M T, (u 1 ,...,u n ), (v 1 ,...,v n ) =M T, (v 1 ,...,v n ) A.
Again use 10.1, this time replacing U and W with V, replacing w j
with u j , and replacing S with I, getting
−1
M T, (u 1 ,...,u n ) = A M T, (u 1 ,...,u n ), (v 1 ,...,v n ) ,
where we have used 10.2. Substituting 10.5 into the equation above
gives 10.4, completing the proof.

Trace

Let’s examine the characteristic polynomial more closely than we
did in the last two chapters. If V is an n-dimensional complex vector
space and T ∈L(V), then the characteristic polynomial of T equals

(z − λ 1 )...(z − λ n ),

where λ 1 ,...,λ n are the eigenvalues of T, repeated according to multi-
plicity. Expanding the polynomial above, we can write the characteristic
polynomial of T in the form

n
n
10.6 z − (λ 1 + ··· + λ n )z n−1 + ··· + (−1) (λ 1 ...λ n ).