Page 122 - Matrices theory and applications

6.3. Similarity Invariants and Jordan Reduction
105
When A 0 = A 1 = I n , Theorem 6.3.2 tells that XI n − B 0 and XI n − B 1
are equivalent, namely that they have the same invariant polynomials, if
there exists P ∈ GL n (k) such that PB 0 = B 1 P, which is the criterion
given by Theorem 6.3.1.
Proof
We prove Theorem 6.3.2. The condition is clearly sufficient.
Conversely, if XA 0 +B 0 and XA 1 +B 1 are equivalent, there exist matrices
P, Q ∈ GL n (A), such that P(XA 0 + B 0 )= (XA 1 + B 1 )Q.Since A 1 is
2
invertible, one may perform Euclidean division of P by XA 1 + B 1 on the
right:
P =(XA 1 + B 1 )P 1 + G,
where G is a matrix whose entries are constant polynomials. We warn the
reader that since M n (k) is not commutative, Euclidean division may be
done either on the right or on the left, with distinct quotients and distinct
remainders. Likewise, we have Q = Q 1 (XA 0 + B 0 )+ H with H ∈ M n (k).
Let us write, then,
(XA 1 + B 1 )(P 1 − Q 1 )(XA 0 + B 0 )= (XA 1 + B 1 )H − G(XA 0 + B 0 ).
The left-hand side of this equality has degree (the degree is defined as the
supremum of the degrees of the entries of the matrix) 2 + deg(P 1 − Q 1 ),
while the right-hand side has degree less than or equal to one. The two
sides, being equal, must vanish, and we conclude that
GA 0 = A 1 H, GB 0 = B 1 H.
There remains to show that G and H are invertible. To do so, let us define
R ∈ M n(A) as the inverse matrix of P (which exists by assumption). We
still have
R =(XA 0 + B 0 )R 1 + K, K ∈ M n(k).

Combining the equalities stated above, we obtain
I n − GK =(XA 1 + B 1 )(QR 1 + P 1 K).
Since the left-hand side is constant and the right-hand side has degree
1+ deg(QR 1 + P 1 K), we must have I n = GK,so that G is invertible.
Likewise, H is invertible.

We conclude this paragraph with a remarkable statement:
Theorem 6.3.3 If B ∈ M n (k),then B and B T are similar.
T
Indeed, XI n − B and XI n − B are transposes of each other, and hence
have the same list of minors, hence the same invariant factors.

2
The fact that A 1 is invertible is essential, since the ring M n(A) is not an integral
domain.