Linear Algebra II (pdf, 500 kB)

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(3) Of course, the statement of the theorem remains true for endomorphisms.
Let f : V → V be an endomorphism of the finite-dimensional F -vector
space V , then Pf(f) = 0 (which is the zero endomorphism in this case).
For evaluating the polynomial at f, we have to interpret f n as the n-fold
composition f ◦ f ◦ · · · ◦ f, and f 0 = idV .
Our next goal is to define the minimal polynomial of a matrix or endomorphism,
as the monic polynomial of smallest degree that has the matrix or endomorphism
as a “root”. However, we need to know a few more facts about polynomials in
order to see that this definition makes sense.
2.3. Lemma (Polynomial Division). Let f and g be polynomials, with g monic.
Then there are unique polynomials q and r such that r = 0 or deg(r) < deg(g)
and such that
f = qg + r .
Proof. We first prove existence, by induction on the degree of f. If deg(f) <
deg(g), then we take q = 0 and r = f. So we now assume that m = deg(f) ≥
deg(g) = n, f = amx m + · · · + a0. Let f ′ = f − amx m−n g, then (since g = x n + . . . )
deg(f ′ ) < deg(f). By the induction hypothesis, there are q ′ and r such that
deg(r) < deg(g) or r = 0 and such that f ′ = q ′ g +r. Then f = (q ′ +amx m−n )g +r.
(This proof leads to the well-known algorithm for polynomial long division.)
As to uniqueness, suppose we have f = qg + r = q ′ g + r ′ , with r and r ′ both of
degree less than deg(g) or zero. Then
(q − q ′ )g = r ′ − r .
If q = q ′ , then the degree of the left hand side is at least deg(g), but the degree of
the right hand side is smaller, hence this is not possible. So q = q ′ , and therefore
r = r ′ , too.
Taking g = x − α, this provides a different proof for Theorem 17.13 of Linear
Algebra I.
2.4. Lemma and Definition. Let A ∈ Mat(n, F ). There is a unique monic polynomial
MA(x) of minimal degree such that MA(A) = 0. If p(x) is any polynomial
satisfying p(A) = 0, then p is divisible by MA (as a polynomial).
This polynomial MA is called the minimal (or minimum) polynomial of A. Similarly,
we define the minimal polynomial Mf of an endomorphism f of a finitedimensional
vector space.
Proof. It is clear that monic polynomials p with p(A) = 0 exist (by the Cayley-
Hamilton Theorem 2.1, we can take p = PA). So there will be one such polynomial
of minimal degree. Now assume p and p ′ were two such monic polynomials of (the
same) minimal degree with p(A) = p ′ (A) = 0. Then we would have (p − p ′ )(A) =
p(A) − p ′ (A) = 0. If p = p ′ , then we can divide p − p ′ by its leading coefficient,
leading to a monic polynomial q of smaller degree than p and p ′ with q(A) = 0,
contradicting the minimality of the degree.
Now let p be any polynomial such that p(A) = 0. By Lemma 2.3, there are
polynomials q and r, deg(r) < deg(MA) or r = 0, such that p = qMA + r.
Plugging in A, we find that
0 = p(A) = q(A)MA(A) + r(A) = q(A) · 0 + r(A) = r(A) .