Calculus of Real and Complex Variables

A.11 Direct Sums

Here is a very useful theorem due to Sylvester. The field of scalars will be F and the entries of A,B come from F. We will have in mind F = ℂ but the results hold more generally. You really only need to be able to completely factor something called the minimal polynomial which will be described below.

Proof: If x ∈ ker

, then Ax ∈ ker and so A ⊆ ker. The following picture may help.

Now let

be a basis of ker and let be a basis for A. Take any z ∈ ker. Then Az = ∑ _i=1^ma_iAy_i and so

( ) m∑ A z − aiyi = 0 i=1

which means z −∑ _i=1^ma_iy_i ∈ ker

and so there are scalars b_i such that

m∑ ∑n z − aiyi = bixi. i=1 j=1

It follows span

⊇ ker and so by the first part (see the picture)

dim (ker(BA )) ≤ n + m ≤ dim (ker(A))+ dim (ker(B )) ■

Of course this result holds for any finite product of square matrices by induction. One way this is quite useful is in the case where you have a finite product ∏ _i=1^lL_i all of which are square matrices of the same size. Then

( ) ∏l ∑l dim ker Li ≤ dim (kerLi) i=1 i=1

and so if you can find a linearly independent set of vectors in ker

of size

∑l dim (kerLi), i=1

then it must be a basis for ker

.

Proof: Suppose ∑ _i=1^r ∑ _j=1^m_ic_ijv_jⁱ = 0. then since it is a direct sum, it follows for each i,

m∑i cijvij = 0 j=1

and now since

is a basis, each c_ij = 0.  ■

Here is a useful lemma.

Proof: Note that since the matrices commute, L_j : ker

ker. Here is why. If L_iy = 0 so that y ∈ ker, then

LiLjy = LjLiy = Lj0 = 0

and so L_j : ker

ker. Next observe that it is obvious that, since the operators commute,

∑p (∏p ) ker(Lp) ⊆ ker Li i=1 i=1

Suppose

p ∑ vi = 0,vi ∈ ker(Li), i=1

but some v_i≠0. Then do ∏ _j≠iL_j to both sides. Since the matrices commute, this results in

∏ Ljvi = 0 j⁄=i

which contradicts the assumption that these L_j are one to one and the above observation that they map ker

to ker. Thus if

∑ vi = 0,vi ∈ ker(Li) i

then each v_i = 0. It follows that

( ∏p ) ker(L1) ⊕+ ⋅⋅⋅+ ⊕ ker(Lp) ⊆ ker Li (*) i=1

(*)

From Sylvester’s theorem and the observation about direct sums in Lemma A.11.3,

which implies all these are equal. Now in general, if W is a subspace of V, a finite dimensional vector space and the two have the same dimension, then W = V . This is because W has a basis and if v is not in the span of this basis, then v adjoined to the basis of W would be a linearly independent set so the dimension of V would then be strictly larger than the dimension of W. It follows from * that

( p ) ker(L )⊕ + ⋅⋅⋅+ ⊕ ker(L ) = ker ∏ L ■ 1 p i=1 i

Here is a situation in which the above holds. ker

^r is sometimes called a generalized eigenspace in case λ_i is an eigenvalue.

Proof: It is obvious the linear transformations

^r_i commute. Now here is a claim.

Claim : Let μ≠λ_i. Then

^m : V _iV _i and is one to one and onto for every m  a positive integer.

Proof: It is clear

^m maps V _i to V _i because v ∈ V _i is equivalent to ^r_iv = 0. Consequently,

(A − λ I)ri (A − μI)mv = (A − μI)m(A − λ I)ri v = (A − μI)m 0 = 0 i i

which shows that

^mv ∈ V _i.

It remains to verify

^m is one to one. This will be done by showing that is one to one. Let w ∈ V _i and suppose w = 0 so that Aw = μw. Then for m ≡ r_i, ^mw = 0 and so by the binomial theorem,

( ) ( ) m m∑ m m −l l m∑ m m− l l (μ− λi) w = l (− λi) μw = l (− λi) A w l=0 l=0

= (A − λ I)m w = (A − λ I)ri w = 0. i i

Therefore, since μ≠λ_i, it follows w = 0 and this verifies

is one to one. Thus ^m is also one to one on V _i. Letting be a basis for V _i, it follows

{ m i m i } (A − μI) u 1,⋅⋅⋅,(A − μI) urk

is also a basis and so

^m is also onto. The desired result now follows from Lemma A.11.4.  ■

By the Cayley-Hamilton theorem for A a n×n complex matrix, it satisfies its characteristic polynomial which was a polynomial of degree n, denoted as q

. Let the minimal polynomial p be the monic polynomial (leading coefficient is 1) of smallest degree such that p = 0. In all of this, A⁰ is defined as I. Thus the minimal polynomial is of the form

p(λ) = λm + am −1λm−1 + ⋅⋅⋅+ a1λ+ a0,m ≤ n

and

p(A ) ≡ Am + am− 1Am −1 + ⋅⋅⋅+ a1A + a0I = 0

where the 0 is the n × n 0 matrix.

Proof: By division of polynomials, Lemma 1.11.2,

q(λ) = p(λ)k (λ) +r (λ )

where either r

= 0 or else it has smaller degree than p. However, in the second case, you would have 0 = k + r = r which is impossible because p has the smallest degree with the property that p = 0. As to this polynomial being unique, suppose is another one. Then the same argument just used shows that p divides and divides p. Since both are monic polynomials of the same degree, they must be the same polynomial. ■

From now on, assume the field of scalars is ℂ so that all polynomials have all roots.

Proof: For the first claim, just factor out

∏m (A − λiI) (A− λjI)rj (A − λiI)ri−1 ≡ (A − λiI)B j⁄=i

If Bv = 0 for all v, then this would contradict that p

is the minimal polynomial. Hence there exists v such that Bv≠0 and then = 0 showing that Bv is an eigenvector. Thus each λ_i is an eigenvalue for A. It remains to show that these are the only eigenvalues for A.

Next suppose

(A− μI) v = 0

for some μ and v≠0. Then

continuing this way yields = ∏ _k=1^mv, a contradiction unless μ = λ_k for some k. ■

Proof: Since ℂⁿ = ker

, the conclusion follows from Theorem A.11.5.  ■