Chapter 6

Direct Sums and Block Diagonal Matri-ces

This is a convenient place to put a very interesting result about direct sums and blockdiagonal matrices. First is the notion of a direct sum. In all of this, V will be a finitedimensional vector space of dimension n and field of scalars F.

Definition 6.0.1 Let {Vi}ri=1 be subspaces of V. Then ∑

ri=1 Vi ≡ V1 + · · ·+Vr denotes all

sums of the form ∑ri=1 vi where vi ∈Vi. If whenever ∑

ri=1 vi = 0,vi ∈Vi, it follows that vi = 0

for each i, then a special notation is used to denote ∑ri=1 Vi. This notation is V1⊕·· ·⊕Vr,

or sometimes to save space⊕r

i=1Vi and it is called a direct sum of subspaces. A subspace

W of V is called A invariant for A ∈L (V,V ) if AW ⊆W.

The next lemma tells how to recognize a direct sum.

Lemma 6.0.2 For the Vi subspaces as above, V1 + · · ·+Vr =V1⊕·· ·⊕Vr if and only if

0 = (V1 + · · ·+Vi−1 +Vi+1 + · · ·+Vr)∩Vi = 0

for each i.

Proof: Suppose the sum is a direct sum. Then if m ∈ Mi ∩∑ j ̸=i M j it follows thatm = mi = ∑ j ̸=i m j where m j ∈M j for all j and so 0 =−mi +∑ j ̸=i m j so all the m j = 0 andmi = 0. Next suppose the condition about the intersection.

Then if ∑i mi = 0 it follows that −mi = ∑ j ̸=i m j and so m = −mi = ∑ j ̸=i m j ∈ Mi ∩∑ j ̸=i M j and so m = 0. Since i was arbitrary, each mi = 0. ■

The important idea is that you seek to understand A by looking at what it does on eachVi. It is a lot like knowing A by knowing what it does to a basis, an idea used earlier.

Lemma 6.0.3 If V = V1⊕·· ·⊕Vr and if β i ={

vi1, · · · ,vi

mi

}is a basis for Vi, then a basis

for V is {β 1, · · · ,β r}. Thus

dim(V ) =r

∑i=1

dim(Vi) =r

∑i=1|β i|

where |β i| denotes the number of vectors in β i. Conversely, if β i linearly independent andif a basis for V is {β 1, · · · ,β r} , then V = span(β 1)⊕·· ·⊕ span(β r)

Proof: Suppose ∑ri=1 ∑

mij=1 ci jvi

j = 0. Since a direct sum, for each i,∑mij=1 ci jvi

j = 0 andnow, since

{vi

1, · · · ,vimi

}is a basis, each ci j = 0 for each j, this for each i.

Suppose now that each β i is independent and a basis is {β 1, · · · ,β r} . Then clearly

V = span(β 1)+ · · ·+ span(β r)

Suppose then that 0 = ∑ri=1 ∑

mij=1 ci jvi

j, the inside sum being something in span(β i). Since{β 1, · · · ,β r} is a basis, each ci j = 0. Thus each ∑

mij=1 ci jvi

j = 0 and so V = span(β 1)⊕·· ·⊕ span(β r). ■

Thus, from this lemma, we can produce a basis for V of the form {β 1, · · · ,β r} , so whatis the matrix of a linear transformation A such that each Vi is A invariant?

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