This is a convenient place to put a very interesting result about direct sums and block diagonal matrices.
First is the notion of a direct sum. In all of this, V will be a finite dimensional vector space of dimension n
and field of scalars F.
Definition 6.0.1Let
{Vi}
_{i=1}^{r}be subspaces of V. Then
∑r
Vi ≡ V1 + ⋅⋅⋅+ Vr
i=1
denotes all sums of the form∑_{i=1}^{r}v_{i}where v_{i}∈ V_{i}. If whenever
∑r
vi = 0,vi ∈ Vi, (6.1)
i=1
(6.1)
it follows that v_{i} = 0 for each i, then a special notation is used to denote∑_{i=1}^{r}V_{i}. This notation is
V1 ⊕ ⋅⋅⋅⊕ Vr,
or sometimes to save space
r
i=1Vi
and it is called a direct sumof subspaces. A subspace W of V is called A invariant for A ∈ℒ
(V,V)
ifAW ⊆ W.
The important idea is that you seek to understand A by looking at what it does on each V_{i}. It is a lot
like knowing A by knowing what it does to a basis, an idea used earlier.
Lemma 6.0.2If V = V_{1}⊕
⋅⋅⋅
⊕ V_{r}and if β_{i} =
{vi,⋅⋅⋅,vi }
1 mi
is abasis for V_{i}, then a basis for V is
{β1,⋅⋅⋅,βr}
. Thus
∑r ∑r
dim (V) = dim (Vi) = |βi|
i=1 i=1
where
|βi|
denotes the number of vectors in β_{i}. Conversely, if β_{i}linearly independent and if a basis for Vis
{β1,⋅⋅⋅,βr}
, then
V = span(β )⊕ ⋅⋅⋅⊕span (β)
1 r
Proof: Suppose ∑_{i=1}^{r}∑_{j=1}^{mi}c_{ij}v_{j}^{i} = 0. then since it is a direct sum, it follows for each
i,
m
∑ i i
cijvj = 0
j=1
and now since
{ i i }
v1,⋅⋅⋅,vmi
is a basis, each c_{ij} = 0 for each j, this for each i.
Suppose now that each β_{i} is independent and a basis is
{β1,⋅⋅⋅,βr}
. Then clearly
V = span(β1)+ ⋅⋅⋅+ span (βr)
Suppose then that 0 = ∑_{i=1}^{r}∑_{j=1}^{mi}c_{ij}v_{j}^{i}, the inside sum being something in span
(βi)
. Since
{β1,⋅⋅⋅,βr}
is a basis, each c_{ij} = 0. Thus each ∑_{j=1}^{mi}c_{ij}v_{j}^{i} = 0 and so V = span
(β1)
⊕
⋅⋅⋅
⊕span
(βr)
.
■
Thus, from this lemma, we can produce a basis for V of the form
{β1,⋅⋅⋅,βr}
, so what is the matrix of
a linear transformation A such that each V_{i} is A invariant?
Theorem 6.0.3Suppose V is a vector space with field of scalars F and A ∈ℒ
(V,V)
. Supposealso
V = V1 ⊕ ⋅⋅⋅⊕ Vq
where each V_{k}is A invariant. (AV_{k}⊆ V_{k}) Also let β_{k}be an ordered basis for V_{k}and let A_{k}denotetherestriction of A to V_{k}. Letting M^{k}denote the matrix of A_{k}with respect to this basis, it follows the matrixof A with respect to the basis
{β1,⋅⋅⋅,βq}
is
( 1 )
| M 0 |
|( ... |)
0 M q
Proof: Let β denote the ordered basis
{β1,⋅⋅⋅,βq}
,
|βk|
being the number of vectors in β_{k}. Let
q_{k} : F^{|βk|
}→ V_{
k} be the usual map such that the following diagram commutes.
Ak
Vk → Vk
q ↑ ∘ ↑ q
k|βk| |βkk|
F → F
M k
Thus A_{k}q_{k} = q_{k}M^{k}. Then if q is the map from F^{n} to V corresponding to the ordered basis β just
described,
( )
q 0 ⋅⋅⋅ x ⋅⋅⋅ 0 T = q x,
k
where x occupies the positions between ∑_{i=1}^{k−1}
|βi|
+ 1 and ∑_{i=1}^{k}
|βi|
. Then M will be the
matrix of A with respect to β if and only if a similar diagram to the above commutes. Thus it is
required that Aq = qM. However, from the description of q just made, and the invariance of each
V_{k},
( ) ( 1 ) ( )
| 0 | | M 0 | | 0 |
|| ...|| || ... || || ... ||
Aq || x || = A q x = q M kx = q|| M k || || x ||
|| .|| k k k || . || || . ||
( ..) ( .. ) ( .. )
0 0 M q 0
It follows that the above block diagonal matrix is the matrix of A with respect to the given ordered basis.
■
The matrix of A with respect to the ordered basis β which is described above is called a block diagonal
matrix. Sometimes the blocks consist of a single number.
Example 6.0.4Consider the following matrix.
( )
1 0 0
A ≡ |( 1 0 − 1 |)
− 2 2 3
Let V_{1}≡span
( ( ) ( ))
| | 1 | | 1 ||
( ( 1 ) ,( 0 ))
0 1
,V_{
2}≡span
(( ) )
|| 0 | |
(( − 1 ) )
2
. Show that ℝ^{3
} = V_{1}⊕V_{2}and and that V_{i}isA invariant. Find the matrix of A with respect to the ordered basis
⊆ V_{2}. The vectors in ∗ clearly are a basis for ℝ^{3}. You can verify this by observing that there
is a unique solution x,y,z to the system of equations
( ) ( ) ( ) ( )
1 1 0 a
x |( 1 |) + y|( 0 |) + z|( − 1 |) = |( b |)
0 1 2 c
for any choice of the right side. Therefore, by Lemma 6.0.2, ℝ^{3} = V_{1}⊕ V_{2}.
If you look at the restriction of A to V_{1}, what is the matrix of this restriction? It satisfies
and the matrix of A restricted to V_{1} is just the 1 × 1 matrix consisting of the number 1. Thus the matrix of
A with respect to this basis is
( )
| 1 0 0 |
( 0 2 1 )
0 0 2
How can you come up with invariant subspaces? In the next section, I will give a systematic way based
on a profound theorem of Sylvester. However, there is also a very easy way to come up with an invariant
subspace. Let v ∈ V an n dimensional vector space and let A ∈ℒ
(V,V)
. Let W ≡span
(v,Av,A2v,⋅⋅⋅)
.
It is left as an exercise to verify that W is a finite dimensional subspace of V . Recall that the span is the
set of all finite linear combinations. Of course W might be all of V or it might be a proper subset of V . The
method of Sylvester will end up typically giving proper invariant subspaces whose direct sum is the whole
space.