Definition 8.3.1In Theorem 8.2.3, the matrix of the linear transformation L ∈ℒ
(V,W )
with respect to the ordered bases β ≡
{v1,⋅⋅⋅,vn}
for V and γ ≡
{w1,⋅⋅⋅,wm }
for W isdefined to be
[L]
where
[L]
_{ij} = d_{ij}. Thus this matrix is defined by L = ∑_{i,j}
[L ]
_{ij}w_{i}v_{i}.When it is desired to feature the bases β,γ, this matrix will be denoted as
[L ]
_{γβ}. When thereis only one basis β, this is denoted as
[L]
_{β}.
If V is an n dimensional vector space and β = {v_{1},
⋅⋅⋅
,v_{n}} is a basis for V, there exists a linear
map
qβ : Fn → V
defined as
∑n
qβ(a) ≡ aivi
i=1
where
( )
a1 n
a = || .. || = ∑ ae ,
( . ) i=1 ii
an
for e_{i} the standard basis vectors for F^{n} consisting of
( )
0 ⋅⋅⋅ 1 ⋅⋅⋅ 0
^{T}. Thus the 1 is in
the i^{th} position and the other entries are 0. Conversely, if q : F^{n}→ V is one to one, onto, and
linear, it must be of the form just described. Just let v_{i}≡ q
(ei)
.
It is clear that q defined in this way, is one to one, onto, and linear. For v ∈ V,q_{β}^{−1}
(v)
is a vector in F^{n} called the component vector of v with respect to the basis
{v_{1},
⋅⋅⋅
,v_{n}}.
Proposition 8.3.2The matrix of a linear transformation with respect to ordered bases β,γas described above is characterized by the requirement that multiplication of the componentsof v by
[L ]
_{γβ}gives the components of Lv.
Proof:This happens because by definition, if v = ∑_{i}x_{i}v_{i}, then
∑ ∑ ∑ ∑ ∑
Lv = xiLvi ≡ [L ]jixiwj = [L]jixiwj
i i j j i
and so the j^{th} component of Lv is ∑_{i}
[L ]
_{ji}x_{i}, the j^{th} component of the matrix times the
component vector of v. Could there be some other matrix which will do this? No, because if such
a matrix is M, then for any x , it follows from what was just shown that
[L ]
x = Mx. Hence
[L]
= M. ■
The above proposition shows that the following diagram determines the matrix of a linear
transformation. Here q_{β} and q_{γ} are the maps defined above with reference to the ordered bases,
{v_{1},
⋅⋅⋅
,v_{n}} and {w_{1},
⋅⋅⋅
,w_{m}} respectively.
L
β = {v1,⋅⋅⋅,vn} V → W {w1,⋅⋅⋅,wm } = γ
q ↑ ∘ ↑ q (8.1)
βn mγ
F → F
[L ]γβ
(8.1)
In terms of this diagram, the matrix
[L]
_{γβ} is the matrix chosen to make the diagram
“commute”. It may help to write the description of
[L ]
_{γβ} in the form
( ) ( )
Lv1 ⋅⋅⋅ Lvn = w1 ⋅⋅⋅ wm [L]γβ (8.2)
(8.2)
with the understanding that you do the multiplications in a formal manner just as
you would if everything were numbers. If this helps, use it. If it does not help, ignore
it.
Example 8.3.3Let
V ≡ { polynomials of degree 3 or less},
W ≡ { polynomials of degree 2 or less},
and L ≡ D where D is the differentiationoperator. A basis for V is β =
{1,x,x2,x3}
and a basisfor W is γ = {1,x, x^{2}}.
What is the matrix of this linear transformation with respect to this basis? Using
8.2,
Now consider the important case where V = F^{n}, W = F^{m}, and the basis chosen is the standard
basis of vectors e_{i} described above.
β = {e1,⋅⋅⋅,en},γ = {e1,⋅⋅⋅,em}
Let L be a linear transformation from F^{n} to F^{m} and let A be the matrix of the transformation
with respect to these bases. In this case the coordinate maps q_{β} and q_{γ} are simply the identity
maps on F^{n} and F^{m} respectively, and can be accomplished by simply multiplying by the
appropriate sized identity matrix. The requirement that A is the matrix of the transformation
amounts to
Lb = Ab
What about the situation where different pairs of bases are chosen for V and W? How are the
two matrices with respect to these choices related? Consider the following diagram which
illustrates the situation.
Fn A Fm
−→2
qβ2 ↓ ∘ qγ2 ↓
V L−→ W
qβ1 ↑ ∘ qγ1 ↑
Fn A Fm
−→1
In this diagram q_{βi} and q_{γi} are coordinate maps as described above. From the diagram,
q−γ11 qγ2A2q−β1qβ1 = A1,
2
where q_{β2}^{−1}q_{β1} and q_{γ1}^{−1}q_{γ2} are one to one, onto, and linear maps which may be accomplished
by multiplication by a square matrix. Thus there exist matrices P,Q such that P : F^{n}→ F^{n} and
Q : F^{m}→ F^{m} are invertible and
P A2Q = A1.
Example 8.3.4Let β ≡
{v1,⋅⋅⋅,vn}
and γ ≡
{w1,⋅⋅⋅,wn}
be two bases for V . Let Lbe the linear transformation which maps v_{i}to w_{i}. Find
[L ]
_{γβ}. In case V = F^{n}and lettingδ =
{e1,⋅⋅⋅,en}
, the usual basis for F^{n}, find
[L ]
_{δ}.
Letting δ_{ij} be the symbol which equals 1 if i = j and 0 if i≠j, it follows that L = ∑_{i,j}δ_{ij}w_{i}v_{j}
and so
[L]
_{γβ} = I the identity matrix. For the second part, you must have
( ) ( )
w1 ⋅⋅⋅ wn = v1 ⋅⋅⋅ vn [L]δ
and so
( )−1( )
[L]δ = v1 ⋅⋅⋅ vn w1 ⋅⋅⋅ wn
where
( )
w1 ⋅⋅⋅ wn
is the n × n matrix having i^{th} column equal to w_{i}.
Definition 8.3.5In the special case where V = W and only one basis is used for V = W, thisbecomes
−1 −1
qβ1 qβ2A2qβ2 qβ1 = A1.
Letting S be the matrix of the linear transformation q_{β2}^{−1}q_{β1}with respect to the standard basisvectors in F^{n},
S−1A2S = A1. (8.3)
(8.3)
When this occurs, A_{1}issaid to be similar to A_{2}and A → S^{−1}AS is called a similaritytransformation.
Recall the following.
Definition 8.3.6Let S be a set. The symbol ∼ is called an equivalence relation on S if itsatisfies the following axioms.
x ∼ x for all x ∈ S. (Reflexive)
If x ∼ y then y ∼ x. (Symmetric)
If x ∼ y and y ∼ z, then x ∼ z. (Transitive)
Definition 8.3.7
[x]
denotes the set of all elements of S which are equivalent to x and
[x]
is called the equivalence class determined by x or just the equivalence class of x.
Also recall the notion of equivalence classes.
Theorem 8.3.8Let ∼ be an equivalence class defined on a set S and let ℋ denote the setof equivalence classes. Then if
[x]
and
[y]
are two of these equivalence classes, either x ∼ yand
[x]
=
[y]
or it is not true that x ∼ y and
[x]
∩
[y]
= ∅.
Theorem 8.3.9In the vector space of n × n matrices, define
A ∼ B
if there exists an invertible matrix S such that
−1
A = S BS.
Then ∼ is an equivalence relation and A ∼ B if and only if whenever V is an n dimensionalvector space, there exists L ∈ℒ
(V,V )
and bases {v_{1},
⋅⋅⋅
,v_{n}} and {w_{1},
⋅⋅⋅
,w_{n}} such that A isthe matrix of L with respect to {v_{1},
⋅⋅⋅
,v_{n}} and B is the matrix of L with respect to{w_{1},
⋅⋅⋅
,w_{n}}.
Proof:A ∼ A because S = I works in the definition. If A ∼ B , then B ∼ A,
because
−1
A = S BS
implies B = SAS^{−1}. If A ∼ B and B ∼ C, then
−1 − 1
A = S BS, B = T CT
and so
−1 −1 −1
A = S T CT S = (TS) CTS
which implies A ∼ C. This verifies the first part of the conclusion.
Now let V be an n dimensional vector space, A ∼ B so A = S^{−1}BS and pick a basis for
V,
β ≡ {v1,⋅⋅⋅,vn}.
Define L ∈ℒ
(V,V )
by
∑
Lvi ≡ ajivj
j
where A =
(aij)
. Thus A is the matrix of the linear transformation L. Consider the
diagram
where q_{γ} is chosen to make the diagram commute. Thus we need S = q_{γ}^{−1}q_{β} which
requires
qγ = qβS− 1
Then it follows that B is the matrix of L with respect to the basis
{qγe1,⋅⋅⋅,qγen} ≡ {w1,⋅⋅⋅,wn}.
That is, A and B are matrices of the same linear transformation L. Conversely, suppose whenever
V is an n dimensional vector space, there exists L ∈ℒ
(V,V)
and bases {v_{1},
⋅⋅⋅
,v_{n}} and
{w_{1},
⋅⋅⋅
,w_{n}} such that A is the matrix of L with respect to {v_{1},
⋅⋅⋅
,v_{n}} and B is
the matrix of L with respect to {w_{1},
⋅⋅⋅
,w_{n}}. Then it was shown above that A ∼ B.
■
What if the linear transformation consists of multiplication by a matrix A and you want to
find the matrix of this linear transformation with respect to another basis? Is there an easy way to
do it? The next proposition considers this.
Proposition 8.3.10Let A be an m × n matrix and let L be the linear transformation which isdefined by
In simple language, to find Lx,you multiply on the left of x by A. (A is the matrix of L withrespect to the standard basis.) Then the matrix M of this linear transformation withrespect to the bases β =
{u1,⋅⋅⋅,un }
for F^{n}and γ =
{w1,⋅⋅⋅,wm }
for F^{m}is givenby
( ) ( )
M = w ⋅⋅⋅ w − 1A u ⋅⋅⋅ u
1 m 1 n
where
( )
w1 ⋅⋅⋅ wm
is the m × m matrix which has w_{j}as its j^{th}column.
Proof: Consider the following diagram.
L
Fn → Fm
qβ ↑ ∘ ↑ qγ
Fn → Fm
M
Here the coordinate maps are defined in the usual way. Thus
( )T ∑n
qβ x1 ⋅⋅⋅ xn ≡ xiui.
i=1
Therefore, q_{β} can be considered the same as multiplication of a vector in F^{n} on the left by the
matrix
( )
u ⋅⋅⋅ u
1 n
. Similar considerations apply to q_{γ}. Thus it is desired to have the
following for an arbitrary x ∈ F^{n}.
A ( u ⋅⋅⋅ u ) x = ( w ⋅⋅⋅ w )M x
1 n 1 n
Therefore, the conclusion of the proposition follows. ■
In the special case where m = n and F = ℂ or ℝ and
{u1,⋅⋅⋅,un}
is an orthonormal basis and
you want M, the matrix of L with respect to this new orthonormal basis, it follows from the above
that
( ) ( )
M = u1 ⋅⋅⋅ um ∗A u1 ⋅⋅⋅ un = U ∗AU
where U is a unitary matrix. Thus matrices with respect to two orthonormal bases are unitarily
similar.
Definition 8.3.11An n × n matrix A, is diagonalizable if there exists an invertible n × nmatrix S such that S^{−1}AS = D, where D is a diagonal matrix. Thus D has zero entrieseverywhere except on the main diagonal.Writediag
(λ1⋅⋅⋅,λn)
to denote the diagonalmatrix having the λ_{i}down the main diagonal.
The following theorem is of great significance.
Theorem 8.3.12Let A be an n×n matrix. Then A is diagonalizable if and only if F^{n}hasa basis of eigenvectors of A. In this case, S of Definition 8.3.11consists of the n×n matrixwhose columns are the eigenvectors of A and D = diag
(λ ,⋅⋅⋅,λ )
1 n
.
Proof: Suppose first that F^{n} has a basis of eigenvectors,
{v1,⋅⋅⋅,vn}
where Av_{i} = λ_{i}v_{i}. Then
let S denote the matrix
( )
v1 ⋅⋅⋅ vn
and let S^{−1}≡
( )
uT1
|| .. ||
( . )
uTn
where
{
T 1 if i = j
ui vj = δij ≡ 0 if i ⁄= j .
S^{−1} exists because S has rank n. Then from block multiplication,
( ) ( )
uT1 uT1
S−1AS = || .. || (Av ⋅⋅⋅Av ) = || .. || (λ v ⋅⋅⋅λ v )
( . ) 1 n ( . ) 1 1 n n
uTn uTn
Next suppose A is diagonalizable so S^{−1}AS = D ≡diag
(λ1,⋅⋅⋅,λn)
. Then the
columns of S form a basis because S^{−1} is given to exist. It only remains to verify that
these columns of S are eigenvectors. But letting S =
( v ⋅⋅⋅ v )
1 n
, AS = SD
and so
( )
Av1 ⋅⋅⋅ Avn
=
( )
λ1v1 ⋅⋅⋅ λnvn
which shows that Av_{i} = λ_{i}v_{i}.■
It makes sense to speak of the determinant of a linear transformation as described in the
following corollary.
Corollary 8.3.13Let L ∈ℒ
(V,V)
where V is an n dimensional vector space and let A be thematrix of this linear transformation with respect to a basis on V. Then it is possible todefine
det(L) ≡ det(A ).
Proof:Each choice of basis for V determines a matrix for L with respect to the basis. If A
and B are two such matrices, it follows from Theorem 8.3.9 that
A = S−1BS
and so
( )
det(A) = det S− 1 det (B )det(S).
But
1 = det(I) = det(S− 1S ) = det(S)det(S−1)
and so
det(A) = det(B ) ■
Definition 8.3.14Let A ∈ℒ
(X,Y )
where X and Y are finite dimensional vector spaces.Definerank
(A)
to equal the dimension of A
(X)
.
The following theorem explains how the rank of A is related to the rank of the matrix of
A.
Theorem 8.3.15Let A ∈ℒ
(X, Y)
. Thenrank
(A)
= rank
(M )
where M is the matrixof A taken with respect to a pair of bases for the vector spaces X, and Y.
Proof:Recall the diagram which describes what is meant by the matrix of A. Here the two
bases are as indicated.
β = {v1,⋅⋅⋅,vn} X A−→ Y {w1,⋅⋅⋅,wm} = γ
qβ ↑ ∘ ↑ qγ
Fn M−→ Fm
Let
{Ax ,⋅⋅⋅,Ax }
1 r
be a basis for AX. Thus
{ − 1 − 1 }
qγM qβ x1,⋅⋅⋅,qγM qβ xr
is a basis for AX. It follows that
{ }
M q−X 1x1,⋅⋅⋅,M q−X1xr
is linearly independent and so rank
(A )
≤rank
(M )
. However, one could interchange
the roles of M and A in the above argument and thereby turn the inequality around.
■
The following result is a summary of many concepts.
Theorem 8.3.16Let L ∈ℒ
(V,V)
where V is a finite dimensional vector space. Then thefollowing are equivalent.
L is one to one.
L maps a basis to a basis.
L is onto.
det
(L)
≠0
If Lv = 0 then v = 0.
Proof:Suppose first L is one to one and let β =
{vi}
_{i=1}^{n} be a basis. Then if
∑_{i=1}^{n}c_{i}Lv_{i} = 0 it follows L
(∑n civi)
i=1
= 0 which means that since L
(0)
= 0, and L is one to
one, it must be the case that ∑_{i=1}^{n}c_{i}v_{i} = 0. Since
{vi}
is a basis, each c_{i} = 0 which shows
{Lvi}
is a linearly independent set. Since there are n of these, it must be that this is a
basis.
Now suppose 2.). Then letting
{vi}
be a basis, and y ∈ V, it follows from part 2.) that there
are constants,
{ci}
such that y = ∑_{i=1}^{n}c_{i}Lv_{i} = L
(∑n civi)
i=1
. Thus L is onto. It has been shown
that 2.) implies 3.).
Now suppose 3.). Then the operation consisting of multiplication by the matrix of L,
[L]
, must
be onto. However, the vectors in F^{n} so obtained, consist of linear combinations of the columns of
[L]
. Therefore, the column rank of
[L]
is n. By Theorem 3.3.23 this equals the determinant rank
and so det
([L])
≡ det
(L)
≠0.
Now assume 4.) If Lv = 0 for some v≠0, it follows that
[L]
x = 0 for some x≠0. Therefore, the
columns of
[L]
are linearly dependent and so by Theorem 3.3.23, det
([L ])
= det
(L )
= 0 contrary
to 4.). Therefore, 4.) implies 5.).
Now suppose 5.) and suppose Lv = Lw. Then L
(v − w )
= 0 and so by 5.), v −w = 0 showing
that L is one to one. ■
Also it is important to note that composition of linear transformations corresponds to
multiplication of the matrices. Consider the following diagram in which
[A]
_{γβ} denotes the matrix
of A relative to the bases γ on Y and β on X,
[B]
_{δγ} defined similarly.
X A−→ Y −B→ Z
qβ ↑ ∘ ↑ qγ ∘ ↑ qδ
n m p
F [A−−]γ−→β F [−B−]−δ→γ F
where A and B are two linear transformations, A ∈ℒ
(X,Y )
and B ∈ℒ
(Y,Z )
. Then
B ∘ A ∈ℒ
(X,Z )
and so it has a matrix with respect to bases given on X and Z, the coordinate
maps for these bases being q_{β} and q_{δ} respectively. Then