What does this show? It shows that if the component vector of v is
( x1 )
|| . ||
x = ( .. )
xn
meaning that v = ∑_{i}x_{i}v_{i}, then the component vector of w has i^{th} component equal to
∑n
Airxr = (Ax )i
r=1
The idea is that acting on a vector v with a linear transformation T yields a new vector w whose
component vector is obtained as the matrix of the linear transformation times the component vector of v.
It is helpful for some of us to think of this in terms of diagrams. On the other hand, some people hate such
diagrams. Use them if it helps. Otherwise ignore them and go right to the algebraic definition
5.2.
Let β = {v_{1},
⋅⋅⋅
,v_{n}} be a basis for V and let {w_{1},
⋅⋅⋅
,w_{m}} = γ be a basis for W. Then let
q_{β} : F^{n}→ V,q_{γ} : F^{m}→ W be defined as
n m
q x ≡ ∑ xv , q y ≡ ∑ y w
β i=1 ii γ j=1 j j
Thus these mappings are linear and take the component vector to the vector determined by the component
vector.
Then the diagram which describes the matrix of the linear transformation L is in the following
picture.
L
β = {v1,⋅⋅⋅,vn} V → W {w1,⋅⋅⋅,wm } = γ
q ↑ ∘ ↑ q (5.3)
βn mγ
F → F
[L]γβ
(5.3)
In terms of this diagram, the matrix
[L]
_{γβ} is the matrix chosen to make the diagram “commute”. It is
the matrix of the linear transformation because it takes the component vector of v to the component vector
for Lv. As implied by the diagram and as shown above, for A =
[L]
_{γβ},
m∑
Lvi = Ajiwj
j=1
Gimmick for finding matrix of a linear transformation
It may be useful to write this in the form
( ) ( )
Lv1 ⋅⋅⋅ Lvn = w1 ⋅⋅⋅ wm A, A is m × n (5.4)
(5.4)
and multiply formally as if the Lv_{i},w_{j} were numbers.
Example 5.2.1Let L ∈ℒ
(Fn,Fm )
and let the two bases be
{ }
e1 ⋅⋅⋅ en
,
{ }
e1 ⋅⋅⋅ em
,e_{i}denoting the column vector of zeros except for a 1 in the i^{th}position. Then from the above, you need tohave
m
∑
Lei = Ajiej
j=1
which says that
( Le ⋅⋅⋅ Le ) = ( e ⋅⋅⋅ e ) A
1 n m×n 1 m m ×m m ×n
and so Le_{i}equals the i^{th}column of A. In other words,
Thus, doing L to a vector x is the same as multiplying on the left by the matrix A.
Example 5.2.2Let
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 basis for W isγ = {1,x, x^{2}}.
What is the matrix of this linear transformation with respect to this basis? Using 5.4,
( 2 ) ( 2 )
0 1 2x 3x = 1 x x [D]γβ.
It follows from this that the first column of
[D]
_{γβ} is
( )
| 0 |
( 0 )
0
The next three columns of
[D]
_{γβ} are
( ) ( ) ( )
1 0 0
|( 0 |) ,|( 2 |) ,|( 0 |)
0 0 3
and so
( 0 1 0 0 )
| |
[D ]γβ = ( 0 0 2 0 ) .
0 0 0 3
Say you have a + bx + cx^{2} + dx^{3}. Then doing D to it gives b + 2cx + 3dx^{2}. The component vector of the
function is
( )
a b c d T
and after doing D to the function, you get for the component vector
( )T
b 2c 3d
This is the same result you get when you multiply by
[D ]
.
( )
( ) a ( )
| 0 1 0 0| || b || | b |
( 0 0 2 0) |( c |) = ( 2c )
0 0 0 3 3d
d
Of course, this is what it means to be the matrix of the transformation.
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 A2 Fm
−→
qβ2 ↓ ∘ qγ2 ↓
V −→L W
qβ1 ↑ ∘ qγ1 ↑
Fn A1 Fm
−→
In this diagram q_{βi} and q_{γi} are coordinate maps as described above. From the diagram,
q−1qγ A2q−1qβ = A1,
γ1 2 β2 1
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
PA2Q = A1.
Example 5.2.3Let β ≡
{v ,⋅⋅⋅,v }
1 n
and γ ≡
{w ,⋅⋅⋅,w }
1 n
be two bases for V . Let L be thelinear transformation which maps v_{i}to w_{i}. Find
[L]
_{γβ}.
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.
Definition 5.2.4In 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 basis vectors inF^{n},
−1
S A2S = A1. (5.5)
(5.5)
When this occurs, A_{1}issaid to be similar to A_{2}and A → S^{−1}AS is called a similarity transformation.
Recall the following.
Definition 5.2.5Let S be a set. The symbol ∼ is called an equivalence relation on S if it satisfiesthe 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 5.2.6
[x]
denotes the set of all elements of S which are equivalent to x and
[x]
is calledthe equivalence class determined by x or just the equivalence class of x.
Also recall the notion of equivalence classes.
Theorem 5.2.7Let ∼ be an equivalence class defined on a set S and let ℋ denote the set ofequivalence classes. Then if
[x]
and
[y]
are two of these equivalence classes, either x ∼ y and
[x]
=
[y]
or it is not true that x ∼ y and
[x]
∩
[y]
= ∅.
Theorem 5.2.8In the vector space of n × n matrices, define
A ∼ B
if there exists an invertible matrix S such that
A = S−1BS.
Then ∼ is an equivalence relation and A ∼ B if and only if 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 withrespect 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
A = S−1BS
implies B = SAS^{−1}. If A ∼ B and B ∼ C, then
A = S −1BS,B = T −1CT
and so
−1 −1 −1
A = S T CT S = (T S) CT S
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
n n
F B−→ F
qγ ↓ ∘ qγ ↓
V L−→ V
qβ ↑ ∘ qβ ↑
Fn A Fn
−→
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 5.2.9Let A be an m×n matrix and consider it as a linear transformation by multiplicationon the left by A. Then the matrix M of this linear transformation with respect to the basesβ =
{u1,⋅⋅⋅,un}
for F^{n}and γ =
{w1,⋅⋅⋅,wm }
for F^{m}is given by
( )−1 ( )
M = w1 ⋅⋅⋅ wm A u1 ⋅⋅⋅ un
where
( )
w1 ⋅⋅⋅ wm
is the m × m matrix which has w_{j}as its j^{th}column. Note thatalso
( ) ( )−1
w1 ⋅⋅⋅ wm M u1 ⋅⋅⋅ un = A
Proof: Consider the following diagram.
A
Fn → Fm
q ↑ ∘ ↑ q
βn mγ
F → F
M
Here the coordinate maps are defined in the usual way. Thus
n
q (x) ≡ ∑ x u = ( u ⋅⋅⋅ u )x
β i=1 i i 1 n
Therefore, q_{β} can be considered the same as multiplication of a vector in F^{n} on the left by the matrix
( )
u1 ⋅⋅⋅ un
. Similar considerations apply to q_{γ}. Thus it is desired to have the following for an
arbitrary x ∈ F^{n}.
( ) ( )
A u1 ⋅⋅⋅ un x = w1 ⋅⋅⋅ wn M x
Therefore, the conclusion of the proposition follows. ■
The second formula in the above is pretty useful. You might know the matrix M of a linear
transformation with respect to a funny basis and this formula gives the matrix of the linear transformation
in terms of the usual basis which is really what you want.
Definition 5.2.10Let A ∈ℒ
(X, Y)
where X and Y are finite dimensional vector spaces. Definerank
(A)
to equal the dimension of A
(X)
.
Lemma 5.2.11Let M be an m × n matrix. Then M can be considered as a linear transformation asfollows.
M (x) ≡ M x
That is, you multiply on the left by M.
Proof:This follows from the properties of matrix multiplication. In particular,
M (ax + by) = aM x + bM y ■
Note also that, as explained earlier, the image of this transformation is just the span of the columns, known
as the column space.
The following theorem explains how the rank of A is related to the rank of the matrix of
A.
Theorem 5.2.12Let A ∈ℒ
(X,Y )
. Thenrank
(A)
= rank
(M )
where M is the matrix of A takenwith respect to a pair of bases for the vector spaces X, and Y. Here M is considered as a lineartransformation by matrix multiplication.
Proof:Recall the diagram which describes what is meant by the matrix of A. Here the two bases are
as indicated.
β = {v ,⋅⋅⋅,v } X A Y {w ,⋅⋅⋅,w } = γ
1 n −→ 1 m
qβ ↑ ∘ ↑ qγ
Fn M−→ Fm
Let
{Ax1,⋅⋅⋅,Axr}
be a basis for AX. Thus
{ }
qγM q−β1x1,⋅⋅⋅,qγM q−β1xr
is a basis for AX. It follows that
{ }
M q−X1x1,⋅⋅⋅,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 5.2.13Let L ∈ℒ
(V,V)
where V is a finite dimensional vector space. Then the followingare equivalent.
L is one to one.
L maps a basis to a basis.
L is onto.
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
( i=1 civi)
= 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
( i=1 civi)
. Thus L is onto. It has been shown that 2.)
implies 3.).
Now suppose 3.). Then L
(V )
= V . If
{v1,⋅⋅⋅,vn}
is a basis of V, then V = span
(Lv1,⋅⋅⋅,Lvn)
. It
follows that
{Lv1,⋅⋅⋅,Lvn }
must be linearly independent because if not, one of the vectors
could be deleted and you would then have a spanning set with fewer vectors than dim
(V)
. If
Lv = 0,
∑
v = xivi
i
then doing L to both sides,
∑
0 = xiLvi
i
which imiplies each x_{i} = 0 and consequently v = 0. Thus 4. follows.
Now suppose 4.) and suppose Lv = Lw. Then L
(v − w )
= 0 and so by 4.), 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
Fn [A] Fm [B ] Fp
−−−γ→β −− δ−→γ
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
B ∘A = qδ[B]δγq−γ1qγ[A]γβ q−β1= qδ[B]δγ[A ]γβ qβ−1.
But this shows that
[B ]
_{δγ}
[A ]
_{γβ} plays the role of
[B ∘A]
_{δβ}, the matrix of B ∘ A. Hence the matrix of
B ∘ A equals the product of the two matrices