159 Linear Algebra B • Lecture 5
Matrix of composition. Changes of bases.
Similarity of matrices1
for all vectors v ∈ V . Applying it to each of the vectors
e01 , . . . , e0n , we arrive at
P = PB,B 0 = [e01 ]B [e02 ] · · · [e0n ] .
Theodore Th. Voronov
MSS/P5 • X63682
[email protected]
www.ma.umist.ac.uk/tv/159.html
This shows that the transition matrix is defined
uniquely and gives a practical method of calculating
it: one has to express the vectors of a ‘new’ basis B 0 in
terms of the ‘old’ basis B, and this gives the columns
of the transition matrix PBB 0 from B 0 to B.
Lectures: Monday 14:00 MSS/C19
Tutorials: Tuesday 11:00 MSS/C19
Example 1. Let V = R2 and consider the bases
e1 =
and
Matrix of composition
e01 =
Theorem 8.4.2 If T1 : U → V and T2 : V → W are
linear transformations, and if B, B 00 , and B 0 are bases
for U , V , and W , respectively, then:
1
,
0
1
,
1
Then
e02 =
P =
e2 =
1
1
0
1
0
.
1
0
1
is the transition matrix from the basis B 0 = {e01 , e02 } to
the basis B = {e1 , e2 }.
[T2 ◦ T1 ]B 0 ,B = [T2 ]B 0 ,B 00 [T1 ]B 00 ,B .
Proof. Directly expanding the definition, for an ar- Theorem. For three bases B, B 0 , B 00 in V holds
bitrary vector u ∈ U we have [T2 ◦ T1 ]B 0 ,B [u]B =
[(T2 ◦ T1 )(u)]B 0 = [T2 (T1 (u))]B 0 = [T2 ]B 0 ,B [T1 (u)]B 00 =
[T2 ]B 0 ,B ([T1 ]B 00 ,B [u]B ) = ([T2 ]B 0 ,B [T1 ]B 00 ,B )[u]B .
We can represent the maps in Theorem 8.4.2 by the
diagram
T2
T1
WB 0 ←−
VB 00 ←−
UB
Theorem 8.4.3 If T : V → V is a linear operator,
and if B is a basis for V , then the following are equivalent.
(a) T is one-to-one.
(b) [T ]B is invertible.
Moreover, when these equivalent conditions hold
[T −1 ]B = [T ]−1
B .
Proof. Immediately follows from the previous theorem.
Transition matrix. Let
B = {e1 , . . . , en }
and
B 0 = {e01 , . . . , e0n }
be bases for a vector space V . Then the transition
matrix from B 0 to B is defined by the formula:
PB,B 0 [v]B 0 = [v]B .
1 Definitions
and theorems numbered as ‘8.3.2’ are taken
from the book: H. Anton, Elementary Linear Algebra, John
Wiley, 2000.
PB,B 0 = PB,B 00 PB 00 ,B 0 .
Proof. Directly from the definition.
Corollary The transition matrix P from B 0 to B is
invertible, and the inverse is the transition matrix from
B to B 0 .
Changes of bases and matrices of a linear transformation
Consider two vector spaces V and W and let
T : V →W
be a linear transformation. Consider bases E =
{e1 , . . . , en } and G = {g1 , . . . , gm } in V and W , respectively. Let A = [T ]GE be the matrix of T w.r.t. the
bases E and G. Consider other bases E 0 = {e01 , . . . , e0n }
0
and G0 = {g10 , . . . , gm
} in V and W . Let A0 = [T ]G0 E 0
stand for the matrix of T w.r.t. the bases E 0 and G0 .
How the matrices A and A0 are related?
Theorem. The matrices of a linear transformation
T in ‘new’ bases E 0 , G0 and ‘old’ bases E, G are related
by the formula
−1
[T ]GE = PGG0 [T ]G0 E 0 PEE
0
where PEE 0 is the transition matrix from E 0 to E and
PGG0 is the transition matrix from G0 to G, or:
A = QA0 P −1
2
Th. Th. Voronov • 159 Linear Algebra B • Lecture 5 • Matrix of composition. Changes of bases. Similarity of matrices
if we denote P = PEE 0 and Q = PGG0 . Here A = [T ]GE
and A0 = [T ]G0 E 0 .
Proof. By the definition of a matrix of a linear transformation, for an arbitrary vector v ∈ V we have
[T (v)]G = [T ]GE [v]E
(1)
Example 2. We continue Example 1. If T : R2 →
R2 is a linear operator given by T (x, y) = (x + y, x − y),
then its matrix in the basis B is
1
1
[T ]B =
−1 1
and
in the bases G, E, and
[T (v)]G0 = [T ]G0 E 0 [v]E 0
(2)
in the bases G0 , E 0 . On the other hand, starting
from (2) and using the definition of the transition matrix twice, we can write
1
1
−1 0
1
1
−1
1
1
1 −1
0
2
=
2
1
1
−2
Similarity. If A and B are square matrices, we say
that B is similar to A if there is an invertible matrix
P such that B = P −1 AP . Matrices of the same linear
operator in two different bases are similar.
A property of square matrices is said to be a similarity invariant or invariant under similarity if
that property is shared by any two similar matrices.
[T (v)]G = PGG0 [T (v)]G0 = PGG0 [T ]G0 E 0 [v]E 0 =
PGG0 [T ]G0 E 0 PE 0 E [v]E .
Comparing with (1), we see that
PGG0 [T ]G0 E 0 PE 0 E = [T ]GE ,
as claimed. (Recall PE 0 E = PEE 0 .)
[T ]B 0 =
Corollary 1. The matrix of a linear operator
T : V →V
We state without proof the following facts from the
first semester course:
Similar matrices B and P −1 AP have the same
• determinant,
• rank,
• nullity,
transforms as follows:
−1
[T ]B = PBB 0 [T ]B 0 PBB
0
if PBB 0 is the transition matrix from a base B 0 to a
base B.
Corollary 2. The coordinate row vector (a row matrix) of a linear functional ξ ∈ V ∗ , i.e.,
ξ : V → R,
transforms by the formula:
−1
(ξ)B = (ξ)B 0 PBB
0
if PBB 0 is the transition matrix from a base B 0 to
a base B.
(Recall that (ξ)B is the row vector
(ξ(e1 ), . . . , ξ(en ).)
Remark. If we write coordinates of a covector ξ ∈ V ∗
as a column instead of a row, then the transformation
law will take the appearance
−1 T
[ξ]B = (PBB
0 ) [ξ]B 0 ,
the column [ξ]B being the transpose of the row (ξ)B .
It is instructive to compare it with the transformation
law for coordinates of a vector v ∈ V :
[v]B = PB,B 0 [v]B 0 .
You see that these laws are different. (Exercise: find
the condition on the transition matrix P = PBB 0 so
that the transformation laws for coordinates of vectors
and covectors will be the same.)
• trace,
• characteristic polynomial,
• eigenvalues;
• are invertible or non-invertible simultaneously.
Example 3. In the list above, let us check the determinant property.
det(B)
=
det(P −1 AP )
=
det(P −1 ) det(A) det(P )
=
(det(P ))−1 det(A) det(P )
=
det(A).
Since two matrices representing the same linear operator T with respect to different bases are similar, if B
and B 0 are two bases in V , then the matrices [T ]B and
[T ]B 0 have the same determinant, rank, nullity, trace,
characteristic polynomial, eigenvalues. This allows
us to use the terms determinant, rank, nullity,
trace, eigenvalues with respect to linear operators without causing confusion. For example, the
determinant of the linear operator T is det([T ]B ) with
respect to some (or any) basis B of V :
det(T ) = det ([T ]B ) .