5
Change of Basis
In many applications, we may need to switch between two or more different bases for a vector
space. So it would be helpful to have formulas for converting the components of a vector with
respect to one basis into the corresponding components of the vector (or matrix of the operator)
with respect to the other basis. The theory and tools for quickly determining these “change of
basis formulas” will be developed in these notes.
5.1
Unitary and Orthogonal Matrices
Definitions
Unitary and orthogonal matrices will naturally arise in change of basis formulas. They are defined
as follows:
A is a unitary matrix ⇐⇒ A is an invertible matrix with A−1 = A† ,
and
A is an orthogonal matrix
⇐⇒
A is an invertible real matrix with A−1 = AT
⇐⇒
A is a real unitary matrix .
Because an orthogonal matrix is simply a unitary matrix with real-valued entries, we will mainly
consider unitary matrices (keeping in mind that anything derived for unitary matrices will also
hold for orthogonal matrices after replacing A† with AT ).
The basic test for determining if a square matrix is unitary is to simply compute A† and see
if it is the inverse of A ; that is, see if
AA† = I
!◮Example 5.1:
Let
3
5
A = 
4
i
5
Then
 3
 5
A† = 
4
− i
5
9/22/2013
.

4
i
5 
3
5

4
5
− i
3
5



Chapter & Page: 5–2
and
Change of Basis
3
5
AA† = 
4
i
5
So A is unitary.

4
3
i
5  5

3
4
− i
5
5
4
5
− i
3
5


 = ··· = I
.
Obviously, for a matrix to be unitary, it must be square. It should also be fairly clear that, if
A is a unitary (or orthogonal) matrix, then so are A∗ , AT , A† and A−1 .
?◮Exercise 5.1:
and A−1 .
Prove that, if A is a unitary (or orthogonal) matrix, then so are A∗ , AT , A†
The term “unitary” comes from the value of the determinant. To see this, first observe that,
if A is unitary then
I = AA−1 = AA† .
Using this and already discussed properties of determinants, we have
1 = det(I) = det(AA† ) = det(A) det(A† ) = det(A) det(A)∗ = |det(A)|2
.
Thus,
A is unitary
H⇒
|det A| = 1 .
And since there are only two real numbers which have magnitude 1 , it immediately follows that
A is orthogonal
H⇒
det A = ±1
.
An immediate consequence of this is that if the absolute value of the determinant of a matrix
is not 1 , then that matrix cannot be unitary.
Why the term “orthogonal” is appropriate will become obvious later.
Rows and Columns of Unitary Matrices
Let

u 11 u 12 u 13 · · · u 1N

u
 21 u 22 u 23 · · · u 2N 
U = 
..
.. 
..
..

 ..
.
.
.
. 
 .
uN1 uN2 uN3 · · · uN N

be a square matrix. By definition, then,

u 11 ∗ u 21 ∗
 u 12 ∗ u 22 ∗

 ∗
†
u 13
u 23 ∗
U = 

..
 ..
.
 .
∗
u 1N u 2N ∗

u N 1∗
u N 2∗ 


u N 3∗ 

.. 
. 
· · · uNN∗
···
···
···
..
.
.
Unitary and Orthogonal Matrices
Chapter & Page: 5–3
More concisely,
and
[U] jk = u jk
Now observe:
U†
jk
= [U]k j
∗
= uk j ∗
.
U is unitary
⇐⇒
U† = U−1
⇐⇒
U† U = I
† U U jk = [I] jk
⇐⇒
⇐⇒
X
U†
m
X
⇐⇒
m
[U]mk = δ jk
“for all ( j, k)”
u m j ∗ u mk = δ jk
“for all ( j, k)”
jm
The righthand side of the last equation is simply the formula for computing the standard matrix
inner product of the column matrices
 


u1 j
u 1k
u 
u 
 2j 
 2k 
 . 
 .  ,
and
 . 
 . 
 . 
 . 
uNj
uNk
and the last line tells us that
this inner product =
In other words, that line states that


u 11




  u 21 

 . ,




 .. 



uN1


u 12
u 
 22 
 . 
 . 
 . 
uN2
(
1
if
j =k
0
if
j 6= k
.



u 13
u 1N 

u 
u  

 23 
 2N  



,
 ..  , . . . ,  ..  
 . 
 . 



uN3
uNN

is an orthonormal set of column matrices. But these column matrices are simply the columns of
our original matrix U . Consequently, the above set of observations (starting with “ U is unitary”)
reduces to
A square matrix U is unitary if and only if its columns form an orthonormal set of
column matrices (using the standard column matrix inner product).
You can verify that a similar statement holds using the rows of U instead of its columns.
?◮Exercise 5.2:
Show that a square matrix U is unitary if and only if its rows form an
orthonormal set of row matrices (using the row matrix inner product). (Hints: Either consider
how the above derivation would have changed if we had used UU† instead of U† U , or use
the fact just derived along with the fact that U is unitary if and only if UT is unitary.)
version: 9/22/2013
Chapter & Page: 5–4
Change of Basis
In summary, we have just proven:
Theorem 5.1 (The Big Theorem on Unitary Matrices)
Let


u 11 u 12 u 13 · · · u 1N
u

 21 u 22 u 23 · · · u 2N 

U =  .
..
..
.. 
..

.
.
.
.
.
. 

uN1 uN2 uN3 · · · uN N
be a square matrix. Then the following statements are equivalent (that is, if any one statement is
true, then all the statements are true).
1.
U is unitary.
2.
The columns of U form an orthonormal set of column matrices (with respect to the usual
matrix inner product). That is,
(
X
1
if j = k
u m j ∗ u mk =
if j 6= k
0
m
.
3.
The rows of U form an orthonormal set of row matrices (with respect to the usual matrix
inner product). That is,
(
X
1
if j = k
u jm ∗ u km =
if j 6= k
0
m
.
?◮Exercise 5.3:
What is the corresponding “Big Theorem on Orthogonal Matrices”?
An Important Consequence and Exercise
You can now verify a result that will be important in our change of basis formulas involving
orthonormal bases.
?◮Exercise 5.4:
Let
be a square matrix, and let


u 11 u 12 u 13 · · · u 1N
u

 21 u 22 u 23 · · · u 2N 

U =  .
..
..
.. 
..

.
.
.
.
.
. 

uN1 uN2 uN3 · · · uN N
S = { e1 , e2 , . . . , e N }
and
B = { b1 , b2 , . . . , b N }
be two sets of vectors (in some vector space) related by
b1 b2 . . . b N = e1 e2 . . . e N U .
Change of Basis for Vector Components:
The General Case
(I.e.,
bj =
X
Chapter & Page: 5–5
for
ek u k j
k
j = 1, 2, . . . , N
.)
a: Show that
S is orthonormal and U is a unitary matrix
H⇒
B is also orthonormal .
b: Show that
S and B are both orthonormal sets
5.2
U is a unitary matrix .
H⇒
Change of Basis for Vector Components:
The General Case
Given the tools and theory we’ve developed, finding and describing the “most general formulas
for changing the basis of a vector space” is disgustingly easy (assuming the space is finite
dimensional).
So let’s assume we have a vector space V of finite dimension N and with an inner product
h · | · i . Let
A = { a1 , a2 , . . . , a N }
and
B = { b1 , b2 , . . . , b N }
.
be two bases for V , and define the corresponding four N × N matrices
MAA
,
MAB
by
, MBB
a j ak
= a j bk
= b j bk
[MAA ] jk =
[MAB ] jk
[MBB ] jk
and
[MBA ] jk =
MBA
and
b j ak
,
,
,
.
(See the pattern?) These matrices describe how the vectors in A and B are related to each
other.
Two quick observations should be made about these matrices:
1.
The first concerns the relation between MAB and MBA . Observe that
∗
∗
= [MBA ]k j = MBA † jk
[MAB ] jk = a j bk = bk a j
.
So
Likewise, of course,
version: 9/22/2013
MAB = MBA †
.
(5.1a)
MBA = MAB †
.
(5.1b)
Chapter & Page: 5–6
2.
Change of Basis
The second observation is that MAA and MBB greatly simplify if the bases are orthonormal. If A is orthonormal, then
[MAA ] jk = a j ak = δ jk = [I] jk .
So
if A is orthonormal .
MAA = I
(5.2a)
By exactly the same reasoning, it should be clear that
if B is orthonormal .
MBB = I
(5.2b)
Now let v be any vector in V and, for convenience, let us denote the components of v with
respect to A and B using α j ’s and β j ’s respectively,
|viA
Remember, this means


α1
α 
 2

= 
 .. 
 . 
αN
v =
and
X
k
α k ak =
|viB
X


β1
β 
 2

= 
 .. 
 . 
βN
βk bk
.
.
(5.3)
k
Our goal is to find the relations between the α j ’s and the β j ’s so that we can find one set
given the other. One set of relations can be found by taking the inner product of v with each a j
and using equations (5.3). Doing so:
*
+
*
+
X
X
aj v = aj α k ak = a j βk bk
k
H⇒
H⇒
H⇒
k
X X aj v =
α k a j ak =
βk a j bk
k
k
X
X
aj v =
a j ak α k =
a j bk βk
k
k
X
X
aj v =
[MAA ] jk αk =
[MAB ] jk βk
k
k
The formulas in the second equation in the last line are simply formulas for the j th entry in the
products of MAA and MAB with the column matrices of αk ’s and β’s . So that equation tells
us that
 
 


β1
α1
h a1 | v i
β 
α 
h a | v i
 2
 2

 2
 = MAA  .  = MAB  . 

..
 . 
 . 


.
 . 

 . 

βN
αN
h aN | v i
Change of Basis for Vector Components:
The General Case
Chapter & Page: 5–7
Recalling what the column matrices of αk ’s and β’s are, we see that this reduces to


h a1 | v i
h a | v i
 2


 = MAA |viA = MAB |viB .
..


.


h aN | v i
(5.4)
?◮Exercise 5.5 (semi-optional):
Using the same assumptions as were used to derive (5.4),
derive that


h b1 | v i
h b | v i
 2


 = MBA |viA = MBB |viB .
(5.5)
.


..


h bN | v i
Equations (5.4) and (5.5) give the relations between the components of a vector with respect
to the two different bases, as well as the relations between these components and the inner products
of the vector with each basis vector. Our current main interest is in the relations between the
different components (i.e., the rightmost two-thirds of (5.4) and (5.5)). For future reference, let
us summarize what the above tells us in that regard.
Lemma 5.2 (The Big Lemma on General Change of Bases)
Let
A = { a1 , a2 , . . . , a N }
and
B = { b1 , b2 , . . . , b N }
be any two bases for an N -dimensional vector space V . Then, for any v in V ,
and
MAA |viA = MAB |viB
(5.6a)
MBA |viA = MBB |viB
(5.6b)
where MAA , MAB , MBB and MBA are the four N × N matrices given by
,
,
[MAA ] jk = a j ak
[MAB ] jk = a j bk
and
.
[MBB ] jk = b j bk
[MBA ] jk = b j ak
Naturally, the above formulas simplify considerably when the two bases are orthonormal.
That will be of particular interest to us. Before going there, however, let us observe an number
of “little results” that immediately follow from the above lemma and its derivation, and which
apply whether or not either basis is orthonormal:
1.
Solving equations (5.6a) and (5.6b) for |viB yields the change of basis formulas
and
|viB = MAB −1 MAA |viA
|viB = MBB −1 MBA |viA
version: 9/22/2013
.
Chapter & Page: 5–8
2.
Change of Basis
Solving equations (5.6a) and (5.6b) for |viA yields the change of basis formulas
|viA = MAA −1 MAB |viB
and
|viA = MBA −1 MBB |viB
3.
.
From equation (5.4) we have that
|viA


h a1 | v i
h a | v i
 2

−1 

= MAA 
..

.


h aN | v i
.
Don’t memorize these formulas. They are nice formulas and can save a little work when
dealing with arbitrary matrices. However, since we will attempt to restrict ourselves to orthonormal (or at least orthogonal) bases, we won’t have a great need for these formulas as written. If
you must memorize formulas, memorize those in the Big Lemma — they are much more easily
memorized. And besides the above change of basis formulas are easily derived from the formulas
in that lemma.
5.3
Change of Basis for Vector Components:
When the Bases Are Orthonormal
The Main Result
Now consider how the formulas in the Big Lemma (lemma 5.2 on page 5–7) simplify when the
bases
A = { a1 , a2 , . . . , a N }
and
B = { b1 , b2 , . . . , b N }
are orthonormal. Then, as noted on page 5–5 (equation set (5.1)),
MAA = I
and
MBB = I
.
Thus, equation (5.6a) in the Big Lemma reduces to
MAB |viB = |viA
,
(5.7a)
and equation (5.6b) in the Big Lemma reduces to
|viB = MBA |viA
.
(5.7b)
Equations (5.7a) and (5.7b) are certainly nicer than the original equations in the Big Lemma,
but look at what we get when they are combined:
|viB = MBA |viA = MBA MAB |viB
| {z }
|viA
That is,
|viB = [MBA MAB ] |viB
for each v ∈ V
,
Change of Basis for Vector Components:
When the Bases Are Orthonormal
Chapter & Page: 5–9
which is only possible if
MBA MAB = I
,
and which, in turn, means that
MBA −1 = MAB
MAB −1 = MBA
and
.
But remember (equation set (5.1) on page 5–5) that
MAB = MBA †
MBA = MAB †
and
.
Combining this with the previous line gives us
MBA † = MBA −1
MAB † = MAB −1
and
.
So MBA and its adjoint MAB are unitary matrices.
What the above tells us it that all the matrices involved are easily computed via the adjoint
once we know one of the matrices.
We will make one more small set of observation regarding matrices MAB and MBA and
the components of the vectors in basis B with respect to basis A . Remember, since A is an
orthonormal basis, the k th component of b j with respect to A is given by h ak | b j i . That is,
X
X X
bj =
ak b j ak =
ak ak b j =
ak [MAB ]k j .
k
k
k
This is just the formula for the matrix product
b1 b2 · · · b N = a1 a2 · · · a N MAB
.
In summary, we have the following:
Theorem 5.3 (The Big Theorem on Change of Orthonormal Bases)
Let V be an N -dimensional vector space with orthonormal bases
A = { a1 , a2 , . . . , a N }
B = { b1 , b2 , . . . , b N }
and
.
Let MAB and MBA be a pair of N × N matrices which are adjoints of each other and which
satisfy any one of the following sets of conditions:
1. [MAB ] jk = a j bk
.
2. [MBA ] jk = b j ak
.
3. b1 b2 · · · b N = a1 a2 · · · a N MAB .
4. a1 a2 · · · a N = b1 b2 · · · b N MBA .
5.
|viA = MAB |viB
for each v ∈ V
.
6.
|viB = MBA |viA
for each v ∈ V
.
Then MAB and MBA are unitary matrices satisfying all of the above conditions, as well as
MAB = MBA † = MBA −1
version: 9/22/2013
and
MBA = MAB † = MAB −1
.
Chapter & Page: 5–10
Change of Basis
Finding the Matrices
If you keep in mind the equation
b1 b2 · · · b N = a1 a2 · · · a N MAB
from the Big Theorem (theorem 5.3, above), and have the formulas for the bk ’s in terms of the
a j ’s ,
bk = βk1 a1 + βk2 a2 + · · · + βk N a N
for j = 1, 2, . . . , N ,
then you can obtain the matrix MAB easily by simply noting that each of these equations can be
written as


βk1


 βk2 

bk = a1 a2 · · · a N 
 ..  .
 . 
βk N
Comparing the last equation with the one a few lines earlier for b1 b2 · · · b N , it should
be clear that the above column matrix on the right must be the k th column in MAB .
See, no computations are needed (provided the bases are orthonormal).
!◮Example 5.2:
basis
Let V be a three-dimensional space of traditional vectors with a standard
A = { i, j, k }
.
Let
B = { b1 , b2 , b3 }
where
1
b1 = √ [i + k]
2
1
,
and
b2 = √ [i − k]
2
b3 = j
“By inspection”, it should be clear that B is also an orthonormal basis for V .
Now observe that the formulas for the bk ’s can be rewritten as
 1 
 1 
and
√
2

 
b1 = i j k  0 
 1 
√
2
,
√
2 



b2 = i j k  0 
 1 
−√
 
0
 
b3 = i j k 1
0
2
,
which can be written even more consisely as
b1 b2
 1
√
2

 0
b3 = i j k 
 1
√
2
1
√
2
0
1
−√
2
0


1


0
.
.
Traditional Rotated and Flipped Bases
Chapter & Page: 5–11
As noted above, we then must have
MAB
 1
√
 2

=  0
 1
√
2
1
√
2
0
1
−√
2
0


1


0
.
Multiple Change of Basis
Suppose, now, we have three orthonormal bases
and
A = { a1 , a2 , . . . , a N }
,
B = { b1 , b2 , . . . , b N }
C = { c1 , c2 , . . . , c N }
Then, in addition to MBA and MAB we also have MAC , MBA , MCA and MCB , all defined
analogously to the way we defined MBA and MAB . Applying the above theorem we see that,
for every v in V ,
MCA |viA = |viC = MCB |viB = MCB MBA |viA
Thus,
MCA |viA = MCB MBA |viA
for every v ∈ V
.
.
From this, it immediately follows that
MCA = MCB MBA
.
(5.8)
Remember, this is assuming the bases are all orthonormal.
5.4
Traditional Rotated and Flipped Bases
Let us briefly restrict ourselves to a two- or three-dimensional traditional vector space V , with
orthonormal bases
or
version: 9/22/2013
A = { a1 , a2 }
and
B = { b1 , b2 }
A = { a1 , a2 , a3 }
and
B = { b1 , b2 , b3 }
.
Chapter & Page: 5–12
Change of Basis
Direction Cosines
Since we are assuming a space of traditional vectors, the scalars are real numbers, the inner
product is the traditional dot product, and the change of bases matrices MAB and MBA are
orthogonal and are transposes of each other. In this case,
.
[MAB ] jk = a j bk = a j · bk = a j kbk k cos θ(a j , bk )
And because the a j ’s and bk ’s are unit vectors, this reduces to
where θ jk = angle between a j and bk
[MAB ] jk = cos θ jk
.
These are called the direction cosines relating the two bases.
Two-Dimensional Case
If V is two dimensional, then the possible geometric relationships between
A = { a1 , a2 }
and
B = { b1 , b2 }
are easily sketched:
b2
a2
a2
b1
ψ
a1
b1
or
ψ
a1
b2
Clearly, we can only have one of the following two situations:
1.
{b1 , b2 } are the vectors that can be obtained
by
“rotating”
{a1 , a2 } through some angle
ψ , in which case the matrix relating b1 b2 to a1 a2 is a “rotation matrix” Rψ
(see problem K in Homework Handout IV ).
2.
{b1 , b2 } are the vectors that can be obtained by “rotating” {a1 , a2 } through some angle
ψ
, and then
“flipping”
the direction of the second vector. In this case, the matrix relating
b1 b2 to a1 a2 is given by a “rotation matrix” Rψ followed by a “flip the direction
of the second vector matrix”. (Equivalently, we could first rotate by a slightly different
angle and then flip the direction of the first rotated vector — or do the “flipping” first to
either a1 or a2 and then rotate.)
or
With a little thought, it should be clear that the matrices for the “flips” in the second case
are simply
"
#
"
#
−1 0
1 0
F1 =
and
F2 =
.
0 1
0 −1
Traditional Rotated and Flipped Bases
Chapter & Page: 5–13
Clearly, then the matrix MBA can be written as either a single rotation matrix, or as the
product of a rotation matrix with one of these “flip” matrices. It’s not at all hard to show that
det(any rotation matrix) = 1
and
det(either flip matrix) = −1
.
Consequently, at least if we are considering orthonormal bases A and B for a two-dimensional,
traditional vector space,
A and B are rotated images of each other
⇐⇒
det (MBA ) = +1 .
Three-Dimensional Case
Likewise, if
A = { a1 , a2 , a3 }
and
B = { b1 , b2 , b3 }
are any two orthonormal bases for a traditional three-dimensional space, then MBA will either
be a matrix for a “rotation by some angle about some vector”, or the product of such a rotation
matrix with one of the flip matrices






−1 0 0
1 0 0
1 0 0






F1 =  0 1 0 , F2 = 0 −1 0 and F3 = 0 1 0  .
0 0 1
0 0 1
0 0 −1
And, again, it can be shown that (still assuming A and B are orthonormal bases)
A and B are rotated images of each other
⇐⇒
det (MBA ) = +1 .
Consequently,
A and B are both right-handed, or are both left-handed
⇐⇒
det(MBA ) = +1
.
It can be shown that the rotation matrix, itself, can be written as a product of three “simple
rotation matrices”
R3,γ R2,β R3,α
where R j,φ corresponds to a rotation of angle φ about the j th basis vector. The angles α , β ,
and γ are the infamous “Euler angles”. Now turn to pages 139, 140 and 141 in Arfken, Weber
and Harris and skim their discussion of rotations and Euler angles.
Warnings:
1.
There are subtle differences between a matrix for a rotation operator (to be discussed in
the next chapter) and the change of basis matrix for “rotated bases”. Both may be called
rotation matrices, but may differ in the signs of some of the corresponding entries.
2.
I have not carefully checked Arken and Weber’s discussion so I will not guarantee whether
their product
R3,γ R2,β R3,α
is actually what we are calling MBA or if the signs of some of the entries in some of the
matrices need to be switched. In other words, I have not verified whether each of Arfken,
Weber and Harris’s “rotation matrices” are describing the operation of rotating a vector
in space, or describing a change of basis when one basis is a rotation of the other.
3.
Different authors use different conventions for the Euler angles — different axes of
rotation and different order of operations. Be careful, the computations based on two
different conventions may well lead to two different and incompatible results.
version: 9/22/2013
Chapter & Page: 5–14
5.5
Change of Basis
Sidenote: Vectors Defined by a Transformation
Law
Suppose we have a huge collection of bases
1 2 3
m
m
B , B , B , ...,
with Bm = bm
1 , b2 , . . . , b1
for our N -dimensional vector space V (for example, this might be the collection of all “rotations”
of some favorite orthonormal basis). Then each v in B has components with respect to each of
these Bm ’s ,
 m
v1
v m 
N
X
 2

|viBm = 
where
v
=
v mj bmj .
.
 . 
.
 
j=1
v mN
Also, for each pair of bases Bm and Bn , there will be a corresponding “change of basis”
matrix MBm Bn such that
for each v ∈ V
|viBm = MBm Bn |viBn
.
(5.9)
This last expression is sometimes called a “transformation law”, even though we are not really
dealing with a true transform of vectors here.
Now suppose that, in the course of tedious calculations and/or drinking, we obtain, corre
sponding to each different basis Bm , a corresponding N -tuple of scalars w1m , w2m , . . . , wmN .
The question naturally arrises as to whether
w11 , w21 , . . . , w1N
,
w12 , w22 , . . . , w2N
,
w13 , w23 , . . . , w3N
, ...
all describe the same vector w but in the various corresponding bases. Obviously (to us at least)
the answer is “yes” if and only if
 m
 n
w1
w1
w m 
 wn 
 2
 2
 .  = MBm Bn  .  for every pair of bases Bm and Bn .
(5.10)
 . 
 . 
 . 
 . 
wmN
wnN
This is because we can set
w =
N
X
w1j b1j
j=1
and use the fact that (5.10) holds to verify that the change of basis formulas
(5.9) will give us,
for each Bm , the originally given N -tuple of scalars w1m , w2m , . . . , wmN as the components of
w with respect to that basis.
This w is called the vector defined by the transformation law (5.9).
“Volumes” of N -Dimensional Hyper-Parallelpipeds (Part II)
5.6
Chapter & Page: 5–15
“Volumes” of N-Dimensional Hyper-Parallelpipeds
(Part II)
Let us now continue our discussion from section 3.6 on “volumes of hyper-parallelepipeds”.
Recollections from Section 3.6
Recall the situation and notation: In a N -dimensional Euclidean space we have a “hyperparallelepiped” P N generated by a linearly independent set of N vectors
{ v1 , v2 , . . . , v N }
.
Our goal is to find a way to compute the N -dimensional volume of this object — denoted by
VN (P N ) — using the components of the vk ’s with respect to any basis
B = { b1 , b2 , . . . , b N }
.
For our computations, we are letting VB be the “matrix of components of the v j ’s with respect
to basis B ” given by
VB = “ |v1 iB |v2 iB |v3 iB

v1,1
v
 1,2

v
· · · |v N iB ” = 
 1,3
 ..
 .
v1,N
where, for j = 1, 2, . . . , N ,
vj =
N
X
v j,k bk
v2,1
v2,2
v2,3
..
.
v2,N
v3,1
v3,2
v3,3
..
.
v3,N
···
···
···
..
.

v N ,1
v N ,2 


v N ,3 

.. 
. 
· · · v N ,N
.
k=1
Recall, also, that we derived one formula for the volume; namely,
VN (P N ) = det(VN )
(5.11)
where N is the orthonormal set generated from {v1 , v2 , . . . , v N } via the Gram-Schmidt procedure (see theorem 3.4 on page 3–27).
General Component Formulas
From lemma 5.2 on page 5–7 on general change of bases formulas, we know
MN N v j N = MN B v j B
.
Since N is orthonormal, however, MN N is just the identity matrix, and the above reduces to
version: 9/22/2013
v j
v j
=
M
N
B
N
B
.
Chapter & Page: 5–16
Thus,
Change of Basis
VN = |v1 iN |v2 iN |v3 iN · · · |v N iN
= MN B |v1 iB |v2 iB |v3 iB · · · |v N iB = MN B VB
.
Combining this with formula (5.11) and a property of determinants (equation (4.3) on page 4–11)
then yields
VN (P N ) = det(VN ) = det (MN B VB ) = (det MN B ) (det VB )
.
Let us simplify matters a little more. Let U be any orthonormal basis. As noted on page
5–11 (equation (5.8)),
MN B = MN U MUB .
But, since U and N are orthonormal bases of “traditional” vectors, MN U must be an orthogonal matrix (and, thus, have det(MN U ) = ±1 ). Consequently,
det (MN B ) = det (MN U MUB )
= det (MN U ) det (MUB ) = ± det (MUB )
.
Combining this with our last formula for VN (P N ) , gives us our most general component formula
for VN (P N ) (and the next theorem).
Theorem 5.4 (general component formula for the volume of a hyper-parallelepiped)
Let P N be the hyper-parallelepiped in an generated by a linearly independent set {v1 , v2 , . . . , v N }
of traditional vectors. Also, using any basis B for the space spanned by the vk ’s , let VB be the
matrix of components of the v j ’s with respect to basis B ,

VB
v1,1
v
 1,2

v
= 
 1,3
 ..
 .
v1,N
v2,1
v2,2
v2,3
..
.
v2,N
v3,1
v3,2
v3,3
..
.
v3,N
···
···
···
..
.

v N ,1
v N ,2 


v N ,3 

.. 
. 
· · · v N ,N
with


vk,1
 vk,2 


 vk,3 

 = |vk iB
 .. 
 . 
.
vk,N
Then, letting VN (P N ) denote the N -dimensional volume of P N ,
VN (P N ) = |det (MUB ) det (VB )|
(5.12)
where U is any orthonormal basis.
Special Cases
B is orthonormal
In particular, if B is an orthonormal basis, then we can let U = B in formula (5.12). Because
of the orthonormality of B we have
det (MUB ) = det (MBB ) = det (I) = 1
,
“Volumes” of N -Dimensional Hyper-Parallelpipeds (Part II)
Chapter & Page: 5–17
and formula (5.12) reduces to

v1,1
v
 1,2

v
VN (P N ) = |det (VB )| = det 
 1,3
 ..
 .
v
v2,1
v2,2
v2,3
..
.
v2,N
1,N
?◮Exercise 5.6:
basis

v N ,1 v N ,2 


v N ,3 

.. 
. 
· · · v N ,N ···
···
···
..
.
v3,1
v3,2
v3,3
..
.
v3,N
.
(5.13)
Let V be a four-dimensional space of traditional vectors with orthonormal
{i, j, k, l}
and let
,
v1 = 3i
v2 = 2i + 4j
,
v3 = 8i − 3j + 5k
and
v4 = 4i + 7j − 2k + 2l
.
Using formula (5.13), compute the “four-dimensional volume” of the hyper-parallelepiped
generated by v1 , v2 , v3 and v4 . (Compare the work done here to that done for the same
problem in exercise set 3.15 on page 3–24.)
B and P N are “Parallel”
Let’s now assume each b j is parallel to v j and “pointing in the same direction”. That is
v jk = 0
if
j 6= k
and
vjj ≥ 0
.
For “convenience”, let
1vk = vkk
so that we can write
v j = 1v j b j
for
j = 1, 2, . . . , N
.
Then

1v1
0
0
 0
1v2
0


0
0
1v3
det (VB ) = det 

..
..
 ..
 .
.
.
0
0
0
and formula (5.12) becomes

0
0 


0  = 1v1 1v2 1v3 · · · 1v N

.. 
. 
· · · 1v N
···
···
···
..
.
VN (P N ) = |det(MUB )| 1v1 1v2 1v3 · · · 1v N
where U is any orthonormal basis.
version: 9/22/2013
.
,
(5.14)
Chapter & Page: 5–18
Change of Basis
Of course, if there is any othogonality, then the above simplifies. In particular, if B is
orthogonal, then you can easily show that
VN (P N ) = kb1 k kb2 k · · · kb N k 1v1 1v2 1v3 · · · 1v N
.
If you compare formula (5.14) with formulas (3.11) through (3.11), starting on page 3–28,
it should be clear that we’ve already derived geometric formulas for the above |det MUB | when
N = 1 , N = 2 and N = 3 . Since “cut and paste” is so easy, we’ll rewrite those geometric
formulas:
V1 (P1 ) = · · · = kb1 k 1v1
V2 (P2 ) =
and
V3 (P3 ) =
q
,
kb1 k2 kb2 k2 − (b1 · b2 )2 1v1 1v2
√
,
A − B + C 1v1 1v2 1v3
where
A = kb1 k2 kb2 k2 kb3 k2
and
,
B = kb1 k2 (b2 · b3 )2 + kb2 k2 (b1 · b2 )2 + kb3 k2 (b1 · b2 )2
C = 2
(b1 · b2 )2 (b1 · b3 )2
(b2 · b1 )2 (b2 · b3 )2
=
2
= ···
kb1 k2
kb2 k2
,
.
As we noted in on page 3–29, these formulas reduce even further if the bk ’s are unit vectors,
and even further if the set of bk ’s is orthonormal. If you’ve not already done so, then do the next
exercise (which is the same as exercise 3.17):
?◮Exercise 5.7:
To what do the above formulas for V1 (P1 ) , V2 (P2 ) and V3 (P3 ) reduce
a: when the bk ’s are unit vectors.
b: when the set of bk is orthonormal.