Class Notes
(808,239)

Canada
(493,100)

University of Toronto Scarborough
(30,807)

Mathematics
(837)

MATB24H3
(13)

Sophie Chrysostomou
(12)

Lecture

# Lecture19-20f.pdf

Unlock Document

University of Toronto Scarborough

Mathematics

MATB24H3

Sophie Chrysostomou

Winter

Description

University of Toronto at Scarborough
Department of Computer & Mathematical Sciences
MAT B24S Fall 2011
MAT B24
Lecture 19-20
Matrices with Complex Numbers
In this lecture we will work with matrices that have entries from C.
Since we know that C is a ﬁeld we know that every nonzero element in C
has an inverse. And from the previous lecture we know that for an nonzero
z
z ∈ C, z−1 = . We use that to reduce matrices with complex numbers
|z|2
as entries.
i 1 3i
EXAMPLE: Find the inverse of the matrix A = 1 + i 1 2i
−1 0 1 + i
solution.
▯ ▯
i 1 3i ▯1 0 0 1−i1 1 −i 3 ▯−i 0 0
A = 1 + i 1 2i ▯0 1 0 −−−→2 1 1−i 1 + i▯0 1−i 0
▯ −1R3 2 ▯ 2
−1 0 1 + i 0 0 1 1 0 −1 − i 0 0 −1
▯ ▯
R2−R1 1 −i 3 ▯−i 0 0 (1−i)2 1 −i 3 ▯ −i 0 0
−−−−→ 0 2+i −2 + i▯i 12i 0 −−−−→ 0 1 −1 + 3i▯1 + i −i 0
R3−R1 0 i −4 − i i 0 −1 −iR3 0 1 −1 + 4i▯ 1 0 i
1 −i 3 ▯ −i 0 0 R1+3i3 1 −i 0 ▯−i + 3 −3 −3
R3−R2 ▯ R 2(−3−i)3 ▯
−−−−→ 0 1 −1 + 3i▯1 + i −i 0 −−−−iR3− → 0 1 0▯ 4i 1 − 4i 1 − 3i
0 0 i −i i i 0 0 1 −1 1 1
▯
1 0 0 ▯i − 1 1 + i i
−−−−i→ 0 1 0 ▯ 4i 1 − 4i 1 − 3i
▯
0 0 1 −1 1 1
1
−i − 1 1 + i i
Thus: A −1 = 4i 1 − 4i 1 − 3i
−1 1 1
Similarly,:
(i) the process of ﬁnding if a set of vectors in C is linearly independent
the same as the process in R .
(ii) given a vector v ∈ C , the process of ﬁnding a coordinate vectorBv
n
relative to a given ordered basisB of C is also the same as the process
in R .
However, you must always use the addition and multiplication as deﬁned
in the ﬁeld C.
Euclidean Inner Product in C n
n
We know that C is a vector space, but is it an inner product space? The
dot product makes R an inner product space. Does it make C also?
If we try it we see that there are some problems.
For example, in C if v = [a,ai] for a ▯= 0 gives
▯v▯ = [a,ai] ▯ [a,ai] = a − a = 0.
Also if b = [a,bi] with a < b then:
2 2 2
▯b▯ = [a,bi] ▯ [a,bi] = a − b < 0
So we would get the magnitude of a nonzero vector to be 0 and the magnitude
of a vector to be negative. That is deﬁnitely contrary to the deﬁnition of an
inner product. So the dot product is not an inner product in C .
2 Regardless, C has an inner product deﬁned on it:
n
DEFINITION: If u = [u ,u ,..1,u 2 and n = [v ,v ,...1v 2 are in C then
the Euclidean inner product of u and v is:
< u,v > = u v 1 1 v + 2 2 + u v n n
This inner product has the following properties:
THEOREM: Let u, v, w ∈ C and let z ∈ C be a scalar.Then:
1. < u,u > ≥ 0, and < u,u > = 0 if and only if u = 0.
2. < u,v > = < v,u >,
3. < (u + v),w > = < u,w > + < v,w >,
4. < w,(u + v) > = < w,u > + < w,v >,
5. < zu,v > = z < u,v >, and < u,zv > = z < u,v >.
proof.
2 2 2
1. < u,u > = u u + 1 1 + .2. 2 u u = |un| n |u | 1 ... +2|u | ≥ 0 n
and it is equal to zero if and only if every |u | = 0, which is true if and
i
only if each ui= 0
2. < u,v > = u v + u1 1+ ..2 2 u v n n = u v 1 1 v +2 2. + u v n n =
u1 1+ u v2 2... + u vn nv u +1v 1 + .2.2+ v u =
Verify the rest as an exercise.
n
This inner product is not commutative as the inner product is in R and
also, the result of an inner product in C is in C.
It seems like this inner product does not satisfy the inner product of
spaces over the reals. So it is puzzling why it is called an inner product. The
reason is that the deﬁnition given here of an inner product is satisﬁed by the
deﬁnition of an inner product over the reals.
3 THEOREM: Let u, v, w ∈ R and let r ∈ R be a scalar.Then:
1. < u,u > ≥ 0, and < u,u > = 0 if and only if u = 0.
2. < u,v > = < v,u > =< v,u >, therefore commutativity holds
3. < (u + v),w > = < u,w > + < v,w > = < w,u > + < w,v > =
< w,(u + v) >,
4. < ru,v > = r < u,v > = r < u,v > = < u,rv >
this inner product is actually the dot product if restricted to the reals.
With this inner product we can get the following deﬁnitions for vectors
n
u,v ∈ C to be
1. perpendicular if < u,v >= 0,
2. parallel if u = zv for some scalar z ∈ C.
3. A unit vector is a vector with magnitude of 1.
4. If W is a subspace of C , there is an orthogonal complement Wthat
can be found the same way as it is found in R by replacing the dot
n
product with the inner product on C .
5. The Gram Schmidt process is the same as well. However, using this
inner product which is not commutative we must be careful. Given
vectors u,v we must ﬁnd
(a) what is the projection of u onto v,
< v,u > < u,v >
is it v or v
< v,v > < v,v >
and
(b) what vector is the orthogonal complement of u onto v.
Is it u − < v,u > v or u − < u,v > v
< v,v > < v,v >
4 solution.
< v,u > < v,u >
< u − v,v > = < u,v > − < v,v >
< v,v > < v,v >
= < u,v > − < v,u >
= < u,v > − < u,v >= 0
Whereas:
< u,v > < u,v >
< u − v,v > = < u,v > − < v,v >
< v,v > < v,v >
= < u,v > − < u,v >
= < u,v > − < v,u >
which is not necessarily zero.
so the orthogonal complement of u onto v isu − < v,u > v
< v,v >
EXAMPLE: Given the basis {[−i,1,1,0],[i,1,i,0],[1,1,00]}ﬁnd an orthog-
onal basis for the subspace W that they span.
solution. Let a1= [−i,1,1,0], a2= [i,1,i,0], 3 = [1,1,00], then:
v1= [−i,1,1,0]
▯ ▯
v = a − < v1,a2> v
2 2 < v1,v1> 1
▯ ▯
((i)(i) + (1)(1) + (1)(i) + 0)
= [i,1,i,0] − ((i)(−i) + 1 + 1) [−i,1,1,0]
▯ ▯
i 1 i 2i
= [i,1,i,0] −3 [−i,1,1,0] = −3+ i,1 − ,3 3 ,0
5 ▯ ▯
< v 1a 3 < v 2a 3
v3 = a 3 < v ,v > v 1 < v ,v > v 2
1 1 2 2
< [−i,1,1,0][1,1,0,0] >
= [1,1,0,0] − [−i,1,1,0]
3
▯ ▯
< − 3+ i,1 − 3 3,0 ,[1,1,0,0] >
− ▯ 1 i 2i ▯ ▯ 1 i 2i ▯
< − 3 + i,1 − 3 3,0 , − 3 i,1 − , 3 3,0 >
▯ 1 i ▯▯ ▯
= [1,1,0,0] − (1 + i)[−i,1,1,0] − − 3− i + 1 + 3 − 1 + i,1 − , 2i,0
3 1 + 1 + 1 +1 + 4 3 3 3
9 9 9
▯ ▯ ▯ ▯▯ ▯
1 − i1 + i 1 + i 232i 1 i 2i
= [1,1,0,0] − , , ,0 − 24 − + i,1 − , ,0
3 3 3 9 3 3 3
▯ ▯ ▯ ▯▯ ▯
= [1,1,0,0] − 1 − ,1 + i,1 + i,0 − (1 − i) − 1 + i,1 − ,2i ,0
3 3 3 4 3 3 3
▯ ▯ ▯ ▯
1 − i1 + i 1 + i 1 2i 1 2i 1 i
= [1,1,0,0] − 3 , 3 , 3 ,0 − 6 + 6 6 − 6 6 + ,6
▯ ▯
3 3 −3 3i
= , , − ,0
6 6 6 6
▯ ▯
1 1 −1 i
= , , − ,0
2 2 2 2
Thus v ,v ,v is an orthogonal basis for W. Since
1 2 3
▯
√ 8
||v1|| = 3, ||2 || = , ||3 || = 1
3
then
▯ ▯ ▯ ▯ ▯
1 3 1 i 2i 1 1 −1 i
q1= √ [−i,1,1,0], q2= − + i,1 − , ,0 , q 3 , , − ,0
3 8 3 3 3 2 2 2 2
form an orthonormal basis for W.
6 Hermitian and Unitary Matrices
n ▯ n
The Euclidean inner product on C is deﬁned by < x,y > = i=1xi i
n n
for all x,y ∈ C From now on we will be using this inner product for C .
In R we had that x ▯ y = x y where the operation on the right is the
inner product and the one on the left is matrix multiplication. Is there a
relation like that with the Euclidean inner

More
Less
Related notes for MATB24H3