# MATB24H3 Lecture Notes - Orthogonal Matrix, Orthogonal Complement, Hermitian Matrix

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Published on 21 Apr 2013

Department

Mathematics

Course

MATB24H3

Professor

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 Cis 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∈C, z−1=z

|z|2. We use that to reduce matrices with complex numbers

as entries.

EXAMPLE: Find the inverse of the matrix A=

i1 3i

1 + i1 2i

−1 0 1 + i

solution.

A=

i1 3i

1 + i1 2i

−1 0 1 + i

100

010

001

−iR1

1−i

2R2

−−−−→

−1R3

1−i3

11−i

21 + i

1 0 −1−i

−i0 0

01−i

20

0 0 −1

R2−R1

−−−−→

R3−R1

1−i3

01+i

2−2 + i

0i−4−i

−i0 0

i1−i

20

i0−1

(1−i)R2

−−−−→

−iR3

1−i3

0 1 −1 + 3i

0 1 −1 + 4i

−i0 0

1 + i−i0

1 0 i

R3−R2

−−−−→

1−i3

0 1 −1 + 3i

0 0 i

−i0 0

1 + i−i0

−i i i

R1+3iR3

R2+(−3−i)R3

−−−−−−−−→

−iR3

1−i0

0 1 0

0 0 1

−i+ 3 −3−3

4i1−4i1−3i

−1 1 1

R1+iR2

−−−−→

100

010

001

−i−1 1 + i i

4i1−4i1−3i

−1 1 1

1

Thus: A−1=

−i−1 1 + i i

4i1−4i1−3i

−1 1 1

Similarly,:

(i) the process of ﬁnding if a set of vectors in Cnis linearly independent

the same as the process in Rn.

(ii) given a vector v∈Cn, the process of ﬁnding a coordinate vector vB

relative to a given ordered basis Bof Cnis also the same as the process

in Rn.

However, you must always use the addition and multiplication as deﬁned

in the ﬁeld C.

Euclidean Inner Product in Cn

We know that Cnis a vector space, but is it an inner product space? The

dot product makes Rnan inner product space. Does it make Cnalso?

If we try it we see that there are some problems.

For example, in C2if v= [a, ai] for a6= 0 gives

kvk2= [a, ai]·[a, ai] = a2−a2= 0.

Also if b= [a, bi] with a < b then:

kbk2= [a, bi]·[a, bi] = a2−b2<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 Cn.

2

Regardless, Cnhas an inner product deﬁned on it:

DEFINITION: If u= [u1, u2, ..., un] and v= [v1, v2, ..., vn] are in Cnthen

the Euclidean inner product of u and v is:

<u,v>=u1v1+u2v2+... +unvn

This inner product has the following properties:

THEOREM: Let u,v,w∈Cnand let z∈Cbe 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.

1. <u,u>=u1u1+u2u2+... +unun=|u1|2+|u2|2+... +|un|2≥0

and it is equal to zero if and only if every |ui|2= 0, which is true if and

only if each ui= 0

2. <u,v>=u1v1+u2v2+... +unvn=u1v1+u2v2+... +unvn=

u1v1+u2v2+... +unvn=v1u1+v2u2+... +vnun=<v,u>

Verify the rest as an exercise.

This inner product is not commutative as the inner product is in Rnand

also, the result of an inner product in Cnis 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

## Document Summary

In this lecture we will work with matrices that have entries from c. Since we know that c is a eld we know that every nonzero element in c has an inverse. And from the previous lecture we know that for an nonzero z c, as entries. We use that to reduce matrices with complex numbers z 1 = z. Example: find the inverse of the matrix a = i. 2 2 + i i 4 i. However, you must always use the addition and multiplication as de ned in the eld c. The dot product makes rn an inner product space. If we try it we see that there are some problems. For example, in c2 if v = [a, ai] for a 6= 0 gives kvk2 = [a, ai] [a, ai] = a2 a2 = 0. Also if b = [a, bi] with a < b then: kbk2 = [a, bi] [a, bi] = a2 b2 < 0.