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## 1.3. Operations on matrices

m x n matrices A = (aij ) ja B = (bij ) are equal, A = B if aij = bij    i, j, that is, if the corresponding entries are equal.

A + B = C,

whose entries are cij = aij + bij .

The product cA of a matrix A = (aij ) and a number c is the matrix (caij ),

cA = (caij ),          ( = Ac)

A - B = (aij - bij ).

The product AB (in this order) of matrices Am x n = (aik ) and Bn x p = (bkj ) is the m x p matrix C = (cij ), where

cij = aik bkj .

 a11 a12 ··· a1n a21 a22 ··· a2n : am1 am2 ··· amn
 b11 b12 ··· b1p b21 b22 ··· b2p : : bn1 bn2 ··· bnp

 = a11 b11 + a12 b21 + ··· + a1n bn1 a11 b12 + ··· + a1n bn2  ··· a21 b11 + a22 b21 + ··· + a2n bn1 a21 b12 + ··· + a2n bn2  ··· : : am1 b11 + am2 b21 + ··· + amn bn1 am1 b12 + ··· + amn bn2  ···

Note that the product AB is defined only when the number of the columns of A and the number of the rows of B coincide.
It is possible that AB is defined while BA is not.

If AB = BA, we say that A and B commute. In this case A and B are n x n matrices and therefore they are square matrices of the same dimension.

Proposition 1 When the operations below are defined, we have

1) A + B = B + A
2) (A + B) + C = A + (B + C)
3) A + O = A
4) A + ( - A) = O
5) (µ)A = (µ A)
6) ( + µ)A = A + µ A
7) (A + B) = A + B
8) 1· A = A
9) (AB)C = A (BC)
10) A(B + C) = AB + AC
11) (A + B)C = AC + BC
12) (AB) = (A)B = A(B)
13) IA = A
14) AI = A

Proof. Follows from the previous definitions of matrix operations.

Notice! If AB = 0, it does not necessarily follow that A = 0 or B = 0:

Example

 A = 1 2 -1 0, 1 0 -1
 B = 2 1 0 0 0 2 1
 AB = 1 2 -1 1 0 -1
 2 1 0 0 2 1
 = 02x2
Also
 IC = C: 1 0 0 0 1 0 0 0 1
 a b c d e f g h i
 = a b c = C d e f g h i
So far we have defined three operations: the product of a matrix and a scalar, the sum of matrices and the product of matrices. We now define

The transpose or the transposed matrix of an m x n matrix A = (aij ) is the n x m matrix AT = (aji ), that is, the rows of AT are the columns of A and the columns of AT are the rows of A.

Proposition 2

1) (AT)T = A
2) (A + B)T = AT + BT
3) (A)T = AT
4) (AB)T = BTAT

Problem: Show that a square matrix A can be written as the sum of a symmetric and a skew symmetric matrix: A = ½(A + AT) + ½(A - AT).

In applications (for example, in signal processing) one often needs real symmetric Toeplitz matrices (for example, autocorrelation matrix)

 r (0) r (1) r (2) ··· r (n - 1) . r (1) r (0) r (1) ··· r (n - 2) r (2) r (1) r (0) ··· r (n - 3) : r (n - 1) r (n - 2) r (n - 3) ··· r (0)

If A = (aij ) is a complex matrix then the conjugate matrix of A is

= (ij ).

A square matrix A is a Hermite matrix (Hermitian), if AT = and a skew Hermite matrix (skew Hermitian) if AT = - .

The Hermite matrix of A is AH = T. If A = AH, then A is Hermitian.

The diagonal elements of a Hermite matrix are real, because aii = ii . The diagonal elements of a skew Hermite matrix are pure imaginary or zero, because aii = - ii . Hermiteness generalizes the notion of symmetricness.

Matrix B is the inverse of a matrix A if

AB = I   ja   BA = I.

Proposition 3. If A has an inverse, the inverse is unique.

Proof. Let B and B' be inverses of a matrix A, that is, AB = BA = I, and AB' = B'A = I.
But it now follows that B = IB = (B'A)B = B' (AB) = B'I = B'.

If the inverse of A exists, it is denoted by A - 1 and we say that A is regular. If A does not have an inverse, it is singular.

It can be shown that for square matrices A and B we have

AB = I => BA = I

Therefore, to prove that B is the inverse of A, it suffices to show that AB = I.

Proposition 4. If A and B are regular and 0, then

1) (A - 1) - 1 = A
2) (A) - 1 = (1/ )A - 1
3) (AB) - 1 = B - 1A - 1
4) (AT) - 1 = (A - 1)T

Proof. 4): Denote B = (A - 1)T. We have to show that ATB = BAT = I

 ATB = AT (A - 1)T = (A - 1A)T = IT = I BAT = (A - 1)TAT = (AA - 1)T = IT = I

A is orthogonal if it is real and AT = A - 1. If A and B are orthogonal, then so is AB.
Proof. (AB)T = BTAT = B-1A-1 = (AB)-1.
An orthogonal mapping(a matrix) preserves norms!

More generally: m x n matrix U is orthogonal if UTU = I (column orthogonal).

A is unitary if T = A - 1, that is, if A - 1 = AH. Compare orthogonal vs. unitary.

Often it is useful to partition matrices into smaller ones, called blocks, for example

 A = x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x
 = A11 A12 A13 A21 A22 A23

Proposition 5. Let matrices A and B be partitioned as follows:

 A = A11 ··· A1n , : : Ap1 ··· Apn
 B = B11 ··· B1m , : : Br1 ··· Brm

where the block Aij is a si x tj matrix and Bij is a ui x vj matrix. Then

 1)
 AT = A11 T ··· Ap1 T : : A1n T ··· Apn T
 2)
 A = A11 ··· A1n : : Ap1 ··· Apn

3) If p = r, n = m and si = ui , tj = vj    i, j, then

 A + B = C11 ··· C1n : : Cp1 ··· Cpn

where Cij = Aij + Bij

4) If n = r and tj = uj    j, then

 AB = C11 ··· C1m : : Cp1 ··· Cpm

where Cij = Aik Bkj .

Proof. Straightforward calculation.

Exercises: E2, E3, E4, E5, E10, E12
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