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A matrix is an ordered set of numbers listed rectangular form.
Example. Let A denote the matrix
[2 5 7 8]
[5 6 8 9]
[3 9 0 1]
This matrix A has three rows and four columns. We say it is a 3 x 4 matrix.
We denote the element on the second row and fourth column with
a2,4.
If a matrix A has n rows and n columns then we say it's a square matrix.
In a square matrix the elements ai,i , with i = 1,2,3,... , are
called diagonal elements. Remark. There is no difference between a 1 x 1
matrix and an ordenary number.
A
diagonal matrix is a square matrix with all de non-diagonal elements 0. The
diagonal matrix is completely denoted by the diagonal elements. Example.
[7 0 0]
[0 5 0]
[0 0 6]
The matrix is denoted by diag(7 , 5 , 6)
A matrix
with one row is called a row matrix
A
matrix with one column is called a column matrix
Matrix A and B are of the same kind if and only if A has as many rows as
B and A has as many columns as B
The n x m matrix A' is the transpose of the m x n matrix A if and only if
The ith row of A = the ith column of A' for (i = 1,2,3,..n) So
ai,j = aj,i'
The transpose of A is denoted T(A) or AT
When all the elements of a matrix A are 0, we call A a 0-matrix. We write
shortly 0 for a 0-matrix.
An identity matrix I is a diagonal matrix with all diagonal element = 1.
A scalar matrix S is a diagonal matrix with all diagonal elements
alike. a1,1 = ai,i for (i = 1,2,3,..n)
If we change the sign of all the elements of a matrix A,
we have the opposite matrix -A. If A' is the opposite of A then
ai,j' = -ai,j, for all i and j.
A square matrix is called symmetric if it is equal to its transpose. Then
ai,j = aj,i , for all i and j.
A square matrix is called skew-symmetric if it is equal to the opposite of
its transpose. Then ai,j = -aj,i , for all i and j.
To add two matrices of the same kind, we simply add the corresponding
elements.
Consider the set S of all n x m matrices (n and m fixed) and A and B are in
S. From the properties of real numbers it's immediate that
- A + B is in S
- the addition of matrices is associative in S
- A + 0 = A = 0 + A
- with each A corresponds an opposite matrix -A
- A + B = B + A
To multiply a matrix with a real number, we multiply each element with this
number.
Consider the set S of all n x m matrices (n and m fixed). A and B are in S; r
and s are real numbers. It is not difficult to see that:
r(A+B) = rA+rB
(r+s)A = rA+sA
(rs)A = r(sA)
(A + B)T = AT + BT
(rA)T = r. AT
Because
in the following, there is an intensive use of the properties of sums, the
reader who is not familiar with these properties must read first Sums in math . Remark. In this
html document, for convenience, we'll write the word sum instead of the sigma
sign.
This multiplication is only possible if the row matrix and the column matrix
have the same number of elements. The result is a ordinary number ( 1 x 1
matrix). To multiply the row by the column, one multiplies corresponding
elements, then adds the results. Example.
[1]
[2 1 3]. [2] = [19]
[5]
This product is defined only if A is a (l x m)
matrix and B is a (m x n) matrix. So the number of columns of A has to be
equal to the number of rows of B. The product C = A.B then is a (l x n)
matrix. The element of the ith row and the jth column of the product is found
by multiplying the ith row of A by the jth column of B.
ci,j = sumk (ai,k.bk,j)
Example.
[1 2][1 3] = [5 7]
[2 1][2 2] [4 8]
[1 3][1 2] = [7 5]
[2 2][2 1] [6 6]
[1 1][2 2] = [0 0]
[1 1][-2 -2] [0 0]
From these examples we see that the product is not commutative and that
there are zero divisors.
If
the multiplication is defined then A(B.C) = (A.B)C holds for all matrices A,B
and C. Proof: We'll show that an element of A(B.C) is equal to the
corresponding element of (A.B)C First we calculate the element of the ith row
and jth column of A(B.C)
Let D denote B.C, then
dk,j = sump bk,p.cp,j (1)
Let E denote A.D then
ei,j = sumk ai,k.dk,j (2)
(1) in (2) gives
ei,j = sumk ai,k.(sump bk,p.cp,j)
<=> ei,j = sumk,p ai,k.bk,p.cp,j
So the element of the ith row and jth column of A(B.C) is
sumk,p ai,k.bk,p.cp,j (3)
Now we calculate the element of the ith row and jth column of (A.B)C
Let D' denote A.B, then
di,p' = sumk ai,k.bk,p (4)
Let E' denote D'C then
ei,j' = sump di,p'.cp,j (5)
(4) in (5) gives
ei,j' = sump (sumk ai,k.bk,p).cp,j
<=> ei,j' = sumk,p ai,k.bk,p.cp,j
So the element of the ith row and jth column of (A.B)C is
sumk,p ai,k.bk,p.cp,j (6)
From (3) and (6) => A(B.C) = (A.B)C
If
the multiplication is defined then A(B+C) = A.B+A.C and (A+B).C = A.C+B.C holds
for all matrices A,B and C. This theorem can be proved in the same way as above.
For each A,
there is always an identity matrix E and an identity matrix E' so that A.E = A
and E'.A = A If A is a square matrix, E = E'.
(A.B)T = BT .AT
This theorem can be proved in the same way as above.
If the
multiplication is defined then for each A
A.0 = 0 = 0.A
r and s are
real numbers and A , B matrices. If the multiplication is defined then (rA)(sB)
= (rs)(AB) This theorem can be proved in the same way as above.
if D = diag(a,b,c) then D.D = ( a2 , b2 , c2)
D.D.D = ( a3 , b3 , c3)
.....
This property can be generalised for D = diag(a,b,c,d,e,...,l). |