Matrices+ 01 - Matrix as Operator: UNIZOR.COM - Math+ & Problems - Matrices

Notes to a video lecture on http://www.unizor.com

Matrices+ 01
Matrix as Operator

In the beginning matrices were just tables of real numbers with m rows and n columns, we called them m⨯n matrices.
We knew how to add two matrices with the same numbers of rows and columns, how to multiply a matrix by a scalar number and how to multiply an m⨯n matrix by an n⨯k matrix.

Here is a short recap.

Addition of matrices
Let A_m⨯n be an m⨯n matrix with elements a_i,j, where the first index signifies the row, where this element is located and the second index is a column.
Let B_m⨯n be another m⨯n matrix with elements b_i,j.
Then a new matrix C=A+B is an m⨯n matrix with elements c_i,j=a_i,j+b_i,j.

Multiplication of a matrix by a scalar
Let s be any real number (scalar) and A_m⨯n be an m⨯n matrix with elements a_i,j.
Multiplication of s by A_m⨯n produces a new matrix B_m⨯n with elements b_i,j=s·a_i,j.
Obviously, since the product of numbers is commutative and associative, the product of a matrix by a scalar is commutative, and the product of a matrix by two scalars is associative.

Product of two matrices
Let A_m⨯n be an m⨯n matrix with elements a_i,j.
Let B_n⨯k be an n⨯k matrix with elements b_i,j.
It's important for a definition of a product of matrices that the number of columns in the first matrix A_m⨯n equals to a number of rows in the second B_n⨯k (in our case this number is n.)
The product C_m⨯k of A_m⨯n and B_n⨯k is m⨯k matrix with elements c_p,q calculated as sums of products of n elements of p^th row of matrix A_m⨯n by n elements of q^th column of matrix B_n⨯k
∀p∈[1,m], ∀q∈[1,k]:
c_p,q = Σ_[1≤t≤n]a_p,t·b_t,q
It's important to note that, generally speaking, multiplication of two matrices IS NOT commutative, see Problem C below.
The product of three matrices, however, is associative, see Problem B below.

For our purposes we will only consider "square" matrices with the same number of rows and columns n and n-dimensional Euclidean vector space of sequences of n real numbers organized in a row (row-vector, which can be viewed as 1⨯n matrix) or in a column (column-vector, which can be viewed as n⨯1 matrix).

What happens if we multiply an n⨯n matrix by a column-vector, which we can consider as an n⨯1 matrix, according to the rules of multiplication of two matrices?
In theory, the multiplication is possible because the number of columns in the first matrix n equals to the number of rows in the second one (the same n) and, according to the rules of multiplication, the resulting matrix should have the number of rows of the first matrix n and the number of columns of the second one, that is 1. So, the result is an n⨯1 matrix, that is a column-vector.

As we see, multiplication of our n⨯n matrix by a column-vector with n rows results in another column-vector with n row.
In other word, multiplication by n⨯n matrix on the left transforms one column-vector into another, that is this multiplication represents an operation in n-dimensional vector space of column-vectors. The n⨯n matrix itself, therefore, acts as an operator in the vector space of column-vectors of n components.

Similar operations can be considered with row-vectors and their multiplication by matrices. The only difference is the order of this multiplication.
In a case of column-vectors we multiplied n⨯n matrix by n⨯1 column-vector getting another n⨯1 column-vector.
In case of row-vectors, we change the order and multiply a 1⨯n row-vector by n⨯n matrix on the right, getting another 1⨯n row-vector.
Therefore, an n⨯n matrix can be considered an operator in a vector space of row-vectors, if we apply the multiplication by matrix from the right.

Let's examine the properties of the transformation of n⨯1 column-vectors by multiplying them by n⨯n matrices.

Problem A
Prove that multiplication by an n⨯n matrix is a linear transformation in the n-dimensional Euclidean space ℝⁿ of n⨯1 column-vectors.
(a) ∀u∈ℝⁿ, ∀ K∈ℝ, ∀A_n⨯n:
A·(K·u) = K·(A·u) = (K·A)·u
(b) ∀u,v∈ℝⁿ, ∀A_n⨯n:
A·(u+v) = A·u + A·v
(c) ∀u∈ℝⁿ, ∀A_n⨯n,B_n⨯n:
(A+B)·u = A·u + B·u

Hint A
The proof is easily obtained straight from the definition of matrix multiplications.


Problem B
Prove that consecutive multiplication by two n⨯n matrices is associative in the n-dimensional Euclidean space ℝⁿ of n⨯1 column-vectors.
∀u∈ℝⁿ, ∀A_n⨯n,B_n⨯n:
A·(B·u) = (A·B)·u

Hint B
It's all about changing of an order of summation.
Let's demonstrate it for n=2.

Given two matrices A and B and a column-vector u.

Matrix A

a1,1	a1,2
a2,1	a2,2

Matrix B

b1,1	b1,2
b2,1	b2,2

Column-vector u

u1

u2

Let v=B·u and w=A·(B·u)=A·v
Then column-vector v's components are
v₁ = b_1,1·u₁ + b_1,2·u₂
v₂ = b_2,1·u₁ + b_2,2·u₂

The components of vector w are
w₁ = a_1,1·v₁ + a_1,2·v₂ =
= a_1,1·(b_1,1·u₁ + b_1,2·u₂) +
+ a_1,2·(b_2,1·u₁ + b_2,2·u₂) =
= a_1,1·b_1,1·u₁ + a_1,1·b_1,2·u₂ +
+ a_1,2·b_2,1·u₁ + a_1,2·b_2,2·u₂ =
= (a_1,1·b_1,1 + a_1,2·b_2,1)·u₁ +
+ (a_1,1·b_1,2 + a_1,2·b_2,2)·u₂

w₂ = a_2,1·v₁ + a_2,2·v₂ =
= a_2,1·(b_1,1·u₁ + b_1,2·u₂) +
+ a_2,2·(b_2,1·u₁ + b_2,2·u₂) =
= a_2,1·b_1,1·u₁ + a_2,1·b_1,2·u₂ +
+ a_2,2·b_2,1·u₁ + a_2,2·b_2,2·u₂ =
= (a_2,1·b_1,1 + a_2,2·b_2,1)·u₁ +
+ (a_2,1·b_1,2 + a_2,2·b_2,2)·u₂

Let's calculate the product of two matrices A·B
Matrix A·B:

a1,1·b1,1+a1,2·b2,1	a1,1·b1,2+a1,2·b2,2
a2,1·b1,1+a2,2·b2,1	a2,1·b1,2+a2,2·b2,2

Notice that coefficients at u₁ and u₂ in expression for w₁ above are the same as elements of the table (A·B) of the first row.
Analogously, coefficients at u₁ and u₂ in expression for w₂ above are the same as elements of the table (A·B) of the second row.
That means that, if we multiply matrix (A·B) by a column-vector u, we will get the same vector w as above.
That proves the associativity for a two-dimensional case.

Problem C
Prove that consecutive multiplication by two n⨯n matrices, generally speaking, IS NOT commutative.
That is, in general,
(A·B)·u) ≠ (B·A)·u

Proof C
To prove it, it is sufficient to present a particular case when a product of two matrices is not commutative.

Consider matrix A

1	2
3	4

Matrix B

−1	2
−3	4

Then matrix A·B will be

−7	10
−15	22

Reverse multiplication B·A will be

5	6
9	10

As you see, matrices A·B and B·A are completely different, and, obviously, their product with most vectors will produce different results.
Take a vector with components (1,0), for example. Matrix A·B will transform it into (−7,−15), while matrix B·A will transform it into (5,9).
End of Proof.