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Matrices: §2.2 Properties of Matrices

Satya Mandal, KU

Satya Mandal, KU Matrices: §2.2 Properties of Matrices



Homework

Goals

We will describe properties of matrices with respect toaddition, scalar multiplications and matrix multiplication andothers. Among what we will see

1. Matrix multiplication does not commute. That means,not always AB = BA.

2. We will define transpose AT of a matrix A and discussits properties.




Homework

Zero MatricesAlgebra of Matrix MultiplicationIdentity MatrixNumber of Solutions

Algebra of Matrices

Let A,B ,C be m × n matrices and c , d be scalars. Then,

A+ B = B + A Commutativity of additionA+ (B + C ) = (A+ B) + C Associativity of addition(cd)A = c(dA) Associativity of scalar multiplicationc(A+ B) = cA+ cB a Distributive property(c + d)A = cA+ dA a Distributive property

These seem obvious, expected and are easy to prove.




Homework


Zero

The m × n matrix with all entries zero is denoted by Omn. Fora matrix A of size m × n and a scalar c , we have

◮ A+ Omn = A (This property is stated as:Omn is theadditive identity in the set of all m × n matrices.)

◮ A+ (−A) = Omn. (This property is stated as:−A is theadditive inverse of A.)

◮ cA = Omn =⇒ c = 0 or A = Omn.

Remark. So far, it appears that matrices behave like realnumbers.Reading Assignment: Example 2 from §2.2 of the textbook.




Homework


Properties of Matrix Multiplication

Let A,B ,C be matrices and c is a constant. Assume all thematrix products below are defined. Then

A(BC ) = (AB)C Associativity Matrix ProductA(B + C ) = AB + AC Distributive Property(A+ B)C = AC + BC Distributive Propertyc(AB) = (cA)B = A(cB)

Proofs would be routine checking, which we would skip.Reading Assignment: Example 3, 4, 5 from §2.2 of thetextbook.




Homework


Definition

For a positive integer, In would denote the square matrix oforder n whose main diagonal (left to right) entries are 1 andrest of the entries are zero. So,

I1 = [1], I2 =

[

1 00 1

]

, I3 =

1 0 00 1 00 0 1




Homework


Properties of the Identity Matrix

Let A be a m × n matrix. Then

◮ AIn = A

◮ ImA = a

◮ If A is a square matrix of size n × n, then

AIn = InA = A.

◮ In is called the Identity matrix of order n. Because ofabove, we say that In is the multiplicative identity for theset of all square matrices of order n.




Homework


Proof.

Proof. We will prove for 3× 3 matrices A. Write

A =

a b c

u v w

x y z

So,

AI3 =

a b c

u v w

x y z

1 0 00 1 00 0 1

=

a b c

u v w

x y z

= A

Similarly, I3A = A. The proof is complete.




Homework


Use of Matrix Algebra to solve

systmes of linear equation

Now that we know some Matrix Algebra, we will use this togive a proof of the following theorem that was stated before:Theorem. For a system of linear equations in n variables,precisely one of the following is true:

◮ The system has exactly one solution.

◮ The system has an infinite number of solutions.

◮ The system has no solution.




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The Proof.

Write the equation in the matrix form Ax = b. If the first orthe last statements are true then there is nothing to prove.So, assume both are false. So, the system has at least twodistint solutions x1, x2 with x1 6= x2. So,

Ax1 = b and Ax2 = b.

With y = x1 − x2 6= 0 we have

Ay = A(x1 − x2) = Ax1 − Ax2 = b− b = 0.




Homework


For any scalar c , we have

A(x1 + cy) = Ax1 + cAy = b+ 0 = b

So, x1 + cy is a solution of the given system Ax = b, for allscalars c , which is infinitely many. The proof is complete.




Homework

DefinitionAlgebra of Transpose

Definition of Transpose of a Matrix

Definition. Given a m × n matrix A, the transpose of A,denoted by AT , is formed by writing the the columns of A asrows (equivalently, writing the rows as columns). So,transpose AT of

A =

a11 a12 a13 · · · a1na21 a22 a13 · · · a2na31 a32 a33 · · · a3n· · · · · · · · · · · · · · ·

am1 am2 am3 · · · amn

an m × n matrix

is given by:




Homework


Transpose of a Matrix: Continued

AT =

a11 a21 a31 · · · am1a12 a22 a32 · · · am2a13 a23 a33 · · · am3· · · · · · · · · · · · · · ·

a1n a2n a3n · · · amn

an n ×m matrix

Reading Assignment: Example 8 from §2.2 of the textbook.




Homework


Properties of Transpose

Let A,B be matrices and c be a scalar. Assume all themultiplications below are defined. Then,

(AT )T = A Double transpose of A is itself(A+ B)T = AT + BT transpose of sum(cA)T = cAT transpose of scalar multiplication(AB)T = BTAT transpose of product.

Again, to prove we check entry-wise equalities.Reading Assignment: Example 9 from §2.2 of the textbook.




Homework

Example: NoncommutativityExample: Non-Cancellation

Let me draw your attention, how algebra of matrices differfrom that of the algebra of real numbers:

◮ Matrix product is not commutative. That meansAB 6= BA, for some matrices A,B . See §2.2 Example 4.

◮ Cancellation property fails. That means there arematrices A,B ,C , with C 6= 0, such that

AC = BC but A 6= B .

See §2.2 Example 5.




Homework


Example of noncommutativity AB 6= BA:

We have[

1 −11 1

] [

0 11 0

]

=

[

−1 11 1

]

[

0 11 0

] [

1 −11 1

]

=

[

1 11 −1

]

Right hand sides of these two equations are not equal. So,commitativity fails for these two matrices.




Homework


Example: AC = BC but A 6= B :

We have[

1 11 1

] [

1 1−1 −1

]

=

[

0 00 0

]

=

[

2 22 2

] [

1 1−1 −1

]

So, cancellation property fails for matrix product.




Homework

Solve for Matrix X

Solve for the matrix X when

A =

−1 13 −12 −1

B =

2 34 −111 −1

◮ (1) 5X + 3A = 2B

◮ (2) 2B + 4A = 5X

◮ (3) X + 2A− 2B = O




Homework

Solution

◮ For (1)

5X + 3A = 2B =⇒ X =2

5B −

3

5A

= .4

2 34 −111 −1

− .6

−1 13 −12 −1

=

1.4 .6−.2 .23.2 .2




Homework

Solution: Continued

◮ For (2), X = 25B + 4

5A =

.4

2 34 −111 −1

+ .8

−1 13 −12 −1

=

0 24 −1.26 −1.2

◮ For (3) X = −2A+ 2B =

−2

−1 13 −12 −1

+ 2

2 34 −111 −1

=

6 42 018 0




Homework

Example

Let

A =

2 3 74 −1 1911 −1 −19

,B =

2 −1 14 3 −111 2 −1

,

C =

2 −1 10 0 00 0 0

Demonstrate AC = BC bur A 6= B .




Homework

Solution:

We have

AC =

2 3 74 −1 1911 −1 −19

2 −1 10 0 00 0 0

=

4 −2 28 −4 422 −11 −11

,

BC =

2 −1 14 3 −111 2 −1

2 −1 10 0 00 0 0

=

4 −2 28 −4 422 −11 −11

So, AC = BC .




Homework

Example

Let f (x) = x2 − 2x + 2

A =

4 −2 08 −4 422 0 0

Compute f (A).

Solution: We have

A2 =

4 −2 08 −4 422 0 0

4 −2 08 −4 422 0 0

=

0 0 −888 0 −1688 −44 0




Homework

Solution: Continued

Let f (A) = A2 − 2A+ 2I3 =

0 0 −888 0 −1688 −44 0

− 2

4 −2 08 −4 422 0 0

+ 2

1 0 00 1 00 0 1

=

−6 4 −872 10 −2444 −44 2




Homework

Homework

§2.2: Exwecise 14a, 14c, 15, 19, 23, 25, 31, 32, 33, 34, 55


PreviewProperties of MatricesZero MatricesAlgebra of Matrix MultiplicationIdentity MatrixNumber of Solutions

Transpose of a MatrixDefinitionAlgebra of Transpose

Dissimilarities with algebra of numbersExample: NoncommutativityExample: Non-Cancellation

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