Review of Linear Algebra: EE 510 Lumped Systems Theory Handout, Lecture notes of Teaching method

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EE 510: Lumped Systems Theory

Fall 2016

Handout 1:

Review of Linear Algebra

Prof. Mohamed Zribi

Updated 2 October 2016 (^1)

Outline of the Handout

1. Systems of Linear Equations

2. Review of Matrices

3. Determinants

4. Vector Spaces

5. Row Space, Column Space and Null Space of a Matrix

6. Solutions of Systems of Linear Equations

7. Eigenvalues and Eigenvectors

8. Similar Matrices

9. Diagonalization of Matrices

10. Jordan Form of Matrices

11. The Cayley-Hamilton Theorem

1.1 Introduction to Systems of Linear

Equations

A linear equation in n variables is such that:

a i : real-number coefficients x i : variables needed to be solved for b : real-number constant term a 1 : leading coefficient x 1 : leading variable Notes: (1) Linear equations have no products or roots of variables and no variables involved in trigonometric, exponential, or logarithmic functions (2) Variables appear only to the first power

(h) 1 1 4 x y

Example 1: Linear or Nonlinear

(a) 3 x + 2 y = 7 (b) 1 2 2

x + y − π z =

(c) x 1 (^) − 2 x 2 (^) + 10 x 3 (^) + x 4 = 0 (d) (sin ) 1 4 2 2 2

π x − x = e

(e) xy + z = 2 (f)^ e^ x −^2 y =^4

(g) sin x 1 (^) + 2 x 2 (^) − 3 x 3 = 0

product of variables

trigonometric function

Linear

Linear Linear

Linear

Nonlinear

Nonlinear

Nonlinear

Nonlinear not the first power

• If you choose x 1 to be the free variable, the parametric representation of the solution set is

Example 2: Parametric representation of a solution set

x 1 + 2 x 2 = 4

• If you solve for x 1 in terms of x 2 , you obtain
• By letting (the variable t is called a parameter), you can represent the solution set as
• The set representation for the solutions: (^) {( 4 − 2 t , t )| t ∈ R }

x 1 (^) = 4 − 2 x 2 (^) (in this form, the variable x 2 is free) x (^) 2 = t

x 1 (^) = 4 − 2 , t x 2 = t , t is any real number

{( s , 2 − 0. 5 s )| s ∈ R }

with a solution: (2, 1), i.e., x 1 =^2 ,^ x 2 =^1

• A system of m linear equations in n variables:

11 1 12 2 13 3 1 1 21 1 22 2 23 3 2 2 31 1 32 2 33 3 3 3

1 1 2 2 3 3

n n n n n n

m m m mn n m

a x a x a x a x b a x a x a x a x b a x a x a x a x b

a x a x a x a x b

A solution of a system of linear equations is a sequence of numbers s 1 , s 2 ,…, s (^) n that can solve each linear equation in the above system.

Example 3: Solution of a system of linear equations in 2 variables

x y

x y

x y

x y

x y

x y

exactly one solution

inifinite number of sol.

no solution

twointersecting lines

twocoincident lines

twoparallel lines

No solution –2 x + y = 3 –4 x + 2 y = 2 Lines are parallel. No point of intersection. No solutions.

Unique solution x + y = 5 2 x - y = 4 Lines intersect at (3, 2) Unique solution: x = 3, y = 2.

Many solution 4 x – 2 y = 6 6 x – 3 y = 9 Both equations have the same graph. Any point on the graph is a solution. Many solutions.

Example 4: Solution of a system of linear equations in 2 variables

No solution

Many solutions

No solution

13

Example 6: Using back substitution to solve the following system

y

x y

Solution: By substituting y =−^2 into Eq. (1), you obtain

x − 2( 2)− = 5 ⇒ x = 1

The system has exactly one solution: x =^1 ,^ y = −^2

• Equivalence: Two systems of linear equations are

called equivalent if they have precisely the same solution set

 Notes: Each of the following operations on a system of linear equations produces an equivalent system

O1: Interchange two equations O2: Multiply an equation by a nonzero constant O3: Add a multiple of an equation to another equation  Gaussian elimination: A procedure to rewrite a system of linear equations in row- echelon form by using the above three operations

Example 8: Solve a system of linear equations (consistent system)

x y z

x y

x y z

Solution: First, eliminate the x-terms in Eqs. (2) and (3) based on Eq. (1) (1) (2) (2) (by O3) 2 3 9 3 5 (4) 2 5 5 17

x y z y z x y z

• → − + =
• = − + = (1) ( 2) (3) (3) (by O3) 2 3 9 3 5 1 (5)

x y z y z y z

× − + → − + =

• = EE 510 Dr. Mohamed Zribi −^ −^ =^ −^17

Example 9: Solve a system of linear equations (inconsistent system)

1 2 3

1 2 3

1 2 3

• − = −

x x x

x x x

x x x

Solution:

1 2 3 2 3 2 3

(1) ( 2) (2) (2) (by O3) (1) ( 1) (3) (3) (by O3) 3 1 5 4 0 (4) 5 4 2 (5)

x x x x x x x

× − + →

× − + →

1 2 3 2 3

(4) ( 1) (5) (5) (by O3) 3 1 5 4 0 0 2

x x x x x

× − + →

So the system has no solution (an inconsistent system)

(a false statement)