Math290-1Notes - Practice Notes PDF

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Math 290-1 Lecture Outline FALL 2016 Martha Precup, 2016 These are notes written for Math 290-1 during Fall of 2016. Please let me know if you notice any mistakes or typos. These are meant to be a brief outline of the material covered in class each day. Many thanks to my colleague Jason Siefken for sharing the style file which typesets the text below so beautifully.

Contents 1 Linear Systems of Equations (§1.1)

3

2 Gauss-Jordan Elimination (§1.2)

4

3 Gauss-Jordan Elimination (§1.2), Solutions of Linear Systems (§1.3)

6

4 Matrix Algebra, Matrix Equations (§1.3)

7

5 Linear Transformations (§2.1)

9

6 More on Linear Transformations (§2.2)

10

7 Matrix Products (§2.3)

12

8 The inverse of a linear transformation (§2.4)

14

9 More on Inverses (§2.4)

15

10 Image and Kernel of a Linear Transformation (§3.1)

16

11 Subspaces of Rn (§3.2)

17

12 Bases and Linear Independence (§3.2)

18

13 More on Bases

20

14 Dimension of a subspace of Rn (§3.3)

21

15 Coordinates (§3.4)

23

16 More on Coordinates (§3.4)

24

17 Determinants (§6.1)

26

18 More on Determinants (§6.2)

27

19 Geometry of Determinants (§6.3)

29

20 Eigenvalues and eigenvectors (§7.2-7.3)

31

21 More on eigenvalues and eigenvectors (§7.2-7.3)

32

1 Linear Systems of Equations (§1.1) Goal: Develop a systematic method for solving linear equations called Gauss-Jordan elimination. EXAMPLE

Find x and y so that x + 2y

=

1

2x + 3 y

=

1.

Solution: x = 1 y = 1. The equations from the example above are called linear since the degree of each term is less than or equal to 1, so their graph is a line. What kind of operations did we do to solve the above? 1. Multiply any equation by a nonzero number. 2. Add (a multiple) of one equation to another. 3. Interchange two of the equations. Solve the system of linear equations x + 3y

=

k

4x + h y

=

8

EXAMPLE

where h, k 2 R (so h and k represent any real numbers). Solution: We get that: x + 3y 4x + h y

= =

x + k ! 8

3y (h  12) y

= =

k 8  4k

We’d like to solve the second equation for y. If h = 12 and k = 6 2 there is no solution. If h = 12 and k = 2 there are infinitely many solutions given by x + 3y = 2 ) x = 2  3 y so a solution is any point of the form (2  3t, t) for t 2 R (a line). If h 6= 12 then ã Å 8  4k 8  4k , x = k3 y= h  12 h  12 for any value of k , so there is a unique solution in this case. In general, given two equations with two unknowns, what kind of solutions to linear systems of equations can we see? • Exactly one solution: the two lines intersect in a single point. • Infinitely many solutions: the two lines are equal. • No solution: the two lines are parallel.

EXAMPLE

What about a system of three linear equations in three unknowns? Aplane in R3 is the set of points which solve an equation of the form ax + b y + cz = d where at least one of the numbers a, b, c is nonzero. Find all solutions to the following linear systems. Describe your solutions in terms of intersecting planes. 1. x + y  z = 0, 4x  y + 5z = 0, 6x + y + 4z = 0. Solution: (0, 0,0), three planes intersecting in a single point. 2. x + y  z = 2, 3 x  5 y + 13z = 18, x  2 y + 5z = 7. Solution: (1  t, 2 t  3, t ) where t is any real number, the three planes intersect in a line. If a linear system has a solution, we say that it is consistent. If not, we say that it is inconsistent .

2 Gauss-Jordan Elimination (§1.2)

EXAMPLE

Systems of equations are useful in all kinds of situations. For example, can you solve the following riddle? Emile and Gertrude are brother and sister. Emile has twice as many sisters as brothers, and Gertrude has just as many brothers as sisters. How many children are in the family? Solution: Let Se and B e respectively denote the number of sisters and brothers Emile has, S g and B g respectively denote the number of sisters and brothers that Gertrude has, and C denote the total number of children in the family. Now we know: Sg Bg

= =

Se  1 Be + 1

Se Sg

= =

2B3 Bg

C

=

Se + Be + 1

=

S g + B g + 1.

Solving this system gives us S g = B g = 3 and Se = 4, and B e = 2. So C = 7; note that the solution to this problem must be a point consisting of positive integers–nothing else makes sense! Note that when solving a system of linear equations, the most important information are the coefficients on the variables. Instead of writing out the entire system, it saves time to write any array of the coefficients. For example: 3 2 3x 21 y 3z = 0 3 21 3 0 9x 2 y z = 5 becomes 49 2 1 55 . 2 7 10 1 2x 7 y 10z = 1

DEFINITION

This is called an augmented matrix; the array to the left of the line is called a coefficient matrix. A matrix is an n ⇥ m array of numbers, where n is the number of rows and m is the number of columns. For example, ⇥ ⇤ ïa11 a12 a13 ò a11 , . a21 a22 a23 where ai j denotes the entry of the matrix in the i-th row and the j-th column. Matrix Terminology: 1. Two matrices are equal if they are the same size and all their entries are equal. 2. A square matrix is an n ⇥ n matrix. 3. A diagonal matrix is a square matrix such that ai j = 0 for all i 6= j . 4. A matrix is upper-triangular ifai j = 0 for all i > j , lower triangular if ai j = 0 for all i < j , and the zero matrix if ai j = 0 for all i and j .

EXAMPLE

Let’s use matrices to solve a linear system. Note that if we manipulate the rows of a matrix like we would the equations, then it doesn’t change the solution. Solve the linear system x 1  2x 2 + x 3 2x 2  8x 3 5x 1  5x 3

= = =

0 8. 10

Solution: x 1 = 1, x 2 = 0 x 3 = 1. What makes a system of equations easy to solve? (P1) The leading coefficient (i.e. the coefficient on the first nonzero variable) in each equation is 1. (P2) The leading variable in each equation does not appear in any other equations. (P3) The leading variables appear in “natural order” (x 1 , x 2 , x 3 , ..., etc.)

DEFINITION

A matrix is in reduced row-echelon form (rref) if all the following hold: • If a row has nonzero entries, then the first non-zero entry is a 1, called a leading 1 (or pivot ), or a pivot of this row. • If a column contains a leading 1, all other entries in that column are 0. • If a row contains a leading 1, then each row above it contains a leading 1 further to the left.

EXAMPLE

We denote the reduced row-echelon form of the matrix A by rref(A). Determine which of the following matrices are in reduced row echelon form. If they are not, indicate which part of the definition above fails. 2 3 2 3 2 1 1. 40 1 4 8 5 Solution: No 0 0 0 5/2 3 2 1 0 0 50 2. 40 1 0 0 5 Solution: Yes 0 0 1 9 2 1 3. 40 0

2 0 0

0 0 1

3 3 0 5 Solution: No 5/2

Fact: Every matrix can be transformed by a sequence of elementary row operations to one and only one reduced row-echelon matrix. This process is callGauss-Jordan Elimination. We call two matrices row equivalent if we can change one into the other using elementary row operations. Elementary Row Operations: (R1) Divide or multiply a row by a nonzero scalar. (R2) Subtract a multiple of a row from another row. (R3) Swap two rows.

DEFINITION

EXAMPLE

Solve x + y  2z = 5 2x + 3 y + 4z = 2 2 3 2 3 10 13 using Gauss-Jordan Elimination. Solution: 4 85 + t 485 where t 2 R. 1 0 A vector is a matrix with only one column. For example, ï ò 1 . 2 The set of all vectors with n components is denoted Rn and called a vector space.

3 Gauss-Jordan Elimination (§1.2), Solutions of Linear Systems (§1.3)

EXAMPLE

Use Gauss-Jordan elimination to row reduce each of the following matrices to reduced row-echelon form. Circle the leading 1’s in the final matrix to identify the pivot columns. Then, find the general solutions of the systems whose augmented matrices are given by these matrices. ò ï 1 3 4 7 1. 3 9 7 6 Solution: ô ñ ò ï ò ï ò ï 1 3 0 5 1 3 4 7 1 3 4 7 1 3 4 7 ! ! ! 0 0 5 15 3 9 7 6 0 0 1 3 0 0 1 3 Rewriting the matrix as a system we get that x 1= 5 + 3 x 2 , x 2 is arbitrary, 2 3of equations 2 3 5 3 and x 3 = 3, so the solution is 4 0 5 + t 41 5 where t 2 R. 0 3

2 0 1 3 61 0 9 2. 4 0 1 3 1 0 9 Solution:

0 0 1 0

2 0 61 40 1

3 1 47 85 5 1 0 1 0

3 9 3 9

0 0 1 0

2 3 1 1 6 4 7 6 0 ! · · ·6 8 5 4 0 5 0

0 1 0 0

9

3

0

0

3 0

0 1

0

0

7 0 7 7 0 5

1

DEF

THEOREM

The last row of the matrix becomes: 0x 1 + 0 x 2 + 0x 3 + 0x 4 = 1 ) 0 = 1, which is NEVER the case! Therefore the system is inconsistent. A ⇥ linear system is⇤ inconsistent if and only if the rref of its augmented matrix contains the row: 0 0 · · · 0 1 . If a linear system is consistent, then it has either • infinitely many solutions (i.e. there is at least one free variable), or

• exactly one solution (all variables correspond to a leading 1). The rank of a matrix A is the number of leading 1’s (or pivots) in rref(A), denoted rank(A). Facts about rank: Let A be a n ⇥ m coefficient matrix. Then, 1. rank(A)  n and rank(A)  m since we cannot have more leading 1’s than rows or columns. 2. The system has exactly one solution if and only ifrank( A) = m since there must be a leading one in every row. 3. If the system has infinitely many solutions, rank(A) < m. 4. Important: If the system has a solution, then m = # of columns of A =

THEOREM

=

# variables in the system of equations # of leading 1’s + # of free variables = rank(A) + # free variables.

1. If a linear system has exactly one solution, then there must be as many equations as variables (i.e., m  n). 2. If a linear system has fewer equations than variables, then it is either inconsistent or has infinitely many solutions.

4 Matrix Algebra, Matrix Equations (§1.3)

EX

Recall that a vector in Rn is an n ⇥ 1-matrix. We frequently represent a vector as a arrow which starts at the origin and terminates at the point in space corresponding to the entries of the vector. This is called the standard representation. Let v~ =

ïò ï ò 1 2 and w ~= . Plot w, ~ v~, 2w, ~ ~ v , and 2w ~  t v~ for t 2 R. 2 1

Matrix Algebra 2

a11 6 a21 • sums: 6 4 ...

an1

a12 a22 ... an2

3 2 a1m b11 a2m 7 6 b21 7+6 ... 5 4 ...

··· ··· ··· ···

anm 2 a11 6 a21 • scalar multiplication: k 6 . 4 ..

bn1

a12 a22 ...

··· ···

b12 b22 .. .

··· ··· 3

bn2

··· ···

3 2 b1m a11 + b11 b2m7 6 a21 + b21 7=6 .. ... 5 4 .

bnm 2 a1m ka11 a2m7 6ka21 7 =6 ... 5 4 ... anm kan1

an1 + bn1

ka12 ka22 .. .

··· an1 an2 · · · kan2 2 3 2 3 v1 w1 6 v27 6 w 27 7 6 7 • dot product: v~ · w ~ =6 4 ... 5 · 4 ... 5 = v1 w 1 + v2 w 2 + · · · + vn w n . vn

··· ··· ··· ···

a12 + b12 a22 + b22 .. . an2 + bn2 3 ka1m ka2m 7 7 ... 5

··· ··· ··· ···

3 a1m + b1m a2m + b2m 7 7 5 ... anm + bnm

ka nm

wn

EXAMPLE

DEFINITION

The product of a matrix and a vector Let A be an n ⇥ m matrix and x~ be an m ⇥ 1 vector. Then A~ x is the n ⇥ 1 vector defined as follows. 3 3 2 2 w ~ 1 · x~  w ~1  ~ · x~7 ~ 2 7 6w 6 w 7 x~ = 6 2. 7 . A~ x=6 . . . 4 .. 5 4 .. .. .. 5 

1.

ï

2.

ï

1 0

1 4

1 2

2 5

w ~n



w ~ n · x~

2 3 ò 1 0 4 5 1 = 3 2

2 3 ï ò x1 ò ï ò ï ò ï ò x1 + 2 x2 + 3 x3 3 2 1 3 4 5 x2 = + x3 + x2 = x1 6 4x 1 + 5 x 2 + 6 x 3 5 4 6 x3

DEFINITION

Fact: Let A be an n ⇥ m matrix and x~ be an m ⇥ 1 vector. If v~1 , v~2 , .., v~n are the columns of A, then 2 3 2 3 x1 | | ··· | 6x 7 2 7 A~ x = 4 v~1 v~2 · · · v~m5 6 4 ... 5 = x 1 v~1 + x 2 v~2 + · · · x m v~m . | | ··· | xm b 2 Rn is a linear combination of the vectors v~1 , v~2 , ..., v~m 2 Rn if there exists scalars A vector ~ x 1 , x 2 , ..., x m so that ~ b = x 1 v~1 + x 2 v~2 + · · · x m v~m . The set of all vectors ~ b such that ~ b is a linear combination of v~1 , v~2 , ..., v~m is called the span of v~1 , v~2 , ..., v~m . We denote this set by span{~ v1 , v~2 , ..., v~m }

ï

1ò

ï ò ï ò 2 1 ? and v~2 = 1 1 1 Our goal is to find scalars x 1 and x 2 so that EXAMPLE

Is ~b =

a linear combination of v~1 =

ï

ò ï ò ï ò ï ò ï 1 1 1 1 2 = 1 = x1 + x2 ) 1 1 1 1

2 1

òï

ò x1 . x2

So we solve the system: ï

1 1

2 1

ò ï 1 1 ! 0 1

ò ï ò ï 1 0 1 2 1 1 ! ! 0 1 0 1 2 2

2 1

and see that x 1 = 3 and x 2 = 2 so ~b =

ò 3 2

ï

ï ò ò ï ò 1 2 1 . and v~2 = is a linear combination of v~1 = 1 1 1

To understand the geometry of the example above, plot each of the vectors. What is the span of~ v1 and v~2 ? Solution: span{~ v1 , v~2 } = R2 .

THEOREM

The vector b~ is a linear combination of v~1 , v~2 , ..., v~m 2 Rn if and only if the matrix equation 2

| 4v~1 |

| v~2 |

··· ··· ···

2

3 x1 | 6x 7 2 7 ~ v~m 5 6 4 ... 5 = b | xm 3

has a solution. Consider the system of equations, 2x + 3 y  z 10 x  z

4x  9 y + 2z

=

1

=

2

=

5

EXAMPLE

and notice that we can rewrite this as 3 2 3 3 2 3 2 2 3 2 3 2 1 3 2 1 2x + 3 y  z 1 4 10x  z 5 = 4 25 or 2 410 5 + y 4 0 5 + z 41 5 = 4 25 5 2 9 4 5 4x  9 y + 2z

or

2

2 3 410 0 4 9

32 3 2 3 1 1 x 15 4 y 5 = 4 25 . 2 5 z

The first two of these arevector forms of the system of equations, and the last of these is called the matrix form of the system of equations. 2 2 3 2 3 3 1 2 1 For which values of k is b~ = 4 k 5 a linear combination of 415 and 405? 1 0 2+k Solution: ~b can be written as a linear combination of the given vectors if and only if k =  1. 3

5 Linear Transformations (§2.1)

EXAMPLE

How does a matrix change an arbitrary vector? •

ï

0 1

•

ï

2 0

•

ï

1 0

ò òï ò ï x 2 1 x 1 = . A is a counterclockwise rotation by 90 . x2 x1 0 ò òï ò ï 2x 2 0 x1 = . A multipies every vector by the scalar 2. 2x 1 2 x2 òï ò ï ò x 0 x1 = 1 . A projects every vector onto the x-axis. 0 0 x2

DEFINITION

Each of the above is an example of a linear transformation. Atransformationfrom Rm to Rn is a function T : Rm ! Rn (so T is a map which takes each vector in Rm to exactly one vector in Rn ). A transformation T : Rm ! Rn is a linear transformation if 1. T (~ v + w) ~ = T (~ v ) + T (w) ~ for all v~, w ~ 2 Rm , and

2. T (k~ v ) = kT (~ v ) for all v~ 2 Rm and scalars k 2 R. IMPORTANT IDEA: Notice that any vector can be written as a linear combination of e~1 =

EXAMPLE

so it’s enough to figure out what T does to each of these vectors.

ï ò ï ò 1 0 and e~2 = 0 1

Let T be the transformation from R3 to R3 given by y1 = 2 x 2 + 1, y2 = x 1 + x 2, and y3 = x 3 + ⇡. Is T linear? We start by checking the first part of the definition. 3 2 3 3 2 2 v1 + w 1 2v2 + 2w 2 + 1 2(v2 + w 2 ) + 1 T 4 v2 + w 2 5 = 4(v1 + w 1 ) + (v2 + w 2 )5 = 4 v1 + w 1 + v2 + w 25 v3 + w 3 v3 + w 3 + ⇡ (v3 + w 3 ) + ⇡ and

2 3 2 3 2 3 3 2 3 2 v1 w1 2v2 + 2w 2 + 2 2v2 + 1 2w 2 T 4 v2 5 + T 4 w 25 = 4 v1 + v2 5 + 4w 1 + w 25 = 4 v1 + w 1 + v2 + w 25 . v3 w3 v3 + w 3 + 2⇡ w3 + ⇡ v3 + ⇡

Since T (~ v + w) ~ 6= T (~ v ) + T (w), ~ T is NOT linear!

~ = 0. ~ Fact: If T : Rm ! Rn is a linear transformation, then T (0)

6 More on Linear Transformations (§2.2)

THEOREM

Warm-Up: Is the transformation f : R ! R given by f (x) = e x linear? Let A be an n ⇥ m matrix and T be the transformation from Rm to Rn given by T( x~) = A~ x . Then T is a linear transformation since, T (~ v+w ~ ) = A(~ v+w ~ ) = Av~ + A~ w and T (k~ v ) = A(k~ v ) = kA(~ v ). This tells us that every matrix transformation is a linear transformation. 2

| Let e~i 2 Rm be the vector with zero’s in every coordinate and 1 in the i-th spot. If A = 4 v~1 | then 2 3 0 6 .. 7 .7 2 3 36 2 6 7 | | ··· | | | ··· | 60 7 6 7 A~ e i = 4 v~1 v~2 · · · v~m5 e~i = 4v~1 v~2 · · · v~m5 61 7 = v~i . 7 | | ··· | | | ··· | 6 60 7 6.7 4 .. 5 0

| v~2 |

··· ··· ···

3 | v~m5, |

DEFINITION

Given a linear transformation T : Rm ! Rn , the matrix of T is 2 3 | | | A = 4T (~ e1 ) T (~e2 ) · · · T (~ e m )5 . | | |

THM

This leads to the following definition.

If T : Rm ! Rn is a linear transformation and A is the matrix of T then T (~ x ) = A~ x for all x~ 2 Rm . 3 x1 6 x 27 7 Proof. Let x~ = 6 4 .. 5 . Then . xm 2

A(~ x)

=

A(x 1 e~1 + x 2 ~e2 + · · · + x m e~m )

=

x 1 A(~ e1 ) + x 2 A(~ e2 ) + · · · + x m A(~ em )

=

x 1 T (~ e1 ) + x 2 T (~e2 ) + · · · + x m T (~e m )

=

T (x 1 e~1 + x 2 ~e2 + · · · + x m e~m ) = T (~ x ).

EXAMPLE

É 2 3 ï ò 1 1 = 4 25 and Find the matrix for T where T : R2 ! R3 is the linear transformation with T 1 2 2 3 2 3 ï ò 1 0 1 1 = 4 25. Solution: The matrix for T is A = 40 2 5. Notice that T is not invertible. T 1 2 0 2 Here are some examples of linear transformations from R2 ! R2 :

x 2 R2 counterclockwise Rotations: Let T : R2 ! R2 be the linear transformation which rotates the vector~

through an angle of ✓ . Let’s find the matrix for T . Using the functions sin ✓ and cos ✓ we get that ï ò ï ò ò ï ò ï 1  sin ✓ 0 cos ✓ T . = and T = cos ✓ 1 sin ✓ 0 Therefore T has matrix

ï

cos ✓ sin ✓

ò  sin ✓ . cos ✓

Orthogonal Projections: Let’s compute the projection of the vector x~ onto the vector w ~ . Let L be the line given by all scalar multiples of w. ~ Now, projw~ (x~ ) = k w ~ where k is some scalar, so we have only to solve for k. We have x~ = proj w~ (~ x ) + x~ ? ) x~? = x~  projw~ (~ x ) ) x~  k w. ~ Therefore, x~? · w ~ = 0 ) (~ x  k w) ~ ·w ~ = 0 ) x~ · w ~  kw ~ ·w ~)k=

EX

and we get that: projw~ (~ x) =

x~ · w ~ , w ~ ·w ~

x~ · w ~ w. ~ w ~ ·w ~

Find a matrix for the linear transformation which gives the orthogonal projection of~ x onto the line ï ïò ò 1 1 1 2 . Solution: The transformation has matrix 5 spanned by w ~= ...