# Millsaps College Linear Algebra

Final Exam - Spring 2011

Textbook: Linear Algebra and Its Applications

3rd Edition by David C. Lay

Instructor: Dr. Yan Wang

## Test 1 (1.1 - 1.6)

Definitions

Consistent Linear System

One or more solution

echelon Form

All nonzero rows are above any rows of all zeros

each leading entry of a row is in a column to the right of the leading entry of the row above it

all entries in a column below a leading entry are zeros

Reduced echelon Form

The leading entry of each nonzero row is one

Each leading 1 is the only nonzero entry in its column

Unique Row Reduced Matrix for each Eschelon matrix

Pivot Column

Column of A containing a pivot position

a location in A that corresponds to a leading 1 in the reduced echelon form of A

Basic Variable

Variable corresponding to a pivot column

Free Variable

Variable not corresponding to a pivot column

Linear Combination of Vectors

Given the vectors v1, v2, v3, ..., vp ∈ R^n and given scalars c1,c2, ..., cp in R, the vector y defined by y=c1v1+c2v2+...+cpvp is called a linear combination of v1, v2, ..., vp with weights c1, c2, ..., cp

Span of Vectors

The collection of all vectors that can be written in the form c1v1+c2v2+...+cpvp with c1, ..., cp scalars

Homogenous System

A Linear system is homogenous if it can be written in the form Ax=0, where A is an mxn matrix and 0 is the zero vector in R^m.

The zero solution of Ax=0 is called the trivial solution.

For a given equation Ax=0, a nontrivial solution is a nonzero vector x that satisfies Ax=0

Happens when there is at least one free variable

Parametric Vector Form

Existence and Uniqueness Theorem

A Linear system is consistent if and only if the rightmost column of the augmented matrix is NOT a pivot column--that is, if and only if an eschelon form of the augmented matrix had no row of the form [0 0 0 ... 0 b], with b nonzero

If a Linear system is consistent then the solution set contains either

a unique solution

no free variables

infinitely many solutions

at least one free variable

Theorem 5 (1.4)

Let A be an mxn matrix. The following are logically equivalent

For each b ∈ R^m, the equation Ax=b has a solution

Each b ∈ R^m is a linear combination of the columns of A

The columns of A span R^m

A has a pivot position in every row

About COEFFICIENT matrices

## Test 2 (1.7 - 2.5)

Linear Independence

An indexed set of vectors {v1, ..., vp} in R^n is said to be linearly independent if the vector equation x1v1+x2v2+...+xpvp=0 has ONLY the trivial solution

Ax=0 has only the trivial solution

{v1, ..., vp} does NOT contain the zero vector

Linear Dependence

The set {v1, ..., vp} is said to be Linearly dependent if there exists weights c1, ..., cp not all zero, such that c1v1+c2v2+...+cpvp=0.

Set of TWO vectors

Scalar multiples

S={v1, ..., vp}

there exists at least one vector in S that is a linear combination of the others

Set contains the zero vector

A is an mxn matrix

n > m

A set that contains more vectors than entries in each vector

Linear Transformation

A transformation (or function or mapping) T from R^n to R^m, denoted by T : R^n -> R^m, is a rule that assigns to each vector x in R^n a vector T(x) in R^m.

The set R^n

DOMAIN of T

The set R^m

The CODOMAIN of T

For x ∈ R^n, the vector T(x) in R^m is called the image of x (under the action of T). The set of all images T(x) is called the range of T

A transformation (or mapping) T is linear if:

T(cu+dv)=cT(u)+dT(v) for all vectors u,v and scalars c,d

T(u+v)=T(u)+T(v) for all u,v ∈ the domain of T

T(cu)=cT(v) for all u and scalars c

T(0)=0

All Matrix transformations are Linear transformations

Existence and Uniqueness Questions

ONTO

A mapping T : R^n -> R^m is said to be onto R^m if each b in R^m is the image of at LEAST one x in R^n

Columns of a matrix A span R^m

ONE-TO-ONE

A mapping T : R^n -> R^m is said to be one-to-one if each b in R^m is the image of at MOST one x in R^n

T(x1) is not equivalent to T(x2) when x1 is not equivalent to x2

T(x)=0 has ONLY the trivial solution

The columns of a matrix A are Linearly independent

The Matrix of a Linear Transformation

Identity Matrix

nxn matrix denoted I

has 1's on the diagonal and zeros elsewhere

ej is the jth column of the identity matrix ∈ R^n for 1<j<n

Assuming that T : R^n -> R^m is a linear transformation, there exists a unique matrix A, such that T(x)=Ax for all x in R^n

A is the mxn matrix whose jth column is the vector T(ej)

A = [ T(e1) ... T(ej) ]

The Standard matrix for the Linear transformation T

Matrix Operations

A(BC)=(AB)C

A(B+C)=AB+AC

(B+C)A=BA+CA

r(AB)=r(A)B=A(rB) for any scalar r

IA=A=AI

WARNINGS

AB is not always equal to BA

If B=C, AB is not always equal to AC

If AB=0, you cannot conclude that A=0 or B=0

Transpose Matrix

The transpose of an mxn matrix is denoted A^T and changes A's rows for its columns

A is mxn, A^T is nxm

Properties

(A^T)^T=A

(A+B)^T = A^T + B^T

for any scalar r, (rA)^T = rA^T

(AB)^T = (B^T)(A^T)

inverse Matrix

An nxn matrix A is said to be invertible is there is an nxn matrix C such that CA=I and AC=I, where I is the nxn identity matrix. C=A^-1 (A inverse)

Singular Matrix

NOT invertible

Non-singular Matrix

invertible Matrix

2x2 inverse

Let A = [a b, c d]. If ad-bc is not equal to 0, then A is invertible

A^-1 = (1/ad-bc) [d -b, -c a]

ad-bc = det A, which is the Determinant of A

Theorem 4 (2.2)

If A is an invertible nxn matrix, then for each b in R^n, the equation Ax=b has the unique solution x = (A^-1)b

Properties

If A is an invertible matrix, then A^-1 is invertible and (A^-1)^-1 = A

If A and B are nxn invertible matrices, then so is AB, and the inverse of AB is the product of the inverses of A and B in the reverse order. That is, (AB)^-1 = (B^-1)(A^-1)

(A^T)^-1 = (A^-1)^T

The Product of invertible Matrices is Invertible and the inverse is the product of their inverses in the reverse order

Algorithm for finding A^-1

Row reduce the augmented matrix [ A I ]. If A is row equivalent to I, then [ A I ] is row equivalent to [ I A^-1]. Otherwise, A does not have an inverse.

The invertible Matrix Theorem

A is an invertible, NxN matrix

For all B ∈ R^N, Ax=b has the unique solution x=(A^-1)b

x |-> Ax maps R^N ONTO R^N

At least ONE solution

A has N pivot columns

The Columns of A Span R^N

N Pivot Positions

b is a Linear combination of the columns of A

x |-> AX is ONE-TO-ONE

At Most One Solution

Ax=0 has only the trivial solution

The columns of A do not contain the Zero Vector

det A is not 0

0 is NOT an eigenvalue of A

The Columns of A are Linearly Independent

Columns of A form a Basis for R^n

Rank A = n

dim Col A = n

Col A = R^n

Nul A = {0}

dim Nul A = 0

A^T is invertible

(A^-1)^T is its inverse

A is row equivalent to I

Let C and D be NxN Matrices

CA=I; AD=I

C=D

Let A and B be nxn matrices. If AB=I, then A and B are both invertible with A=B^-1 and B=A^-1

A Linear Transformation T : R^n -> R^n is said to be invertible if there exists a transformation S : R^n -> R^n such that S(T(x))=x for all x in R^n AND T(S(x))=x for all x in R^n

S is the inverse transformation of T, written T^-1

Let T : R^n -> R^n be a Linear transformation and let A be the standard matrix for T. Then T is invertible if and only if A is an invertible matrix. In that case, the linear transformation S given by S(x)=(A^-1)x is the unique function satisfying the previous conditions

## Test 3 (2.5 - 4.1)

LU Factorization

Reduce A to an eschelon form U by a sequence of row replacements ONLY is possible.

Place entries in L such that the same sequence of row operations reduces L to I.

Row Reduce [L b] to find y.

Row Reduce [ U y ] to find x

Determinants

Cofactor Expansion

det A = a11C11+a12C12+...a1nC1n, where Cij = (-1)^(i+j) det Aij

The determinant of an nxn matrix can be computed by cofactor expansion across any row or down any column

Shortcut

If A is a triangular matrix, then det A is the product of the entries on its main diagonal

Properties

A is a square matrix and B is the resultant matrix from one row operation

Replacement

det(B)=det(A)

interchange

det(B) = -det(A)

Scalar Multiplication

det(B) = k^(n) det A

A square matrix A is invertible if and only if det A is not 0

if A is an nxn matrix, then det A^T = det A

A and B are nxn matrices: det AB = (det A)(det B)

Vector Subspaces

A subspace of a vector space V is a subset H of V that has three properties

The zero vector of V is in H

H is closed under vector addition

for each u and v ∈ H, the sum u+v is in H

H is closed under multiplication by scalars.

for each u ∈ H and each scalar c, the vector cu is in H

## Test 4 (4.2 - 7!)

Null Space

Subspace of R^n

The set of all solutions to a system Ax=0 of m homogenous Linear equations in n unknowns is a subspace of R^n

Related to Eigenspace

To check if a vector u is in the null space of A, simply solve the equation Au (not a matrix equation, but matrix A times a vector u). If the answer is 0, then u is in Nul A

Nul A is the set of all solutions to the homogenous equation Ax=0.

Nul A = { x : x is ∈ R^n and Ax=0}

Dimension of Nul A

Number of Free Variables in the matrix equation Ax=0

Column Space

Col A is the set of all Linear combinations of the columns of A. If A = [a1 a2 ... an], then

Col A = Span {a1, ..., an}

dim Col A

number of basic variables in the matrix equation Ax=0

Rank

Let A be an mxn matrix. Rank A is equivalent to

The Dimension of Col A

Number of Basic Variables in A

The number of pivot positions in matrix A

Rank Theorem

Rank A + dimension Nul A = n

Linear Transformation

A Linear Transformation T from a vector space V into a vector space W is a rule that assigns to each vector x in V a unique vector T(x) in W such that

T(u+v) = T(u) + T(v) for all u,v ∈ V

T(cu)=cT(u) for all u ∈ V and scalars C

The Kernel (or NULL SPACE) of a Linear transformation

the set of all u ∈ V such that T(u)=0.

The range of T is the set of all vectors in W of the form T(x) for some x in V

Basis

Let H be a subspace of a vector space V. An indexed set of vectors B={b1, ..., bp} in V is a BASIS for H if:

B is a Linearly independent set

the subspace spanned by B coincides with H

H = Span {b1, ..., bp}

Spanning Set Theorem

Let S = {v1, ..., vp} be a set in V, and let H = Span {v1, ..., vp}

If one of the vectors in S--say vk--is a linear combination of the remaining vectors in S, then the set formed from S by removing vk still spans H

If H is not {0} (the null set), some subset of S is a basis for H.

vp is a Linear combination of v1,v2,...,v(p-1) if [ v1 v2 ... v(p-1)] x = vp

Span {v1,...,vp}=Span{v1...v(p-1)}

The pivot columns of A form a basis for Col A

Let S be a basis for a vector space V. Then

S is a spanning set that is as small as possible because deleting any vector from S results in a non-spanning set of V

Linear Independence

S is also a Linearly independent set that is as large as possible because adding any vector to S results in a linearly dependent set

Span

Coordinate Systems

The Unique Representation Theorem

Let B = {b1, ..., bn} be a basis for a vector space V. Then for each x ∈ V, there exists a unique set of scalars c1, ..., cn such that x=c1b1+...+cnbn

Suppose B={b1, ..., bn} is a basis for V and x is ∈ V. The coordinates of x relative to the basis B (or the B-coordinates of x) are the weights c1, ..., cn such that x=c1b1+...+cnbn.

If c1,..., cn are the B-coordinates of x, then the vector in R^n [x]B = [c1,c2,...,cn] is the coordinate vector of x relative to B, or the B coordinate vector of x.

The mapping x |-> [x]B is the coordinate mapping (determined by B)

Let B = {b1, ..., bn} be a basis for a vector space V. Then the coordinate mapping x |-> [x]B is a one-to-one linear transformation from V onto R^n

Dimension of a Vector Space

If a vector space V has a basis B = {b1, ..., bn}, then any set in V containing more than n vectors must be linearly dependent

If a vector space V has a basis of n vectors then every basis of V must consist of exactly n vectors

The Basis Theorem

Let V be a p-dimensional vector space, p at least 1. Any Linearly independent set of exactly p elements in V is automatically a basis for V. Any set of exactly p elements that spans V is automatically a basis for V.

Eigenvalues

An eigenvalue of an nxn matrix A is a nonzero vector x such that Ax=λx for some scalar λ.. A scalar λ is called an eigenvalue of A if there is a nontrivial solution x of Ax=λx; such an x is called an eigenvector corresponding to λ

An eigenvector must be nonzero, but an eigenvalue may be zero

λ is an eigenvalue of A if and only if the equation (A - λI)x=0 has a nontrivial solution

The scalar equation det(A - λI) = 0 is called the characteristic equation of A

A scalar λ is an eigenvalue of an nxn matrix A if and only if λ satisfies the characteristic equation det(A - λI)=0

The eigenvalues of a triangular matrix are the entries on its main diagonal

0 is an eigenvalue of a square matrix A if and only if A is not invertible

If A is an nxn matrix, then det(A -λI) is a polynomial of degree n in λ, which is called the characteristic polynomial of A. The algebraic multiplicity of an eigenvalue λ is its multiplicity as a root of the characteristic equation.

Eigenvectors

Let λ be an eigenvalue of a square matrix A. Then the set consisting of the zero vector and all the eigenvectors corresponding to λ iscalled the eigenspace of A corresponding to λ. Note that the eigenspace of A corresponding to λ is Nul (A - λI)

If v1, ..., vr are eigenvectors that correspond to distinct eigenvalues λ1, ..., λr of an nxn matrix A, then the set {v1,...,vr} is linearly independent

Diagonalization Theorem

An nxn matrix A is diagonalizable if and only if A has n Linearly independent eigenvectors.

A square matrix is said to be diagonalizable if A is similar to a diagonal matrix, that is, if A=PDP^-1 for some invertible matrix P and some Diagonal matrix D.

A=PDP^-1, with D a diagonal matrix, if and only if the columns of P are n Linearly independent eigenvectors of A. In this case, the diagonal entries of D are eigenvalues of A that correspond , respectively to the eigenvectors in P.

An nxn matrix with n distinct eigenvalues is diagonalizable

An nxn matrix is diagonalizable if and only if the sum of the dimensions of the distinct eigenspaces equals n, and this happens if and only if the dimension of the eigenspace for each eigenvalue equals the multiplicity of that eigenvalue

Steps for Diagonalization (the eigen process)

Let A be an nxn matrix.

Solve det(A - λI)=0 for λ to find all eigenvalues.

Check the algebraic multiplicity

Find corresponding eigenvectors for eigenvalues of A by solving (A - λI)x=0

All eigenvaectors and the zero vector from an eigenspace for each eigenvalue

The dimension of the eigenspace is equal to the number of free variables in (A -λI)x=0

If the dimension of the eigenspace for each eigenvalue is not equal to the multiplicity of that eigenvalue, then A is NOT diagonalizable

If the dimension of the eigenspace for each eigenvalue equals the multiplicity of that eigenvalue, then A is Diagonalizable

A=PDP^-1, where P is formed by the Eigenvectors and D is formed by the Eigenvalues corresponding the the eigenvectors

Symmetric Matrices

A symmetric matrix is a matrix such that A^T = A. Such a matrix is necessarily square

If A is symmetric, then any two eigenvectors from different eigenspaces are orthogonal

A matrix A is said to be orthogonally diagonalizable if there are an othogonal matrix P (with P^-1 = P^T) and a Diagonalizable matrix D such that A = PDP^T=PDP^-1

An nxn matrix A is orthogonally diagonalizable if and only if A isa symmetric matrix

An orthogonal matrix is a square matrix U such that U^-1 = U^T. Clearly the set of all column vectors from an orthogonal matrix forms an orthogonal set of unit vectors.

The Spectral Theorem for Symmetric Matrices

An nxn symmetric matrix has the following properties

A has n real eigenvalues, counting multiplicities

The dimension of the eigenspace for each eigenvalueλ equals the multiplicity of λ as a root of the characteristic equation

The eigenspaces are mutually orthogonal

A is orthogonally diagonalizable

Length, Distance, and Perpendicularity ∈ R^n

Let u and v be vectors ∈ R^n with u=[u1,u2,...,un] and v=[v1,v2,...,vn]. Then the inner product (or DOT PRODUCT) of u and v, written u·v is u1v1+u2v2+...+unvn

u·v=(u^T)(v)

The length or norm of v is the nonnegative scalar ||v|| defined by ||v||=sqrt(v·v) and ||v||^2 = v·v

Normalization

A vector whose length is 1 is called a unit vector. Given a vector v, the new vector u=v / ||v|| is a unit vector in the same direction as v

||av||=a||v||

For u and v ∈ R^n, the distance between u and v, written as dist(u,v) is the length of the vector u-v. ||u-v||

Two vectors are orthogonal to each other if

u·v=0

u^T * v

||u+v||^2=||u||^2+||v||^2

If a vector z is orthogonal to every vector in a subspace W of R^n, then z is said to be orthogonal to W. The set of all vectors z that are orthogonal to W is called the orthogonal complement of W and is denoted by W⊥

W=Span{[1,0]}

W⊥=Span{[0,1]}

Orthagonality

Orthogonal Sets

A set of vectors {u1,...,up} ∈ R^n is said to be an orthogonal set if each pair of distinct vectors from the set is orthogonal, that is, ui·uj=0 when i is not equal to j

An orthogonal basis for a subspace W of R^n is a basis for W that is also an orthogonal set

Let {u1,...,up} be an orthogonal basis for a subspace W of R^n. For each y in W, the weights in the linear combination y=c1u1+...+cpup are given by cj=y·uj/uj·uj

We can apply the Gram-Schmidt process to produce an orthogonal basis for any nonzero subspace or R^n

Orthogonal Projection

Given a nonzero vector u ∈ R^n, we can decompose any vector y in R^n as y = y-hat + z where y-hat=αu for some scalar α and z is some vector orthogonal to u. The vector y-hat is called the orthogonal projection of y onto u, and the vector z is called the component of y orthogonal to u.

it can be shown that y-hat = (y·u/u·u)u

The orthogonal decomposition theorem

Let W be a subspace of R^n. Then each y ∈R^n can be written uniquely in the form y = y-hat + z where y-hat ∈W and z ∈W⊥ In fact, if {u1,...,up} is any orthogonal basis of W, then y-hat = (y·u1/u1·u1)u1+(y·u2/u2·u2)u2+...+(y·up/up·up)up and z = y - y-hat

The length of z, ||z||, is the distance between y and the vector space W.

The vector y-hat ∈ the Orthogonal decomposition theorem is called the orthogonal projection of y onto W and often is written as proj_[w]_y