Linear Alg Matrices Ch 4.5-4.7

Front 1

basis

Back 1

a set of vectors S = {v1, v2,.....vn} in a vector space V is a BASIS for V when:

1. S spans V

2. S is linearly independent

(has enough vectors to span V, but not so many that it becomes linearly dependent/one could be written as a combination of the others)

Front 2

Does every vector space have a basis consisting of a finite number of vectors?

Back 2

no, but our book will only discuss these

Front 3

finite dimensional

Back 3

vector space V has a basis with a finite number of vectors

Front 4

example of infinite dimensional vector space V

Back 4

vector space P of all polynomials

Front 5

How would you show that {(1,0,0),(0,1,0),(0,0,1)} is a basis for R3?

Back 5

show that it is a spanning set for R3

show that it is linearly independent

Front 6

standard basis for R3

Back 6

{(1,0,0),(0,1,0),(0,0,1)}

Front 7

Is a basis a special subset of spanning sets?

Back 7

yes. All bases are spanning sets, but not all spanning sets are bases

Front 8

if a coefficient matrix has a nonzero determinant, what does this mean about the system's solution?

Back 8

it has a unique solution

you can use this to prove a set is a spanning set

Front 9

How do you show that a set is a spanning set?

Back 9

use vector x as a general vector that can represent any vector. write the equation:

1. c1v1 + c2v2 = x

2. write as a system of linear equations

3. show that the coefficient matrix has a nonzero determinant, which means that it has a unique solution (basically means that these vectors can be used to make up any vector)

Front 10

a few different ways to show that something is linearly independent

Back 10

take the general system of equations with the general vector, and set it equal to zero so the system is homogenous. solve and show that only the trivial solution exists

OR show that the vectors are NOT scalar multiples of each other

Front 11

uniqueness of basis representation

Back 11

if S = {v1, v2, ..... vn} is a basis for a vector space V, then every vector in V can be written in one and only one way as a linear combination of S

Front 12

if the determinant is nonzero, the system is ______ and _______ it has _________

Back 12

consistent and invertible and it has a unique solution

Front 13

How do you prove uniqueness of basis representation?

Back 13

show that c1v1 + c2v2 + c3v3 = (u1, u2, u3), where v represents a basis and u represents coordinates of any vector

From the system of linear equations, show that the determinant is nonzero in order to prove it is invertible/has a unique solution

Front 14

bases and linear dependence

Back 14

if S = {v1, v2, v3,.....vn} is a basis for vector space V, then every set containing more than n vectors in V is linearly dependent

Front 15

number of vectors in a basis

Back 15

if a vector space V has one basis with n vectors, then every basis for V has n vectors

Front 16

definition of the dimension of a vector space

Back 16

if a vector space V has a basis consisting of n vectors, then the number n is the dimension of V, denoted by dim(V) = n

Front 17

When V consists of only the zero vector, how to do define the dimension of V?

Back 17

zero

Front 18

Dimension of R^n with standard operations

Back 18

n

Front 19

Dimension of Pn with standard operations

Back 19

n+1

Front 20

Dimension of Mm,n with the standard operations

Back 20

m * n

Front 21

technique for finding the dimension of a subspace

Back 21

find a set of linearly independent vectors that spans the subspace. The number of vectors in this basis gives you the dimension

Front 22

every 2x2 symmetric matrix has the form....

Back 22

a{1,0,0,0} + b{0,1,1,0} + c{0,0,0,1}

Front 23

2 different ways to test basis in n-Dimensional space

Back 23

Let V be a vetor space of dimension n

1. If S = {v1, v2, ....vn} is a linearly independent set of vectors in V, then S is a basis for V

2. If S = {v1, v2,....vn} spans V, then S is a basis for V

Of note: dimension of V MUST be n in order to only prove one of these. if dim(V) is NOT n, you must show that both are true

Front 24

row space of a matrix

Back 24

row space of A is the subspace of R^n spanned by the row vectors of A

Front 25

column space of a matrix

Back 25

column space of A is the subspace of R^m spanned by the column vectors of A

Front 26

If a matrix is row-equivalent to another matrix with the same dimensions, are their row spaces the same?

Back 26

yes. they have the same row space

Front 27

Do elementary row operations change the row space of a matrix? Do elementary row operations change the column space of a matrix?

Back 27

row space does not change; column space does

Front 28

Basis for the row space of a matrix:

If a matrix A is row-equivalent to matrix B in row-echelon form, what are the implications for basis of a row space?

Back 28

the nonzero row vectors of B form a basis for A

Front 29

How do you find the basis for a row space?

Back 29

1. Write each vector as a row

2. Put in row-echelon form (doesn't have to be Gauss-Jordan, just Gauss)

3. The nonzero row vectors of the final matrix for the basis of the original matrix

Front 30

How do you find the basis for a subspace spanned by S = {v1, v2, v3.....} in Rn?

Back 30

1. Write each vector as a row

2. Put in row-echelon form

3. Non-zero row vectors are the basis

Front 31

What are the two ways to find the basis for the column space of a matrix?

Back 31

Option 1: Use the fact that the column space is equal to A^T and apply the techniques for row space to A^T

Option 2: Recognize that although the column spaces change, the dependency relationship between columns does not

Front 32

How do you find the basis for the column space using dependency relationships between columns?

Back 32

1. Reduce to row-echelon form

2. Notice the columns with leading 1's. These are linearly dependent.

3. Those corresponding column numbers in the original matrix form the basis

Front 33

Do the row space and column space of a matrix have the same dimension?

Back 33

yes

Front 34

rank of a matrix

Back 34

dimension of the row (or column) space of a matrix A.

Denoted by rank(A)

Front 35

How do you find the rank of a matrix?

Back 35

find a basis for row space or column space. Number of vectors in the basis is the rank

(convert to row-echelon. find nonzero rows)

Front 36

nullspace of A

Back 36

If A is an m x n matrix, then the set of all solutions of the homogenous system of linear equations Ax = 0 is a subspace of R^n called the nullspace of A

Also called the solution space of Ax = 0

Front 37

nullity of A

Back 37

dimension of the nullspace of A

Front 38

How do you find the nullspace of a matrix?

Back 38

1. Put the coefficient matrix in reduced row echelon form

2. Write the parametric equations

3. Write as vectors

4. Factor out coefficients. The remaining vectors form the basis for the nullspace

Front 39

Once a matrix is in reduced row echelon form, which columns determine the rank of a matrix? Which ones determine its nullity?

Back 39

rank: columns with leading ones

nullity: columns that correspond to free variables

Front 40

dimension of the solution space

Back 40

If A is an m x n matrix of rank r, then the dimension of the solution space of Ax = 0 is n - r. That is, dimension n = rank(A) + nullity(A)

Front 41

Why isn't the set of all solution vectors of the non homogenous system Ax = b ever a subspace?

Back 41

does not contain the zero vector, so it can't be a subspace

Front 42

solutions of a non homogenous linear system

Back 42

if xp is a particular solution of the non homogenous set Ax = b, then every solution of this system can be written in the form x = xp + xh, where xh is a solution of the corresponding homogenous system Ax = 0

Front 43

How do you find the solution set of a non homogenous system?

Back 43

1. Put matrix in reduced row echelon form

2. Write parametric representation

3. Write as vectors and factor out coefficients

4. set those final vectors equal to xp + su1 + tu2 +...., where xh = su1 + tu2 represents an arbitrary vector in the solution space of Ax = 0

Front 44

How can the column space of a matrix be used to determine whether a system of linear equations is consistent?

Back 44

The system Ax = b is consistent if and only if b is in the column space of A. That is, the system is consistent if and only if b is in the subspace of R^n spanned by the columns of A

Front 45

If you're finding the basis for the nullspace of a matrix and the 0 vector is the remaining column, what's the answer?

Back 45

there is no basis for the matrix.

Front 46

Coordinate representation relative to a basis / coordinate matrix

Back 46

Let B {v1, v2....vn} be an ordered basis for vector space V and let x be a vector in V such that:

x = c1v1 + c2v2 + ... + cnvn

The scalars are the coordinates of x relative to B. The coordinate matrix/coordinate vector of x relative to B is the matrix in R^n whose components are those coordinates.

Front 47

How do you write the answer for finding the coordinate matrix in R^3 relative to the standard basis? What about relative to a nonstandard basis?

Back 47

[x]s =[enter coordinates as a column]

[x]B =[enter coordinates as a column]

//coordinates are the scalars representing a linear combination

Front 48

change of basis

Back 48

you're give the coordinates relative to one basis and asked to find relative to another. Basically, multiply it out to get the actual value of the original vector, and then solve for it as a linear combination of the vectors in the new basis

Front 49

inverse of a transition matrix

Back 49

If P is the transition matrix from base B' to base B in R^n, then P is invertible and the transition matrix from B to B' is P^-1

Front 50

simpler way of explaining what the transition matrix is

Back 50

Once you've multiple out the scalars to form a linear combination and first write the matrix to solve for the scalars from the system of equations, that matrix is the transition matrix

Front 51

How do you find a transition matrix from B to B'?

Back 51

1. Write 2 matrices from vector sets, where each individual vector is written as a COLUMN

2. Form the matrix [ B' B]

3. Use Gauss-Jordan elimination to rewrite as [I P^-1]

4. P^-1 is the transition matrix from B to B'

Front 52

Simple way to find transition matrix from standard basis to nonstandard basis

Back 52

Transition matrix from B to B' where B is a standard basis and B' is a nonstandard basis:

P^-1 = (B')^-1 (inverse of nonstandard basis)

Front 53

Simple way to find transition matrix from nonstandard basis to standard basis

Back 53

Transition matrix from B to B' where B is a nonstandard basis and B' is a standard basis:

P^-1 = B (the nonstandard basis)