Exam 3

Question 1

Definition of Null Space

Answer 1

The null space of an m x n matrix A, written as Nul A. is the set of all solutions of the homogeneous equation Ax=0.

In set notation, Nul A = {x : x is in Rⁿ and Ax = 0}

Question 2

Theorem 2 (4.2)

Answer 2

The null space of an m x n matrix A is a subspace of Rⁿ. Equivalently, the set of all solutions to a system Ax = 0 of m homogeneous linear equations in n unknowns is a subspace of Rⁿ

Question 3

Definition of a Column Space

Answer 3

The column of space an m x n matrix A, written as Col A, is the set of all linear combinations of the columns of A. If A = [a₁…a_n ], then
Col A = Span {a₁…a_n}

Question 4

Theorem 3 (4.2)

Answer 4

The column space of an m x n matrix A is a subspace of R^m

Question 5

A linear transformation T from a vector space V into a vector space W is a rule that

Answer 5

assigns to each vector x in V a unique vector T(x) in W, such that

1. T (u+v) = T(u) + T(v)
   for all u,v in V and
2. T(cu) = cT(u) for all u in V and all scalars c

Question 6

Theorem 4 (4.3)

Answer 6

An indexed set {v₁…v_p} of two or more vectors. with v₁ not equal to 0, is linearly dependent if and only if some v_j (with j > 1) is a linear combination of the preceding vectors v₁…v_j-1

Question 7

Definition of a Basis

Answer 7

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

1. B is a linearly independent set and
2. the subspace spanned by B coincides with H; that is, H = Span {b₁…b_p}

Question 8

Theorem 5 (Spanning Set Theorem) 4.3

Answer 8

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

1. If one of the vectors in S - say, v_k - is a linear combination of the remaining vectors in S, then the set formed from S by removing v_k still spans H.
2. If H does not equal {0}, some subset of S is a basis for H.

Question 9

Theorem 6 (4.3)

Answer 9

The pivot columns of a matrix A form a basis for Col A

Question 10

Theorem 7 (Unique Representation Theorem) (4.4)

Answer 10

Let B = {b₁..b_n} be a basis for a vector space V. Then for each x in V, there exists a unique set of scalars c₁…c_n such that x = c₁b₁ +….+ c_nb_n

Question 11

Definition for coordinate system

Answer 11

Suppose B = {b₁…b_n} is a basis for V and x is in V. The coordinates of x relative to the basis B (or the B-coordinates of x) are the weights c₁..c_n such that x = c₁b₁ +….+c_nb_n

Question 12

Theorem 8 (4.4)

Answer 12

Let B = {b₁..b_n} 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 Rⁿ

Question 13

Theorem 9 (4.5)

Answer 13

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

Question 14

Theorem 10 (4.5)

Answer 14

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

Question 15

Definition of finite-dimensional/infinite-dimensional

Answer 15

V is spanned by a finite set, and the dimension of V written as dimV is the number of vectors in a basis for V. The dimension of he zero vector space {0} os defined to be zero. If V is not spanned by a finite set, then V is said to be infinite-dimensional.

Question 16

Theorem 11 (4.5)

Answer 16

Let H be a subspace of a finite-dimensional vector space V. Any linearly independent set in H can be expanded to a basis for H. H is finite dimensional and dim H is less than or equal to dim V

Question 17

Theorem 12 (The Basis Theorem) (4.5)

Answer 17

Let V be a p-dimensional vector space, p greater than or equal to 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.

Question 18

Dimension of Nul A and Col A

Answer 18

Dimension of Nul A is the # of free variables in the equation Ax = 0, and the dimension of Col A is the number of pivot columns in A

Question 19

Theorem 13 (4.6)

Answer 19

If two matrices A and B are row equivalent, then their row spaces are the same. If B is in echelon form, the nonzero rows of B form a basis for the row space of A as well as for that of B.

Question 20

Definition of Rank

Answer 20

The rank of A is the dimension of the column space of A

Question 21

Theorem 14 (The Rank Theorem) (4.6)

Answer 21

The dimensions of the column space and the row space of an m x n matrix A are equal. This common dimension, the rank of A, also equals the number of pivot positions in A and satisfies the equation rank A + dim Nul A = n

Question 22

IMT

Answer 22

Let A be a square n x n matrix. Then the following statements are equivalent. For A, these are all either true or false:

• A is an invertible matrix
• A is row equivalent to the nxn identity matrix
• A has n pivot positions
• The equation Ax=0 has only the trivial solution
• The columns of A form a linearly independent set
• The linear transformation x |-> Ax maps Rn onto Rn
• There is an n x n matrix C such that CA = I
• There is an n x n matrix D such that AD = I
• A^T is an invertible matrix

new ones:

• The columns of A form a basis of Rⁿ
• Col A = Rⁿ
• dim Col A = n
• rank A = n
• Nul A = {0}
• dim Nul A = 0

Question 23

Theorem 1 (3.1)

Answer 23

The determinant of an n x n matrix A can be computed b a cofactor expansion across any row or down any column. The expansion across the “i”th row using the cofactors in C_ij = (-1)^i+j det A_ij is

det A = a_i1C_i1 + … + a_inC_in

The cofactor expansion down the “j”th column is
det A = a_1jC_1j + …. + a_njC_nj

Question 24

Theorem 2 (3.1)

Answer 24

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

Question 25

Theorem 3 (3.2)

Answer 25

Row Operations

Let A be a square matrix:

1. if a multiple of one row of A is added to another tow to produce a matrix B, then det B = det A
2. if two rows of A are interchanged to produce B, then det B = - det A
3. if one row of A is multiplied by k to produce B, then det B = k * det A

Question 26

Theorem 4 (3.2)

Answer 26

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

Question 27

Theorem 5 (3.2)

Answer 27

If A is a n x n matrix, then det A^T = det A

Question 28

Theorem 6 (Multiplicative Property)

Answer 28

If A and B are n x n matrices, then
det AB = (det A)(det B)

Question 29

Definition of Eigenvector

Answer 29

An eigenvector of an n x n 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.

Question 30

Theorem 1 (5.1)

Answer 30

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

Question 31

Theorem 2 (5.1)

Answer 31

If v₁…v_r are eigenvectors that correspond to distinct eigenvalues λ₁…λ_r of an n x n matrix A, then the set {v₁…v_r} is linearly independent

Question 32

IMT (continued 5.2)

Answer 32

Let A be an n x n matrix. Then A is invertible if and only if

• the number 0 is not an eigenvalue of A
• the determinant of A is not zero

Question 33

Theorem 3 (5.2)

Answer 33

Let A and B n x n matrices

1. A is invertible if and only if det A does not equal 0
2. det A B = (det A)(det B)
3. det A^T = det A
4. If A is triangular, then det A is the product of the entries on the main diagonal of A
5. A row replacement operation on A does not change the determinant. A row interchange changes the sign of the determinant. A row scaling also scales the determinant by the same scalar factor

Question 34

Definition of Characteristic Equation

Answer 34

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

Question 35

Theorem 4 (5.2)

Answer 35

If n x n matrices A and B are similar, then they have the same characteristic polynomial and hence the same eigenvalues (with the same multiplicities)

Question 36

Definition of Similarity

Answer 36

Matrices are not similar even though they have the same eigenvalues. Similarity is not the same as row equivalence. Row operations on a matrix usually changes its eigenvalues

Question 37

Theorem 5 (5.3) The Diagonalization Theorem

Answer 37

An n x n matrix A is diagonalizable if and only ifA has n linearly independent eigenvectors.

In fact, 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.

Question 38

Theorem 6 (5.3)

Answer 38

An n x n matrix with n distinct eigenvalues is diagonalizable

Question 39

Theorem 7 (5.3)

Answer 39

Let A be an n x n matrix whose distinct eigenvalues are λ₁…λ_p

1. For 1 less than/equal to k less than/equal to p, the dimension of the eigenspace for λ_k is less than or equal to the multiplicity of the eigenvalue λ_k
2. The matrix A is diagonalizable if and only if the sum of the dimensions of the eigenspaces equals n, and this happens if and only if (i) the characteristic polynomial factors completely into linear factors and (ii) the dimension of the eigenspace for each λ_k equals the multiplicity of λ_k
3. If A is diagonalizable and B_k is a basis for the eigenspace corresponding to λ_k for each k, then the total collection of vectors in the sets B₁…B_p forms an eigenvector basis for Rⁿ