Midterm 2 Definitions and Theorems

invertible

nxn (square) matrix A iff there exists a nxn (square) matrix C such that AC = CA = I_n

inverse of A

A^-1

(A^-1)^-1 =

A

(AB)^-1

B^-1A^-1

(A^T)^-1

(A^-1)^T

Theorem on Invertible Matrices

If A and B are invertible nxn matrices, then A^-1, AB, and A^T are invertible.

Invertible Matrix Theorem (IMT)

Let A be an mxn matrix. Then the following are equivalent:

• a. A is invertible

• b. A is row equivalent to I_n

• c. A has n pivot positions → REF(A) has n pivots

• d. Ax = 0 has only the trivial solution (x = 0)

• e. The columns of A are linearly independent

• f. The linear map T_A(x) = Ax is one-to-one

• g. For any b ϵ Rⁿ, Ax = b has exactly 1 solution

• h. The columns of A span Rⁿ

• i. T_A maps onto Rⁿ

• j. There is an nxn matrix C such that CA = I_n

• k. There is an nxn matrix D such that AD = I_n

• l. A^T is invertible

• m. The determinant of A, det(A), =/= 0

• n. The columns of A form a basis of Rⁿ

Determinant Theorem

There is a unique function det:{nxn (square) matrices} → R with the following properties:

• if you add a multiple of a row to another row, the determinant doesn’t change

• if you swap 2 rows, the determinant gets multiplied by -1

• you can factor out a non-zero scalar from a row (then the determinant gets multiplied by it)

Triangular Matrices and the Determinant Theorem

If A is upper or lower triangular, then det(A) = the product of the diagonal entries

upper triangular matrix

square matrix in which all entries below the main diagonal are 0

lower triangular matrix

square matrix in which all entires above the main diagonal are 0

Transpose Determinant Theorem

det(A^T) = det(A), so determinant rules work for columns as well as rows

det(A^T) =

det(A)

det [a b]

. [c d]

ad - bc

cofactor of [a_ij]

(-1)^{i + j} times the determinant of the submatrix of A made by deleting the ith row and jth column

Algorithm for Finding Inverse

• form the super augmented matrix [A I_n]

• row reduce [A I_n] to REF(A)

  • if A has fewer than n pivots (develops a zero row) → A is not invertible

  • otherwise, proceed

• go to RREF(A), so [A I_n] becomes [I_n A^-1]

Cofactor Expansion Algorithm

• Choose a row or column of A

• For each entry in it, compute its cofactor

• Multiply each entry by its cofactor

• Add up

Output = det(A)

Multiplicity of Determinants Theorem

det(AB) = det(A) * det(B)

Cramer’s Rule

Let A be an invertible nxn (square) matrix, b ϵ Rⁿ. Suppose Ax = b. Then x_i = det(A_i[b])/det(A).

adjugate of A

adj(A) = transpose of the cofactor matrix

formula for A^-1 using adj(A)

A^-1 = 1/det(A) * adj(A)

parallelopiped spanned by v₁, … v_n ϵ Rⁿ

Π_A = {t₁v₁ + … + t_nv_n | 0 <= t_i <= 1 for all i}

vol(parallelopiped) / vol(Π_A)

|det(A)|

Theorem: Effect of Volume on a Linear Map

Let T:Rⁿ→R^m be a linear map. Let S ⊂ Rⁿ with vol(S). Let A be the standard matrix of T.

• Then vol(T(S)) = |det(A)| * vol(S)

• T(S) = {T(v) | v ϵ S} or the set of all images T(v) where v is in S

vol(ellipsoid)

|det(A)| * 4/3π

vol(parallelogram)

|det([vertex 2], [vertex 3])| when anchored at (0,0)

vol(triangle)

1/2|det([vertex 2], [vertex 3])| when anchored at (0,0)

vector space

a set V with 2 operations:

• v + w

• λv, λ ϵ R

that has an element 0 ϵ V such that 0 + v = v

A U B

the set of all elements in A or B or both

A ∩ B

set of all elements in both A and B

polynomial

function p:R→R of the form p(t) = a₀ + a₁t + a₂t² + … + a_ntⁿ, a_i ϵ R

zero polynomial

function where all constants = 0 with degree -∞

H ⊂ V. H is a subspace of V if:

0 ϵ H

H is closed under vector addition and scalar multiplication

Theorem on Span and Subspace

span(v₁, … v_p) ⊂ V is a subspace of V

Theorem on Linear Maps and Subspaces

Let V and W be vector spaces. T:V→W is a linear map. Then,

• ker(T) = {v ϵ V | T(v) = 0} ⊂ V is a subspace of the domain

• range(T) = {T(v) | v ϵ V} ⊂ W is a subspace of the codomain

• T is one-to-one if and only if ker(T) = {0}

• T is onto if and only if range(T) = W

ker(T_A) =

nul(A) ⊂ Rⁿ

range(T_A) =

col(A) ⊂ R^m

basis

a linearly independent spanning set of V

Theorem on Bases and Vector Spaces

Every vector space has a basis.

finite dimensional

a vector space that has a finite spanning set

Theorem on Bases and Finite Dimensional Vector Spaces

Every finite dimensional vector space has a basis.

Theorem on Vector Spaces and Linear Dependence

Let V be a vector space. If v₁, … v_p ϵ V, they are linearly dependent if and only if at least one of them is a linear combination of the preceding vectors.

Theorem on Linear Combinations, Vector Spaces, and Span

Let v₁, … v_p ϵ V. Let H = span(v₁, … v_p). Suppose one of the vectors is a linear combination of the others. Then H will still be the span of the others.

Algorithm for Finding A Basis of a Finite Dimensional Vector Space

Let V be a finite dimensional vector space. Given V has a finite spanning set S = {v₁, … v_p}.

• If S is linearly independent, it is a basis → finished

• If S is linearly dependent, then one of its vectors is a linear combination of the others

  • Suppose v_p is a linear combination of (v₁, … v_p-1), then V = span (v₁, … v_p-1) = S’

  • If S’ is linearly independent → finished

  • Otherwise, we can throw away one vector and still span V

  • Process will stop at the basis