Linear Algebra 1553 Notes, Study notes of Linear Algebra

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Textbook: Interactive Linear Algebra
1. Systems of Linear Equations: Algebra
2. Systems of Linear Equations: Geometry
2.1. Span of columns and Solution set
A = m x n
Lives in
Dimension
Span of columns of A
Rm
Number of pivots
Solution set of Ax = 0
Rn
Number of free variables
2.2. Linear Independence and Dependence
- Check for linear independence
+ Row reduce for Ax = 0 โ‡’ verify that x = y = z = 0 OR pivot in every column
+ Wide matrix โ‡’ cannot be linearly independent
2.3. Subspaces
- Notation: V = { x,y in R2 | x + y = 1}
- Check for subspace:
+ Contain : sub. x = y = z = .. = 0 into subspace equation
0
โ†’
+ Closed under addition: Let u = (a b) and v = (c d) => u + v = (a+c, b+d)
+ Closed under multiplication: Let u = (a b) โ‡’ cu = c(a b)
- Common types of subspaces:
+ Col(A): spans of columns of A
+ Null(A): solution set of Ax = 0
+ Find A from Col(A) and Nul(A)
โ— Eg. Col(A) = (x y), Nul(A) = (3 1) โ‡’ x1 = 3x2 โ‡’ x1 - 3x2 = 0
โ‡’ A = (x -3x
y -3y)
+ Subspace contains the vector = Span of that vector
- Not subspaces:
+ y = |x|: (1 1) and (-1 -1)
+ x2 + y2 โ‰ค 1: (1 0) and (5 0)
+ y โ‰ฅ 2x
+ xy = 0: (1 0) and (0 1)
2.4. Basis and Dimension
- Dim(V) = number of vectors in any basis of V
+ If V = Rn (n vectors) โ‡’ any basis of V has n vectors in it
+ If V = Col(A), dim(V) = dim(ColA) = rank(A) = number of pivot columns
+ If V = Nul(A), dim(V) = dim(NullA) = nullity(A) = number of free variables
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**1. Systems of Linear Equations: Algebra

  1. Systems of Linear Equations: Geometry** 2.1. Span of columns and Solution set

A = m x n Lives in Dimension

Span of columns of A Rm^ Number of pivots

Solution set of Ax = 0 Rn^ Number of free variables

2.2. Linear Independence and Dependence

  • Check for linear independence
    • Row reduce for Ax = 0 โ‡’ verify that x = y = z = 0 OR pivot in every column
    • Wide matrix โ‡’ cannot be linearly independent

2.3. Subspaces

  • Notation: V = { x,y in R^2 | x + y = 1}
  • Check for subspace:
    • Contain 0 : sub. x = y = z = .. = 0 into subspace equation

โ†’

  • Closed under addition: Let u = (a b) and v = (c d) => u + v = (a+c, b+d)
  • Closed under multiplication: Let u = (a b) โ‡’ cu = c(a b)
  • Common types of subspaces:
    • Col(A): spans of columns of A
    • Null(A): solution set of Ax = 0
    • Find A from Col(A) and Nul(A) โ— Eg. Col(A) = (x y), Nul(A) = (3 1) โ‡’ x 1 = 3x 2 โ‡’ x 1 - 3x 2 = 0 โ‡’ A = (x -3x y -3y)
    • Subspace contains the vector = Span of that vector
  • Not subspaces:
    • y = |x|: (1 1) and (-1 -1)
    • x^2 + y^2 โ‰ค 1: (1 0) and (5 0)
    • y โ‰ฅ 2x
    • xy = 0: (1 0) and (0 1)

2.4. Basis and Dimension

  • Dim(V) = number of vectors in any basis of V
    • If V = Rn^ (n vectors) โ‡’ any basis of V has n vectors in it
    • If V = Col(A), dim(V) = dim(ColA) = rank(A) = number of pivot columns
    • If V = Nul(A), dim(V) = dim(NullA) = nullity(A) = number of free variables
  • Verify basis: i) Vectors are in V ii) Vectors span V iii) Vectors are linearly independent OR if dimV = m, any m vectors either span V OR are linearly independent
  • Compute basis for a subspace:
    • Basis of Col(A): Row reduction โ‡’ Original pivot columns โ— Another basis of the same span / column space: any non-zero linear combination of vectors in the basis
    • Basis of a span: find basis of Col(A)
    • Basis of Nul(A): Parametric form of general solution โ— Another vector โ‰  0 and in Nul(A): any non-zero linear combination of vectors in the basis
    • Basis of a subspace: โ— Convert equation to the form Ax = 0 => V = Nul(a b c) โ‡’ Find parametric form of general solution

2.5. Rank Theorem

  • rank(A) + nullity(A) = number of columns of A 3. Linear Transformations and Matrix Algebra

3.1. Matrix Transformation

  • Let A be an mร—n matrix, and let T(x) = Ax be the associated matrix transformation.
    • Domain(T) = Rn
    • Codomain(T) = Rm
    • Range(T) = Col(A) = n-D object in Rm
    • Dim(T) = dim(Range(T)) = dim(Col(A)) = rank(A) = number of pivot columns
  • T : Rn^ โ†’ Rm^ โ‡’ A(x) = m x n

3.2. One-to-one and Onto Transformation

Non-examples of Tranformation

4. Determinants

4.1. Cofactor Expansion Let A be an n ร— n matrix with entries aij

det(A) = with ๐‘— = 1

๐‘› โˆ‘ ๐‘Ž๐‘–๐‘—๐ถ๐‘–๐‘— ๐ถ๐‘–๐‘— = (โˆ’ 1) ๐‘–+๐‘— ๐‘‘๐‘’๐‘ก(๐ด๐‘–๐‘—)

5. Eigenvalues and Eigenvectors

5.1. Definition

  • An eigenvector of A is a nonzero vector v in R n such that Av = ฮปv, for some scalar ฮป
  • An eigenvalue of A is a scalar ฮป such that the equation Av = ฮปv has a nontrivial solution
    • ฮป = 1 โ‡” (A - I) has a non-trivial solution โ‡’ A - I is not invertible
  • Dim(ฮป-eigenspace) = number of free variable

Matrix ฮป

Reflection 1, -

Projection 0, 1

Identity 1

Shear 1

Rotation by theta < 180 No real ฮป

Rotation by 180 aka reflection 1 (counterclockwise) OR -1 (clockwise)

5.2. Characteristic Polynomial

  • f(ฮป) = ฮป^2 - Tr(A)ฮป + det(A)
    • Trace(A) = sum of diagonal entries

5.3. Diagonalization

  • Diagonalizable matrices โ‡” algebraic multiplicity = geometric multiplicity

5.4. Complex Eigenvalues

6. Orthogonality

6.1. Orthogonal Complements

  • dim Col A = dim Row A
  • Row A = Col AT
  • (Col A)โŠฅ^ = Nul AT
  • (Nul A)โŠฅ^ = Col AT

6.2. Orthogonal Projection

  • ATAv = ATx โ‡’ xW = Av 7. Summary

A = m x n Lives in Dimension

Span of columns of A Rm

Number of pivots

Codomain of T

Solution set of Ax = 0 Rn

Number of free variables

Domain of T

A = n x n has pivots in every column โ‡” Col(A) = Rn โ‡” Dim(Col(A)) = n = rank A โ‡” Nul(A) = {0} โ‡” Nullity A = 0 โ‡” Columns of A are linearly independent โ‡” Columns of A spans Rn โ‡” Columns of A form a basis for Rn โ‡” Ax = 0 has only the trivial solution โ‡” Ax = b is consistent for all b in Rn โ‡” Ax = b has a unique solution for each b in Rn โ‡” T is one-to-one โ‡” T is onto โ‡” A is invertible (A-1^ exists) โ‡” T is invertible โ‡” det A โ‰  0 (if det A = 0, A-1^ is undefined)

8. Commonly made mistakes - If u, v, w are linearly independent, u, u - v, u - v + 2w are also linearly independent - If v 1 , v 2 , v 3 are linearly independent vectors in Rn, x1v1 + x2v2 + x3v3 = b has exactly one solution only if b is in span {v 1 , v 2 , v 3 } - If u, v are vectors in subspace W, 2u - 5v is also a vector in W