#import "@preview/ctheorems:1.1.2": *
#show: thmrules
#let theorem = thmbox("theorem", "Theorem", fill: rgb("#eeffee"), padding: (top: 0em, bottom: 0em)).with(numbering: none)
#let remark = thmbox("remark", "Remark", padding: (top: 0em, bottom: 0em)).with(numbering: none)
#let definition = thmbox("definition", "Definition", inset: (x: 1.2em, top: 1em), padding: (top: 0em, bottom: 0em)).with(numbering: none)
#let example = thmplain("example", "Example", padding: (top: 0em, bottom: 0em)).with(numbering: none)
#let proof = thmproof("proof", "Proof", padding: (top: 0em, bottom: 0em))


= Vector Spaces and Associated Matrices
== Row Space and Column Space
- Let $A = mat(a_11, dots, a_(1n);dots.v, dots.v, dots.v; a_(m 1), dots, a_(m n))$
- $r_i = (a_(i 1) dots a_(i n))$
- Row Space of $A$ is the vector space spanned by rows of $A$ : $"span"{r_1, dots, r_m}$
- $c_j = vec(a_(1j),dots.v,a_(m j))$
- Column space of $A$ is vector space spanned by columns of $A$ : $"span"{c_1, dots, c_n}$

#example[Let $A = mat(1,2,3;4,5,6)$.
- $r_1 = (1,2,3), r_2 = (4,5,6)$.
- $c_1 = (1,4)^T, c_2 = (2,5)^T, c_3 = (3,6)^T$
]
#example[To find the basis of $A$
- Row Space of $A$
  - Gaussian elimination on the row-vectors of $A$ and the non-zero rows of $R$ are basis
- Col space of $A$
  - Gaussian elimination on row vectors of $A$ and the pivot columns of $A$ are the basis
]

=== Row Equivalence
Let $A$ and $B$ be matrices of the same size. $A$ and $B$ are row equivalent if one can be obtained from another by ERO


#theorem[If $A$ and $B$ are row equivalent, then $A$ and $B$ have same row spaces]

#remark[Let $R$ be REF of $A$
- Row space of $A$ = row space of $R$
- Nonzero rows of $R$ are linearly independent
  - Nonzero rows of $R$ form basis for row space of $A$
  - No of nonzero rows $R$ = dim of row space of $A$
]

=== Row Operations to Columns
- Let $A$ and $B$ be Row Equivalent Matrices : $A -> A_1 -> dots -> A_(k-1) -> B$
- Exists $E_1, dots, E_k$ s.t. $E_k dots E_1 A = B$
- $M = E_k dots E_1$, $M$ is invertible, $M A = B, A = M^(-1) B$

#theorem[Row Equivalence Preserves the linear relations on the columns
- Suppose $A = mat(a_1, dots, a_n)$ and $B = mat(b_1, dots, b_n)$
- $a_j = c_1 a_1 + dots + c_n a_n arrow.l.r.double b_j = c_1 b_1 + dots + c_n b_n$
]
#remark[Let $R$ be REF of $A$
- Pivot columns of $R$ form basis for column space of $R$
- Columns of $A$ which correspond to pivot columns of $R$ form a basis for column space of $A$
- No of pivot columns of $R$ is the dim of column space of $A$
]

#example[Find basis for a vector space V

Let $V = "span"{v_1, v_2, v_3, v_4, v_5, v_6}, - v_1 = (1,2,2,1), v_2 = (3,6,6,3), "etc"$
1. Method 1 (Row Form)
  - View each $v_i$ as a row vector
    - $mat(1, 2, 2, 1; 3, 6, 6, 3; 4,9,9,5;-2,-1,-1,1;5,8,9,4;4,2,7,3) attach(arrow.r.long, t: "Gaussian Elimination") mat(1,2,2,1;0,1,1,1;0,0,1,1;0,0,0,0;0,0,0,0;0,0,0,0)$
  - $V$ has basis${(1,2,2,1), (0,1,1,1), (0,0,1,1)}$
  - $dim(V) = 3$
2. Method 2 (Column Form)
  - View each $v_i$ as a column vector 
    - $mat(1,3,4,-2,5,4;2,6,9,-1,8,2;2,6,9,-1,9,7;1,3,5,1,4,3) attach(arrow.r.long, t: "Gaussian Elimination") mat(1,3,4,-2,5,4;0,0,1,3,-2,-6;0,0,0,0,1,5;0,0,0,0,0,0)$
  - Using Pivot columns (1,3,5), pick the corresponding vectors, $(v_1, v_3, v_5)$
  - V has basis${(1,2,2,1), (4,9,9,5), (4,2,7,3)}$
  - $dim(V) = 3$
]

#remark[Finding Basis TLDR
+ Method 1: View each $v_1, dots, v_k$ as a row vector
  - Find row echelon form $R$ of $vec(v_1,dots.v,v_k)$
  - Nonzero rows of $R$ form basis of $V$
+ Method 2: View each $v_1, dots, v_k$ as a column vector
  - Find row echelon form $R$ of $mat(v_1,dots,v_k)$
  - Find pivot columns of $R$
  - Corresponding columns $v_j$ form basis for $V$
]

#example[Extend $S$ to a basis for $RR^n$
- Find REF of $S$ as row vectors
- Insert the $e_i$ vectors for each $i$ non-pivot column
]

#theorem[Consistency

Let $A$ be a $m times n$ matrix
- Column space of $A$ is ${A v | v in RR^n}$
- Linear system $A x = b$ is consistent
  - #sym.arrow.l.r.double $b$ lies in the column space of $A$
]

#theorem[Rank

- Let $A$ be a matrix, then $dim("row space of " A) = dim("col space of " A)$
- This is the rank of $A$, denoted by $"rank(A)"$
]

#remark[
Let $A$ be an $m times n$ matrix
- $"rank"(A) = "rank"(A^T)$
- $"rank"(A) = 0 arrow.l.r.double A = 0$
  - $"rank"(A) <= min{m,n}$
  - Full rank if $"rank"(A) = min{m,n}$
- Square Matrix $A$ is full rank #sym.arrow.l.r.double $A$ is invertible.
]

#remark[Rank and Consistency of Linear System

$A x = b$ be linear system
- $A x = b$ is consistent
  - $arrow.l.r.double b in "span"{c_1,dots,c_n}$
  - $arrow.l.r.double "span"{c_1,dots,c_n} = "span"{c_1,dots,c_n,b$
  - $arrow.l.r.double dim("span"{c_1,dots,c_n}) = dim("span"{c_1,dots,c_n,b})$
  - $arrow.l.r.double "rank"(A) = "rank"(A|b)$
- $"rank"(A) <= "rank"(A | b) <= "rank"(A) + 1$
]

If $A B x = b$ has a solution, then $A x = b$ also has a solution


#theorem[Let $A = m times n$ matrix and $B = n times p$

- Column space of $A B subset.eq $ column space of $A$
- Row space of $A B subset.eq $ row space of $B$
- $"rank"(A B) <= min{"rank"(A), "rank"(B)}$
]

#definition[Nullspace and Nullity

- Nullspace of $A$ is the solution space of $A x = 0$
  - $v in RR^n | A v = 0$
- Vectors in nullspace are viewed as column vectors

- Let $R$ be a REF of $A$
  - $A x = 0 arrow.l.r.double R x = 0$
  - $"nullity"(A) = "nullity"(R)$
]

#theorem[Dimension Theorem
Let $A$ be a $m times n$ matrix
- $"rank"(A) + "nullity"(B) = n$ ($n$ is the number of columns)
]