Vector Spaces and Subspaces

Spaces of Vectors

1. The standard ｎ-dimensional space ℝ⃗ⁿ contains all real column vectors with ｎ components.

2. If 𝑣⃗ and 𝑤⃗ are in a vector space 𝑆⃗, every combination 𝑎𝑣⃗ + 𝑏𝑤⃗ must be in 𝑆⃗.

3. The "vectors" in 𝑆⃗ can be matrices or functions of 𝑥. The 1-point space 𝑍 consists of 𝒙⃗ = 𝟎⃗.

4. A subspace of ℝ⃗ⁿ is a vector space inside ℝ⃗ⁿ . Example: The line y = 3x inside R² .


5. The column space of 𝐴⃣ contains all combinations of the columns of 𝐴⃣: a subspace of ℝ⃗ᵐ .


6. The column space contains all the vectors 𝐴⃣x⃗. So 𝐴⃣x⃗ = 𝑏⃗ is solvable when b⃗ is in 𝐶(𝐴⃣).

Subspaces

𝑆⃗ ⊂ ℝ⃗ⁿ
   𝟎⃗∈𝑆⃗
   𝑥⃗,𝑦⃗(∈𝑆⃗) ⟹ 𝑎𝑥⃗+𝑏𝑦⃗ (∈𝑆⃗)
  
∀ 𝑎𝑥⃗+𝑏𝑦⃗  ⟺ 𝑆⃗ᵢ(𝑥⃗,𝑦⃗) : Span 

Nullspace

The nullspace 𝑁(𝐴) consists of all solutions to 𝐴⃣·𝑥⃗= ０⃗. These vectors 𝑥⃗ are in ℝ⃗ⁿ. This includes 0⃗

Elimination (from A⃣ to U⃣ [upper] to R⃣ [Reduced]) does not change the nullspace: 𝑁(A⃣) = 𝑁(U⃣)= 𝑁(R⃣).

The reduced row echelon form R⃣ = rref(A) has ∀pivots= 1, with zeros above and below. 4 If column j of R⃣ is free (no pivot), there is a "special solution" to A⃣·x⃗⃗= 0⃗ with X(j) = 1.

Number of pivots of R⃣ = number of nonzero rows in R⃣ = rank r. There are n - r free columns.


Every matrix with m < n has nonzero solutions to Ax= 0 in its nullspace.

The nullspace 𝑁(A⃣) is a subspace of ℝ⃗ⁿ. It contains all solutions to A⃣x⃗= 0⃗.
 Elimination on A⃣ produces a row reduced R⃣ with pivot columns and free columns.

Every free column leads to a special solution. That free variable is 1, the others are 0.

The rank 𝚛 of A⃣ is the number of pivots. All pivots are 1's in R⃣ = rref(A⃣).


The complete solution to A⃣x⃗ = 0⃗ is a combination of the 𝚗－𝚛 special solutions

A always has a free column if n>m, giving a nonzero solution to A⃣x⃗=0⃗.

Complete solution to A⃣x⃗ = b⃗: x⃗ = (one particular solution x⃗ₚ) + (any x⃗ₙ in the nullspace). 
 Elimination on [A b⃗] leads to [R d⃗]. Then A⃣x⃗=b⃗ is equivalent to R⃣x⃗=d⃗. 
 A⃣x⃗=b⃗ and R⃣x⃗=d⃗ are solvable only when all zero rows of R⃣ have zeros in d⃗. 
 When R⃣x⃗ = d⃗ is solvable, one very particular solution x⃗ₚ has all free variables equal to zero. 
 A⃣ has full column rank r = n when its nullspace 𝑁(A⃣) = zero vector: no free variables. 
  A has full row rank r = m when its column space 𝐶(A⃣) is ℝ⃗ᵐ : A⃣x⃗= b⃗ is always solvable. 
 The four cases are 𝚛 = 𝚖 = 𝚗 (A⃣ is invertible) and 𝚛 = 𝚖 < 𝚗 (every A⃣x⃗= b⃗ is solvable) 
 and 𝚛 = 𝚗 < 𝚖 (A⃣x⃗= b⃗ has 1 or ０ solutions) and 𝚛 < 𝚖, 𝚛 < 𝚗 (０ or ∞ solutions). 

The particular solution solves A⃣x⃗ₚ = b⃗ The nullspace solution, the n - r special solutions solve: A⃣x⃗ₙ = b⃗ The complete solution x⃗ₛ= x⃗ₚ＋x⃗ₙ

Every matrix A⃣ with full column rank (r=n) has all these properties All column of A⃣ are pivot columns These are no free variables or special solutions. The nullspace 𝑁(A⃣) contains only the zero vector x⃗=0⃗ If A⃣x⃗ = b⃗ has a solution (it might not) then it has only one solution)

Every matrix A⃣ with row rank (r=m) has all these properties

1. All rows have pivots, and R⃣ has no zero rows.

2. A⃣x⃗ = b⃗ has a solution for every right side b⃗

3. The column space is the whole space ℝ⃗ᵐ

4. There are n-r = n-m special solutions in the nullspace of A.

5. The rank 𝚛 is the number of pivots. The matrix R⃣ has 𝚖 - 𝚛 zero rows.
2. A⃣x⃗ = b⃗ is solvable if and only if the last m - r equations reduce to 0 = 0.
3. One particular solution x⃗ₚ has all free variables equal to zero.
4. The pivot variables are determined after the free variables are chosen.
S. Full column rank r = 𝑛 means no free variables: one solution or none.
6. Full row rank r = m means one solution if m = n or infinitely many if m < n.

Independent columns of A⃣: The only solution to A⃣x⃗= 0⃗ is x⃗= 0⃗. The nullspace is Z⃣.
 Independent vectors: The only zero combination c₁v₁ + · · · + CₖVₖ= 0 has all c’s= 0.
 A matrix with m < n has dependent columns: At least n-m free variables/ special solutions.

The vectors v₁, ..., vₖ span the space S⃗ if S⃗= all combinations of the v's.

The vectors v₁,…, vₖ are a basis for S⃗ if they are independent and they span S⃗.
 The dimension of a space S⃗ is the number of vectors in every basis for S⃗.
 If A is 4 by 4 and invertible, its columns are a basis for ℝ⃗⁴ . The dimension of ℝ⃗⁴ is 4.