Subspaces

Definition

A subspace of a vector space V is a subset W of V that is itself a vector space under the same operations of addition and scalar multiplication. To be a subspace, W must satisfy three conditions:

  1. The zero vector of V is in W
  2. W is closed under vector addition: for all u, v in W, u + v is in W
  3. W is closed under scalar multiplication: for all v in W and scalar c, cv is in W

Note: Conditions 2 and 3 can be combined into a single condition: W is closed under linear combinations.

Testing if a Set is a Subspace

To determine if a subset W of a vector space V is a subspace:

  1. Check if the zero vector is in W
  2. Check if W is closed under addition: For arbitrary vectors u, v in W, verify that u + v is in W
  3. Check if W is closed under scalar multiplication: For an arbitrary vector v in W and scalar c, verify that cv is in W

In practice, for subsets defined by equations, we can check:

  • If W is defined by a system of homogeneous linear equations, it is always a subspace
  • If W is defined by a system of non-homogeneous linear equations (with non-zero constants), it is never a subspace
  • If W is defined by nonlinear equations, check all three conditions carefully

Common Types of Subspaces

1. Trivial Subspaces

  • The zero subspace {0} containing only the zero vector
  • The entire vector space V itself

2. Solution Spaces

  • Null Space of a matrix A (denoted null(A) or ker(A)): the set of all solutions to Ax = 0
  • Column Space of a matrix A (denoted col(A) or im(A)): the span of the columns of A
  • Row Space of a matrix A (denoted row(A)): the span of the rows of A
  • Left Null Space of a matrix A: the null space of A^T

3. Geometric Subspaces

  • Lines through the origin in ℝ²
  • Planes through the origin in ℝ³
  • Hyperplanes through the origin in ℝⁿ

4. Function Subspaces

  • The set of all polynomials of degree at most n
  • The set of all continuous functions on an interval
  • The set of all differentiable functions on an interval

Basis and Dimension of a Subspace

A basis for a subspace W is a linearly independent set of vectors that spans W. The dimension of W, denoted dim(W), is the number of vectors in a basis.

To find a basis for a subspace:

  1. For a subspace defined by a spanning set {v₁, v₂, …, vₙ}:
    • Remove any linearly dependent vectors
    • The remaining vectors form a basis
  2. For a subspace defined by equations Ax = 0:
    • Find the general solution to the system
    • Extract the vectors that multiply the free variables
    • These vectors form a basis for the solution space

Examples

Example 1: Testing if a Set is a Subspace

Determine if S = {(x, y, z) ∈ ℝ³ x + y - 4 = 0, z = 0} is a subspace of ℝ³.

Step 1: Check if the zero vector (0,0,0) is in S

  • Substituting into the equations: 0 + 0 - 4 = -4 ≠ 0 and 0 = 0
  • Since (0,0,0) doesn’t satisfy x + y - 4 = 0, S is not a subspace

Example 2: Finding a Basis for a Subspace

Find a basis for the subspace S = {(x, y, z, t) ∈ ℝ⁴ x - 3y + z = 0}.

Step 1: Rewrite the equation to express one variable in terms of the others x = 3y - z

Step 2: Express the general solution in parametric form For any values of y, z, and t: (x, y, z, t) = (3y - z, y, z, t) = y(3, 1, 0, 0) + z(-1, 0, 1, 0) + t(0, 0, 0, 1)

Step 3: The vectors that multiply the parameters form a basis Basis = {(3, 1, 0, 0), (-1, 0, 1, 0), (0, 0, 0, 1)}

Step 4: Verify that these vectors are linearly independent The matrix with these vectors as columns has full rank, so they are linearly independent

Step 5: Find the dimension dim(S) = 3

Example 3: Determining the Rank of a Matrix with Parameters

For h ≠ 0: R₃ = R₃ - ((h² + 1)/(2h))R₂: $\begin{pmatrix} 1 & 1/2 & 0 & 0 \ 0 & h & 1 & h + 1 \ 0 & 0 & h - \frac{(h^2 + 1)}{2h} & -(h+1)\frac{(h^2 + 1)}{2h} \end{pmatrix}$

Simplifying the (3,3) entry: $h - \frac{(h^2 + 1)}{2h} = h - \frac{h^2}{2h} - \frac{1}{2h} = h - \frac{h}{2} - \frac{1}{2h} = \frac{h}{2} - \frac{1}{2h}$

Step 2: Analyze the rank based on different values of h

Case 1: If h = 0 The matrix becomes: $\begin{pmatrix} 1 & 1/2 & 0 & 0 \ 0 & 0 & 1 & 1 \ 0 & 0 & 0 & 0 \end{pmatrix}$ This has 2 non-zero rows, so rank(A₀) = 2

Case 2: If $\frac{h}{2} - \frac{1}{2h} = 0$ This occurs when $h^2 = 1$, so h = ±1

For h = 1: $\begin{pmatrix} 1 & 1/2 & 0 & 0 \ 0 & 1 & 1 & 2 \ 0 & 0 & 0 & 0 \end{pmatrix}$ So rank(A₁) = 2

For h = -1: $\begin{pmatrix} 1 & 1/2 & 0 & 0 \ 0 & -1 & 1 & 0 \ 0 & 0 & 0 & 0 \end{pmatrix}$ So rank(A₋₁) = 2

Case 3: For all other values of h: The (3,3) entry is non-zero, so we have 3 non-zero rows, and rank(Aₕ) = 3

Step 3: Conclusion

  • rank(A₀) = 2
  • rank(A₁) = rank(A₋₁) = 2
  • rank(Aₕ) = 3 for all other values of h

Intersection and Sum of Subspaces

Intersection of Subspaces

If U and W are subspaces of a vector space V, then their intersection U ∩ W is also a subspace of V.

To find a basis for U ∩ W:

  1. Find bases for U and W
  2. Form a system of equations whose solution space is U ∩ W
  3. Solve the system to find a basis for the intersection

Sum of Subspaces

The sum of two subspaces U and W is defined as: U + W = {u + w | u ∈ U, w ∈ W}

The sum U + W is also a subspace of V.

To find a basis for U + W:

  1. Find bases {u₁, u₂, …, uₘ} for U and {w₁, w₂, …, wₙ} for W
  2. Form the set {u₁, u₂, …, uₘ, w₁, w₂, …, wₙ}
  3. Remove linearly dependent vectors to obtain a basis for U + W

Dimension Formula

For subspaces U and W: dim(U + W) = dim(U) + dim(W) - dim(U ∩ W)

Direct Sum

Two subspaces U and W form a direct sum, denoted U ⊕ W, if:

  1. V = U + W
  2. U ∩ W = {0}

In a direct sum, every vector v in V can be uniquely written as v = u + w where u ∈ U and w ∈ W.

If U and W form a direct sum, then: dim(U ⊕ W) = dim(U) + dim(W)

Examples

Example 4: Finding the Intersection and Sum of Subspaces

In ℝ³, let U = span{(1,1,0), (0,1,1)} and W = span{(1,0,0), (1,1,1)}.

Step 1: Find a basis for U ∩ W

  • Vector v is in U ∩ W if and only if v ∈ U and v ∈ W
  • If v ∈ U, then v = a(1,1,0) + b(0,1,1) = (a, a+b, b)
  • If v ∈ W, then v = c(1,0,0) + d(1,1,1) = (c+d, d, d)
  • For v to be in both, we need: a = c+d a+b = d b = d
  • From the third equation, b = d
  • Substituting into the second: a+d = d, so a = 0
  • From the first: 0 = c+d, so c = -d
  • So v = d(-1,1,1) + d(1,0,0) = d(0,1,1)
  • Thus U ∩ W = span{(0,1,1)}

Step 2: Find a basis for U + W

  • Combined set: {(1,1,0), (0,1,1), (1,0,0), (1,1,1)}
  • Check for linear independence: (1,1,1) = (1,1,0) + (0,0,1) = (1,1,0) + (0,1,1) - (0,1,0)
  • So (1,1,1) is linearly dependent on the other vectors
  • A basis for U + W is {(1,1,0), (0,1,1), (1,0,0)}
  • dim(U + W) = 3

Step 3: Verify the dimension formula

  • dim(U) = 2, dim(W) = 2, dim(U ∩ W) = 1
  • dim(U) + dim(W) - dim(U ∩ W) = 2 + 2 - 1 = 3 = dim(U + W) ✓

Orthogonal Complements

For a subspace W of ℝⁿ, the orthogonal complement of W, denoted W⊥, is: W⊥ = {v ∈ ℝⁿ | v·w = 0 for all w ∈ W}

Properties of orthogonal complements:

  1. W⊥ is a subspace of ℝⁿ
  2. (W⊥)⊥ = W
  3. dim(W) + dim(W⊥) = n
  4. If W is defined by a system of equations Ax = 0, then W⊥ is the row space of A

Common Mistakes to Avoid

  1. Forgetting to check if the zero vector is in the set
  2. Assuming that any subset defined by equations is a subspace
  3. Incorrectly calculating the dimension of a subspace
  4. Errors in finding a basis for the intersection or sum of subspaces
  5. Confusing the concepts of span, basis, and subspace

Solving Subspace Exercises

Strategy 1: Testing if a Set is a Subspace

  1. Check if the set contains the zero vector
  2. For sets defined by equations:
    • If all equations are homogeneous linear equations, it’s a subspace
    • If any equation is non-homogeneous or nonlinear, check all conditions carefully

Strategy 2: Finding a Basis for a Subspace

  1. For a subspace defined by spanning vectors:
    • Test for linear independence
    • Remove linearly dependent vectors
  2. For a subspace defined by equations:
    • Solve the system to find the general solution
    • Extract coefficient vectors of the free variables

Strategy 3: Finding Dimension and Basis for Parametric Matrices

  1. Perform row reduction keeping the parameter as a variable
  2. Identify cases where the rank changes based on parameter values
  3. Find a basis for each case separately

Connection to Other Topics

Subspaces connect to many other topics in linear algebra:

  1. Linear Transformations: The kernel and image of a linear transformation are subspaces
  2. Eigenspaces: For each eigenvalue λ of a matrix A, the set of all eigenvectors with that eigenvalue, along with the zero vector, forms a subspace
  3. SVD and PCA: These techniques involve finding important subspaces of data
  4. Orthogonal Projections: Projecting onto a subspace is a fundamental operation in many applications
  5. Invariant Subspaces: Subspaces that remain unchanged under specific transformations

    Exercises

  6. Determine, for each value of $h$, the rank of the following matrix: \(A_h = \begin{pmatrix} 0 & h & 1 & h + 1 \\ 2 & 1 & 0 & 0 \\ -h^2 - 1 & 0 & h & 0 \end{pmatrix}\)
  7. Check if the following sets are subspaces and, if there exist, find two bases for each subspace:
    • $S = {(x, y, z) \in \mathbb{R}^3 x + y - 4 = 0, z = 0}$
    • $S = {(x, y, z) \in \mathbb{R}^3 x + y - 4 = 0, z = 1}$
    • $S = {(x, y, z, t) \in \mathbb{R}^4 x - 3y + z = 0}$