Understanding Vector Spaces and Their Transformations
1. Row Space of A
The elimination process transforms matrix A into a jagged array U. The row space of U is easily determined: its dimension is the rank r, and the nonzero rows form a basis. Each elementary operation preserves the row space because each row of U is a combination of the original rows of A. Therefore, row(A) = row(U), implying they share the same dimension r and basis. We efficiently obtain the basis using the nonzero rows of U.
Example: Subspace Basis in R4
Determine a basis for the subspace of R4 generated by the vectors v1 = (1, 1, 0, 1)T, v2 = (1, 2, 2, 1)T, and v3 = (3, 4, 2, 3)T.
We arrange these vectors as rows of matrix A:
A = (1 1 0 1 / 1 2 2 1 / 3 4 2 3)
Gaussian elimination yields:
A = (1 1 0 1 / 1 2 2 1 / 3 4 2 3) = (1 1 0 1 / 0 1 2 0 / 0 1 2 0) = (1 1 0 1 / 0 1 2 0 / 0 0 0 0) = U
With two pivots in U, the rank r is 2. The subspace dimension is two. The first two rows of U form a basis for row(U), and consequently for row(A). Thus, a basis for the subspace is w1 = (1, 1, 0, 1)T and w2 = (0, 1, 2, 0)T.
2. Nullspace of A
The nullspace of A consists of vectors x such that Ax = 0. Gaussian elimination transforms Ax = 0 into Ux = 0 without changing the solutions. Hence, null(A) = null(U). Of the m equations in Ax = 0, only r are independent, corresponding to the r nonzero rows of U. The nullspace dimension is n – r, representing the free variables in Ux = 0, associated with columns of U lacking pivots. To find a basis, set each free variable to 1 and the others to 0, then solve Ux = 0 for the basic variables using back-substitution. The resulting vectors form a basis for null(A).
Example: Subspace Basis in R4
Determine a basis for the subspace W of R4, consisting of vectors x = (x1, x2, x3, x4)T such that:
–x1 + x2 – x3 + 2x4 = 0
2x1 – 2x2 + x3 = 0
5x1 – 5x2 + 3x3 – 2x4 = 0
Here, W = {x ∈ R4 | Ax = 0}, where A = (-1 1 -1 2 / 2 -2 1 0 / 5 -5 3 -2).
Gaussian elimination gives U = (-1 1 -1 2 / 0 0 -1 4 / 0 0 0 0).
With n = 4 and r = 2, the dimension of W is n – r = 2. The basic variables are x1 and x3, while x2 and x4 are free. The system Ux = 0 becomes:
x1 – x2 + x3 -2x4=0
–x3 + 4x4 = 0
Setting x2 = 1 and x4 = 0 yields v1 = (1, 1, 0, 0)T. Setting x2 = 0 and x4 = 1 yields v2 = (-2, 0, 4, 1)T. Vectors v1 and v2 form a basis for W.
3. Column Space of A
While the columns of A are the rows of AT, it’s more insightful to consider the column space in terms of m, n, and r. Gaussian elimination alters the columns, so A and U don’t have the same column space. However, if certain columns of U form a basis for col(U), the corresponding columns of A form a basis for col(A). This is because Ax = 0 if and only if Ux = 0. These equivalent systems share the same solutions. A linear dependence among columns of A, expressed by Ax = 0, corresponds to a linear dependence Ux = 0 among columns of U with the same coefficients. Thus, linear independence among columns of A implies the same for U, and vice-versa. To find a basis for col(A), we find one for col(U), which consists of the r pivot columns. Therefore, the dimension of col(A) equals the rank r, which is also the dimension of row(A). The number of independent rows equals the number of independent columns. A basis for col(A) is formed by the corresponding r columns of A that correspond to the pivot columns of U.
Example
For A = (1 2 0 1 / 0 1 1 0 / 1 2 0 1), Gaussian elimination leads to U = (1 2 0 1 / 0 1 1 0 / 0 0 0 0).
The pivot columns of U are the first and second. Thus, a basis for col(A) is formed by the first two columns of A: (1, 0, 1)T and (2, 1, 2)T.
4. Left Nullspace of A
This is the nullspace of AT. Since dimension of column space + dimension of nullspace = number of columns, applying this to AT (which has m columns) and using row rank = column rank = r, we get dim(null(AT)) = m – r. A basis for null(AT) can be found similarly to the nullspace of A.
Exercise: Basis for Four Fundamental Subspaces
Determine a basis for the four fundamental subspaces of A = (2 4 0 2 / 6 -2 1 0 / 8 2 1 2 / -4 1 -3 -5).
Basis from a Generating System
To find a basis for a subspace given a generating system, find the row space basis of the matrix whose rows are the generating vectors.
Example: Basis for Subspace in R3
Find a basis for the subspace of R3 spanned by v1 = (1, 2, 1)T, v2 = (3, 2, 1)T, v3 = (4, 0, 0)T, and v4 = (-1, 2, 1)T.
Form matrix A with these vectors as rows and perform Gaussian elimination:
A = (1 2 1 / 3 2 1 / 4 0 0 / -1 2 1) → (1 2 1 / 0 -4 -2 / 0 -8 -4 / 0 4 2) → (1 2 1 / 0 -4 -2 / 0 0 0 / 0 0 0)
The row space has dimension two, indicating only two independent vectors among the four. A basis for the subspace is {(1, 2, 1)T, (0, -4, -2)T}.
Basis from a Free System
To complete a basis from a set of free vectors, extend the corresponding row space until the desired dimension is reached.
Example: Completing a Basis in R4
Complete a basis of R4 from v1 = (1, 1, 0, 1)T, v2 = (1, 2, 2, 1)T, and v3 = (3, 0, 2, 3)T.
Arrange the vectors as rows and perform Gaussian elimination:
(1 1 0 1 / 1 2 2 1 / 3 0 2 3) → (1 1 0 1 / 0 1 2 0 / 0 -3 2 0) → (1 1 0 1 / 0 1 2 0 / 0 0 8 0)
Adding a row with a pivot in the fourth column, such as (0, 0, 0, 1)T, completes a basis for R4.
Equations of a Subspace from a Generating System
To find the equations of a subspace from a generating system, arrange the generators as rows of a matrix and add a row with a generic vector. Reduce the matrix using Gaussian elimination, and the conditions on the generic vector’s components give the subspace equations.
Example: Equations of a Subspace
Obtain the equations of the subspace spanned by v1 = (1, 0, 3)T and v2 = (-2, 3, -1)T.
Form the matrix and reduce:
(1 0 3 / -2 3 -1 / x y z) → (1 0 3 / 0 3 5 / 0 y z-3x) → (1 0 3 / 0 3 5 / 0 0 z-3x-(5/3)y)
For (x, y, z)T to be in the subspace, the last row must be zero. Therefore, the subspace equation is z – 3x – (5/3)y = 0.
2.4 Operations on Subspaces
2.4.1 Intersection of Subspaces
The intersection W = W1 ∩ W2 of subspaces W1 and W2 of a vector space V is the set of vectors belonging to both W1 and W2. W is itself a subspace, and it always contains at least the zero vector. Representing each subspace with a homogeneous system of equations, the intersection satisfies all equations simultaneously.
Example: Intersection of Subspaces
For W1 = {(x, y, z)T | x + y + z = 0} and W2 = {(x, y, z)T | x – z = 0}, the intersection is W1 ∩ W2 = {(x, y, z)T | x + y + z = 0, x – z = 0}.
2.4.2 Sum of Subspaces
The sum W = W1 + W2 of subspaces W1 and W2 of V is the subspace generated by their union. To describe the sum, find a basis for each subspace and retain the linearly independent vectors to form a basis for the sum. Every vector w in W can be expressed as w = w1 + w2, where w1 ∈ W1 and w2 ∈ W2. If the intersection is trivial (W1 ∩ W2 = {0}), the sum is a direct sum, denoted W = W1 ⊕ W2. In this case, the decomposition w = w1 + w2 is unique.
Dimension Formula
For finite-dimensional subspaces:
dim(W1 + W2) = dim(W1) + dim(W2) – dim(W1 ∩ W2)
For direct sums:
dim(W1 ⊕ W2) = dim(W1) + dim(W2)
Examples: Linear Mappings and Matrices
[1] Linear Mapping from R4 to R3
For T: R4 → R3, T(x1, x2, x3, x4) = (x1 + x2 + x3, x2 + x4, x1 + x2 + x3 + x4), the matrix in canonical bases is:
M(T) = (1 1 1 0 / 0 1 0 1 / 1 1 1 1)
[2] Linear Mapping between Polynomial Spaces
For T: P4[x] → P3[x], T(p(x)) = p‘(x), the matrix in canonical bases is:
M(T) = (0 1 0 0 0 / 0 0 2 0 0 / 0 0 0 3 0 / 0 0 0 0 4)
[3] Linear Mapping with Different Bases
For T: U → V, where U = R3, V = R2, with bases BU = {(1, 0, 0)T, (0, 1, 0)T, (0, 0, 1)T}, BV = {(1, 0)T, (0, 1)T}, B‘U = {(1, 0, 0)T, (1, 0, 1)T, (1, 1, 0)T}, and B‘V = {(1, 1)T, (0, 1)T}, if M(T, BU, BV) = (-1 0 1 / 2 -1 1), then:
M(T, B‘U, B‘V) = (1 0 / -1 1)-1(-1 0 1 / 2 -1 1)(1 1 1 / 0 0 1 / 0 1 0)
Theorem 3: Dimensionality Relationship
For a linear map T: U → V between finite-dimensional vector spaces over a field K:
dim(U) = dim(ker(T)) + dim(Im(T))
If A is the matrix of T with respect to chosen bases, then dim(ker(T)) = dim(null(A)) = n – r and dim(Im(T)) = dim(col(A)) = r, where r is the rank of A.