Vector Space Axioms, Linear Independence & Basis Dimension

Linear algebra is the foundational mathematical language of modern computing, high-dimensional data processing, scientific simulation, computer graphics, and machine learning neural networks. At its core, linear algebra formalizes the geometry of vector spaces, the mechanics of linear transformations, and the structural decomposition of multi-dimensional mappings.

1. Foundational Architecture of Vector Spaces#

In classical elementary mathematics, vectors are frequently introduced as spatial arrows possessing length and directional orientation in two-dimensional or three-dimensional physical space. In advanced computer science, robotics, and scientific computing, this geometric intuition is generalized into an abstract algebraic framework known as an axiomatic vector space.

The Axiomatic Definition over a Field#

A vector space V over a scalar field F (most commonly the field of real numbers R or the field of complex numbers C) is a mathematical set whose elements are called vectors, equipped with two fundamental operations:

1. Vector Addition: A binary operation + : V × V → V that assigns to every pair of vectors u, v in V a unique vector u + v in V.
2. Scalar Multiplication: An external operation · : F × V → V that assigns to every scalar c in F and every vector v in V a unique vector c · v in V.

For the algebraic structure (V, +, ·) to qualify strictly as a vector space, it must satisfy eight universal axioms across all vectors u, v, w in V and all scalars a, b in F:

1. Associativity of Vector Addition:

u + (v + w) = (u + v) + w

Groupings do not alter the resulting vector sum.

1. Commutativity of Vector Addition:

u + v = v + u

The sequence in which vectors are aggregated does not alter the cumulative displacement.

1. Existence of an Additive Identity Element:

There exists an element 0 in V (the zero vector) such that:

v + 0 = v for all v in V.

1. Existence of Additive Inverse Elements:

For every vector v in V, there exists a unique vector -v in V such that:

v + (-v) = 0.

1. Compatibility of Scalar Multiplication with Field Multiplication:

a · (b · v) = (a · b) · v

Multiplying successively by scalars a and b is equivalent to multiplying by their scalar product (a · b).

1. Identity Element of Scalar Multiplication:

1 · v = v for all v in V, where 1 is the multiplicative identity of the scalar field F.

1. Distributivity of Scalar Multiplication over Vector Addition:

a · (u + v) = a · u + a · v

Scaling a combined vector sum scales each constituent vector summand proportionally.

1. Distributivity of Scalar Multiplication over Field Addition:

(a + b) · v = a · v + b · v

Summing two scalars before scaling a vector yields the identical result as scaling the vector by each scalar individually and summing the outputs.

Closure Properties#

Implicit in these operations are the two closure criteria:

• Additive Closure: If u, v in V, then u + v must reside in V.
• Multiplicative Closure: If c in F and v in V, then c · v must reside in V.

Illustrative Counterexamples: Non-Vector Spaces#

To understand what constitutes a vector space, it is illuminating to examine mathematical sets that fail the axiomatic criteria:

• The Integer Lattice Z^n: The set of vectors with purely integer coordinates is closed under addition. However, it fails closure under real scalar multiplication: for v = (1, 2) in Z^2 and scalar c = 0.5 in R, c · v = (0.5, 1.0) does not belong to Z^2. Hence, Z^n is not a real vector space.
• The Unit Sphere S^(n-1) = {x in R^n : ||x|| = 1}: This set contains vectors of unit length. It fails additive closure: (1, 0) + (0, 1) = (1, 1), whose Euclidean norm is sqrt(2) ≠ 1. Moreover, the zero vector 0 does not have unit length and is absent from the set.
• The General Linear Group GL(n, R): The set of invertible n×n real matrices is closed under matrix multiplication. However, it fails additive closure: the identity matrix I and -I are both invertible, but their sum I + (-I) = 0 is the singular zero matrix, which is non-invertible.

2. Linear Combinations, Span, and Subspaces#

The fundamental algebraic mechanism for synthesizing vectors within a space is the linear combination.

Formal Formulation of Linear Combinations#

Given a finite set of vectors S = {v1, v2, ..., vk} in a vector space V, a vector w in V is called a linear combination of S if there exist scalars c1, c2, ..., ck in F such that:

w = c1 · v1 + c2 · v2 + ... + ck · vk

The Linear Span#

The span of a set of vectors S, denoted Span(S), is the collection of all possible linear combinations that can be formed using vectors in S:

Span(S) = { sum(i=1 to k, ci · vi) : ci in F, vi in S, k >= 1 }

Geometrically, the span represents the complete geometric manifold generated by stretching and combining the vectors in S:

• In R^3, the span of a single non-zero vector is an infinite 1-dimensional straight line passing through the origin.
• The span of two non-collinear vectors in R^3 is an infinite 2-dimensional flat plane passing through the origin.
• If a third vector lies outside that plane, the span expands to fill all 3-dimensional space R^3.

The Subspace Criteria#

A subset W of a vector space V is termed a linear subspace of V if W is itself a vector space under the inherited operations of addition and scalar multiplication.

Because the eight algebraic axioms automatically hold for any subset of V, verifying that W is a subspace requires checking only three conditions:

1. Zero Vector Inclusion:

The zero vector 0 of V must belong to W (0 in W).

1. Closure under Vector Addition:

For all vectors u, v in W, the sum u + v must belong to W.

1. Closure under Scalar Multiplication:

For all scalars c in F and vectors v in W, the product c · v must belong to W.

+-----------------------------------------------------------------------------+
|                      GEOMETRIC SUBSPACE TAXONOMY IN R^3                     |
+-----------------------------------------------------------------------------+
| Dimension | Geometric Object      | Subspace Conditions Satisfied?         |
+-----------+-----------------------+-----------------------------------------+
|    0D     | Origin {0}            | Yes: Contains 0, closed under + and ·   |
|    1D     | Line through Origin   | Yes: Closed under all scalar scaling    |
|    1D     | Line NOT through 0    | NO: Lacks 0 vector; fails closure       |
|    2D     | Plane through Origin  | Yes: Closed under vector combinations   |
|    2D     | Plane NOT through 0   | NO: Fails zero vector test (offset)     |
|    3D     | Entire space R^3      | Yes: Trivial improper subspace          |
+-----------------------------------------------------------------------------+

3. Linear Independence: Geometric & Algebraic Formulations#

A critical objective in multi-dimensional analysis is determining whether a collection of vectors contains redundant directional information.

Rigorous Definition of Linear Independence#

A finite set of vectors S = {v1, v2, ..., vk} in V is linearly independent if the only scalar coefficients c1, c2, ..., ck satisfying the homogeneous equation:

c1 · v1 + c2 · v2 + ... + ck · vk = 0

are the trivial coefficients:

c1 = c2 = ... = ck = 0.

Conversely, if there exist scalar coefficients not all zero such that the linear combination sums to the zero vector, the set S is linearly dependent. In that case, at least one vector can be expressed as a linear combination of the remaining vectors.

Matrix Null Space Criterion#

In Euclidean space R^n, testing the linear independence of k vectors v1, v2, ..., vk in R^m is equivalent to constructing an m×k matrix A whose columns are the vectors:

A = [ v1 | v2 | ... | vk ]

The homogeneous vector equation c1 · v1 + ... + ck · vk = 0 can be written in compact matrix-vector form:

A · c = 0, where c = [c1, c2, ..., ck]^T.

The set S is linearly independent if and only if the null space of A contains only the zero vector:

Null(A) = {0}.

This occurs if and only if Gaussian elimination transforms A into a row echelon form where every column is a pivot column (i.e., rank(A) = k).

The Determinant Test for Square Systems#

When dealing with n vectors in an n-dimensional space R^n, the column matrix A is square (n×n). The determinant det(A) provides a single scalar metric for independence:

• If det(A) ≠ 0: The matrix A is non-singular and invertible. The vectors v1, v2, ..., vn are strictly linearly independent and span R^n.
• If det(A) = 0: The matrix A is singular. The vectors are linearly dependent, and the spatial volume of the parallelotope spanned by them collapses to zero.

Function Spaces & The Wronskian Determinant#

Linear independence extends beyond discrete numeric tuples into infinite-dimensional function spaces. For sufficiently differentiable functions f1(x), f2(x), ..., fn(x), the Wronskian determinant W(x) is defined as:

W(f1, f2, ..., fn)(x) = det([

[ f1(x), f2(x), ..., fn(x) ],

[ f1'(x), f2'(x), ..., fn'(x) ],

[ ..., ..., ..., ... ],

[ f1^(n-1), f2^(n-1), ..., fn^(n-1) ]

])

Theorem: If the Wronskian W(x0) ≠ 0 at a single point x0 in an interval I, then the functions f1, f2, ..., fn are linearly independent on I.

4. Basis, Dimension, and Coordinate Representation#

When a set of vectors is both irreducibly non-redundant and capable of reconstructing every element in a vector space, it is called a basis.

Mathematical Definition of a Basis#

A subset B = {b1, b2, ..., bn} of vector space V is a basis for V if and only if it satisfies two simultaneous criteria:

1. Linear Independence: B is linearly independent.
2. Spanning Capability: Span(B) = V.

A basis represents both a minimal spanning set (removing any vector reduces the span) and a maximal linearly independent set (adding any vector introduces linear dependency).

The Dimension Theorem#

Theorem: If a vector space V possesses a finite basis consisting of n vectors, then every other basis for V must contain exactly n vectors.

This invariant integer n is defined as the Dimension of the vector space, denoted dim(V).

• For Euclidean space R^n, the standard canonical basis is E = {e1, e2, ..., en}, where ei is the column vector with a 1 at row i and 0 elsewhere. Hence, dim(R^n) = n.
• For the space of polynomials of degree at most d, P_d(x), the standard basis is {1, x, x^2, ..., x^d}, which contains d + 1 elements. Hence, dim(P_d(x)) = d + 1.

Unique Coordinate Systems#

Let B = {b1, b2, ..., bn} be an ordered basis for V. For any arbitrary vector v in V, there exists a unique sequence of scalar coordinates [c1, c2, ..., cn] such that:

v = c1 · b1 + c2 · b2 + ... + cn · bn.

We denote the coordinate vector of v relative to basis B as:

[v]_B = [ c1, c2, ..., cn ]^T in F^n.

This establishes an isomorphism between any abstract n-dimensional real vector space V and the concrete numerical space R^n.

Change of Basis Transformations#

In computer vision, robotics, and physics, changing the coordinate reference frame is frequent. Suppose we have two ordered bases for R^n:

• Basis B = {b1, b2, ..., bn}
• Basis C = {c1, c2, ..., cn}

To convert coordinate vectors from basis B to basis C, we apply the change of basis matrix P_(C <- B):

[v]_C = P_(C <- B) · [v]_B

The j-th column of P_(C <- B) is the coordinate vector of the j-th basis vector of B expressed relative to basis C:

P_(C <- B) = [ [b1]_C | [b2]_C | ... | [bn]_C ]

When transforming between the standard basis E and a custom basis B, the matrix P_(E <- B) = [ b1 | b2 | ... | bn ] has the basis vectors of B as its raw columns. The inverse transformation is given by:

P_(B <- E) = (P_(E <- B))^(-1).

5. The Four Fundamental Subspaces & The Rank-Nullity Theorem#

Every linear operator represented by an m×n matrix A : R^n → R^m gives rise to four fundamental vector subspaces that reveal the complete geometry of the linear system.

Domain Space R^n                               Codomain Space R^m
+----------------------------+                 +----------------------------+
|                            |                 |                            |
|       Row Space            |  --- A x --->   |       Column Space         |
|        Row(A)              |                 |         Col(A)             |
|     Dimension: r           |                 |      Dimension: r          |
|                            |                 |                            |
|----------------------------|                 |----------------------------|
|                            |                 |                            |
|       Null Space           |  --- A x = 0 -> |      Left Null Space       |
|        Null(A)             |                 |         Null(A^T)          |
|     Dimension: n - r       |                 |      Dimension: m - r      |
|                            |                 |                            |
+----------------------------+                 +----------------------------+
   Orthogonal Complement:                         Orthogonal Complement:
   Row(A) _|_ Null(A)                             Col(A) _|_ Null(A^T)

The Four Subspaces Defined#

For any real matrix A in R^(m×n) with rank r:

1. The Column Space Col(A) ⊆ R^m:

- The subspace spanned by the column vectors of A.

- Represents all possible output vectors b such that A · x = b is solvable.

- Dimension = rank(A) = r.

1. The Null Space Null(A) ⊆ R^n:

- The subspace of all domain vectors x that map to the zero vector: Null(A) = { x in R^n : A · x = 0 }.

- Represents the kernel of the transformation.

- Dimension = nullity(A) = n - r.

1. The Row Space Row(A) ⊆ R^n:

- The subspace spanned by the row vectors of A, or equivalently, Col(A^T).

- Dimension = rank(A) = r.

1. The Left Null Space Null(A^T) ⊆ R^m:

- The subspace of all codomain vectors y such that A^T · y = 0, or y^T · A = 0^T.

- Dimension = m - r.

Orthogonal Complements#

A cornerstone theorem of linear algebra states that the domain and codomain decompose into pairs of mutually orthogonal complementary subspaces:

• In the domain R^n: The row space and null space are orthogonal complements:

Row(A) ⊥ Null(A), and Row(A) ⊕ Null(A) = R^n.

Every vector x in R^n can be uniquely decomposed as x = x_row + x_null, where x_row in Row(A) and x_null in Null(A), with x_row · x_null = 0.

• In the codomain R^m: The column space and left null space are orthogonal complements:

Col(A) ⊥ Null(A^T), and Col(A) ⊕ Null(A^T) = R^m.

The Rank-Nullity Theorem#

The fundamental relationship governing any linear map represented by an m×n matrix A is the Rank-Nullity Theorem:

dim(Col(A)) + dim(Null(A)) = n

or equivalently:

rank(A) + nullity(A) = n

Notice that the sum equals n (the number of columns / dimension of the domain), regardless of the number of rows m.

6. Software Engineering & Systems Applications#

Linear independence, vector space dimensions, and subspace projections are not abstract mathematical curiosities; they form the operational foundation of high-performance computing systems.

1. Computer Graphics: 4D Homogeneous Coordinates#

In 3D graphics rendering pipelines (DirectX, Vulkan, OpenGL), spatial geometry involves translations, rotations, and perspective projections. Rotations and scaling are linear transformations in R^3:

• Rotation: x_new = R · x_old
• Scaling: x_new = S · x_old

However, spatial translation is an affine transformation:

x_new = x_old + t

Because x_old + t cannot be represented as a 3×3 matrix multiplication on R^3 (the zero vector would map to t ≠ 0, violating linearity), graphics architectures embed 3D Euclidean space into a 4-dimensional projective vector space using homogeneous coordinates:

x_homogeneous = [ x, y, z, 1 ]^T

In this 4D vector space, translation becomes a standard linear transformation represented by a 4×4 shear matrix:

[ [ 1, 0, 0, tx ],

[ 0, 1, 0, ty ],

[ 0, 0, 1, tz ],

[ 0, 0, 0, 1 ] ] · [ x, y, z, 1 ]^T = [ x + tx, y + ty, z + tz, 1 ]^T

By elevating the dimension from 3 to 4, translations, rotations, and perspective divisions compose into unified 4×4 matrix chains that execute concurrently on modern GPU SIMD hardware.

2. Digital Signal Processing: Discrete Fourier Basis#

In audio engineering, telecommunications, and JPEG/MP3 compression, signal analysis transforms time-domain sampled vectors into frequency-domain components. For a discrete signal vector x in C^N, the Discrete Fourier Transform (DFT) projects x onto an orthonormal basis of complex exponential vectors:

f_k = [ 1, e^(-i · 2π · k · 1 / N), e^(-i · 2π · k · 2 / N), ..., e^(-i · 2π · k · (N-1) / N) ]^T for k = 0, 1, ..., N-1.

Because these N basis vectors are mutually orthogonal in C^N under the Hermitian inner product, any digitized waveform can be uniquely decomposed into frequency components. The Fast Fourier Transform (FFT) performs this change of basis in O(N log N) operations rather than naive O(N^2) matrix multiplication.

3. Distributed Storage: Reed-Solomon Error Correction#

In distributed cloud storage (Google Colossus, Apache Hadoop HDFS, Ceph), data loss across storage drives is mitigated using Reed-Solomon erasure codes. A data block of k shards is encoded into n shards (n > k) by multiplying by an n×k generator matrix G:

Encoded_Data = G · Data

The encoded data forms a k-dimensional subspace within R^n. If up to (n - k) drives fail, the surviving shards correspond to a k×k submatrix of G. Because G is constructed such that every k×k submatrix is linearly independent (using Vandermonde or Cauchy matrices), the submatrix is invertible, allowing exact data reconstruction via matrix inversion.

7. Common Architectural Pitfalls & Misconceptions#

When designing numerical software or reasoning about linear systems, several subtle pitfalls frequently arise:

• Pitfall 1: Confusing Linear Independence with Orthogonality.

Two vectors can be linearly independent without being orthogonal. Orthogonality (inner product = 0) is a much stronger condition. Any orthogonal set of non-zero vectors is guaranteed to be linearly independent, but the converse is not true. For example, v1 = (1, 0) and v2 = (1, 0.001) are linearly independent (spanning R^2), but they are nearly parallel and severely non-orthogonal.

• Pitfall 2: Assuming Elementary Row Operations Preserve the Column Space.

Performing Gaussian elimination on matrix A yields an echelon matrix U. The row operations preserve the row space (Row(A) = Row(U)) and the null space (Null(A) = Null(U)). However, they do NOT preserve the column space! Col(A) ≠ Col(U). To construct a basis for Col(A), one must identify the pivot columns in U and extract the corresponding original columns from matrix A.

• Pitfall 3: Misapplying the Rank-Nullity Theorem to the Row Count.

Engineers frequently write rank(A) + nullity(A) = m (row count) instead of n (column count). The domain of the linear map A : R^n → R^m is n-dimensional; nullity measures the dimensions lost from the domain. Always remember: Rank + Nullity = Columns.

• Pitfall 4: Believing that det(A) = 0 Implies A Maps All Vectors to Zero.

A zero determinant indicates that A is singular and collapses at least one dimension. Non-zero vectors residing in the null space map to zero, but vectors in the row space map to non-zero vectors in the column space.

8. Summary & Checkpoint Exercises#

To solidify your mastery of vector space axioms, linear independence, bases, and subspace dimensionality, solve the following four checkpoint problems. Each problem includes a complete step-by-step mathematical solution.

Problem 1 (Subspace Verification Test):

Determine whether the subset W ⊆ R^3 defined by W = { (x, y, z) in R^3 : 2x - 3y + z = 0 } forms a linear subspace of R^3. Contrast this with the subset U = { (x, y, z) in R^3 : 2x - 3y + z = 4 }.

Solution:

We apply the three-step subspace verification test to W:

1. Zero Vector: Substitute (0, 0, 0): 2(0) - 3(0) + 0 = 0 - 0 + 0 = 0. The zero vector belongs to W.
2. Additive Closure: Let u = (x1, y1, z1) and v = (x2, y2, z2) both belong to W, meaning 2x1 - 3y1 + z1 = 0 and 2x2 - 3y2 + z2 = 0.

Consider their sum u + v = (x1 + x2, y1 + y2, z1 + z2).

Evaluate the linear condition:

2(x1 + x2) - 3(y1 + y2) + (z1 + z2) = (2x1 - 3y1 + z1) + (2x2 - 3y2 + z2) = 0 + 0 = 0.

Hence, u + v belongs to W.

1. Multiplicative Closure: Let c in R and v = (x, y, z) in W.

c · v = (cx, cy, cz).

Evaluate the condition:

2(cx) - 3(cy) + cz = c · (2x - 3y + z) = c · 0 = 0.

Hence, c · v belongs to W.

All three conditions are satisfied. W is a 2-dimensional plane passing through the origin and constitutes a valid linear subspace of R^3.

Now examine U where 2x - 3y + z = 4:

Test the zero vector (0, 0, 0): 2(0) - 3(0) + 0 = 0 ≠ 4.

The zero vector does not belong to U. Therefore, U fails the zero vector test immediately and is NOT a subspace of R^3 (it is an affine plane offset from the origin).

Problem 2 (Linear Independence via Row Echelon Form):

Determine whether the set of vectors S = { v1, v2, v3 } in R^3 is linearly independent, where:

v1 = [ 1, 2, 1 ]^T, v2 = [ 2, 5, 4 ]^T, v3 = [ 1, 4, 5 ]^T.

Solution:

Assemble the vectors as the columns of matrix A:

A = [

[ 1, 2, 1 ],

[ 2, 5, 4 ],

[ 1, 4, 5 ]

]

Apply Gaussian elimination to reduce A to row echelon form:

• Operation 1: Row2 <- Row2 - 2 · Row1

Row2 becomes: [ 2 - 2(1), 5 - 2(2), 4 - 2(1) ] = [ 0, 1, 2 ]

• Operation 2: Row3 <- Row3 - 1 · Row1

Row3 becomes: [ 1 - 1(1), 4 - 1(2), 5 - 1(1) ] = [ 0, 2, 4 ]

Matrix after first column elimination:

[

[ 1, 2, 1 ],

[ 0, 1, 2 ],

[ 0, 2, 4 ]

]

• Operation 3: Row3 <- Row3 - 2 · Row2

Row3 becomes: [ 0 - 2(0), 2 - 2(1), 4 - 2(2) ] = [ 0, 0, 0 ]

The resulting row echelon form is:

U = [

[ 1, 2, 1 ],

[ 0, 1, 2 ],

[ 0, 0, 0 ]

]

Analysis:

There are only 2 pivot columns (Column 1 and Column 2). Column 3 is a free column without a pivot.

Therefore, rank(A) = 2 < 3.

The homogeneous system A · c = 0 has non-trivial solutions.

To find the dependency relation, solve U · c = 0 with free variable c3 = t:

From Row 2: c2 + 2 · c3 = 0 => c2 = -2t

From Row 1: c1 + 2 · c2 + c3 = 0 => c1 + 2(-2t) + t = 0 => c1 - 3t = 0 => c1 = 3t

Setting t = 1 yields: c1 = 3, c2 = -2, c3 = 1.

Verification:

3 · v1 - 2 · v2 + 1 · v3 = 3[ 1, 2, 1 ]^T - 2[ 2, 5, 4 ]^T + [ 1, 4, 5 ]^T

= [ 3 - 4 + 1, 6 - 10 + 4, 3 - 8 + 5 ]^T = [ 0, 0, 0 ]^T.

Conclusion: The set S is linearly dependent. The vector v3 can be expressed as: v3 = -3 · v1 + 2 · v2.

Problem 3 (Determining Bases for the Four Fundamental Subspaces):

Given the 3×4 matrix A:

A = [

[ 1, 2, 0, 1 ],

[ 2, 4, 1, 4 ],

[ 3, 6, 1, 5 ]

]

Find the dimension and an explicit basis for each of the four fundamental subspaces: Col(A), Null(A), Row(A), and Null(A^T). Verify the Rank-Nullity Theorem.

Solution:

Step 1: Reduce A to Reduced Row Echelon Form (RREF).

• Row2 <- Row2 - 2 · Row1:

[ 2 - 2, 4 - 4, 1 - 0, 4 - 2 ] = [ 0, 0, 1, 2 ]

• Row3 <- Row3 - 3 · Row1:

[ 3 - 3, 6 - 6, 1 - 0, 5 - 3 ] = [ 0, 0, 1, 2 ]

Matrix becomes:

[

[ 1, 2, 0, 1 ],

[ 0, 0, 1, 2 ],

[ 0, 0, 1, 2 ]

]

• Row3 <- Row3 - Row2:

[ 0, 0, 0, 0 ]

The matrix is now in RREF:

R = [

[ 1, 2, 0, 1 ],

[ 0, 0, 1, 2 ],

[ 0, 0, 0, 0 ]

]

Step 2: Analysis of Pivots.

• Pivot columns: Column 1 and Column 3.
• Free variable columns: Column 2 and Column 4.
• Rank r = 2.
• Number of columns n = 4, number of rows m = 3.

Step 3: Basis for Column Space Col(A).

The basis for Col(A) consists of the original columns of matrix A corresponding to the pivot columns (columns 1 and 3):

Basis for Col(A) = { [ 1, 2, 3 ]^T, [ 0, 1, 1 ]^T }.

Dimension: dim(Col(A)) = r = 2.

Step 4: Basis for Row Space Row(A).

The basis for Row(A) consists of the non-zero rows of the echelon matrix R:

Basis for Row(A) = { [ 1, 2, 0, 1 ]^T, [ 0, 0, 1, 2 ]^T }.

Dimension: dim(Row(A)) = r = 2.

Step 5: Basis for Null Space Null(A).

Solve R · x = 0 where x = [x1, x2, x3, x4]^T with free variables x2 = s and x4 = t:

From row 1: x1 + 2s + t = 0 => x1 = -2s - t

From row 2: x3 + 2t = 0 => x3 = -2t

General solution vector:

x = [ -2s - t, s, -2t, t ]^T = s · [ -2, 1, 0, 0 ]^T + t · [ -1, 0, -2, 1 ]^T

Basis for Null(A) = { [ -2, 1, 0, 0 ]^T, [ -1, 0, -2, 1 ]^T }.

Dimension: dim(Null(A)) = n - r = 4 - 2 = 2.

Step 6: Basis for Left Null Space Null(A^T).

Solve A^T · y = 0 where y = [y1, y2, y3]^T:

A^T = [

[ 1, 2, 3 ],

[ 2, 4, 6 ],

[ 0, 1, 1 ],

[ 1, 4, 5 ]

]

Reduce A^T to row echelon form:

• Row2 <- Row2 - 2 · Row1: [ 0, 0, 0 ]
• Row4 <- Row4 - Row1: [ 0, 2, 2 ]

Swap rows to order:

Row 1: [ 1, 2, 3 ]

Row 2: [ 0, 1, 1 ]

Row 3: [ 0, 2, 2 ] -> Row3 - 2·Row2 = [ 0, 0, 0 ]

Row 4: [ 0, 0, 0 ]

From Row 2: y2 + y3 = 0 => y2 = -y3

From Row 1: y1 + 2y2 + 3y3 = 0 => y1 + 2(-y3) + 3y3 = 0 => y1 + y3 = 0 => y1 = -y3

Let y3 = 1: y = [ -1, -1, 1 ]^T.

Basis for Null(A^T) = { [ -1, -1, 1 ]^T }.

Dimension: dim(Null(A^T)) = m - r = 3 - 2 = 1.

Step 7: Verification of Rank-Nullity and Orthogonality.

• Rank-Nullity: dim(Col(A)) + dim(Null(A)) = 2 + 2 = 4 = n (Columns of A).
• Orthogonality Row(A) ⊥ Null(A):

Dot product 1: [ 1, 2, 0, 1 ] · [ -2, 1, 0, 0 ] = -2 + 2 + 0 + 0 = 0.

Dot product 2: [ 1, 2, 0, 1 ] · [ -1, 0, -2, 1 ] = -1 + 0 + 0 + 1 = 0.

Dot product 3: [ 0, 0, 1, 2 ] · [ -2, 1, 0, 0 ] = 0 + 0 + 0 + 0 = 0.

Dot product 4: [ 0, 0, 1, 2 ] · [ -1, 0, -2, 1 ] = 0 + 0 - 2 + 2 = 0.

• Orthogonality Col(A) ⊥ Null(A^T):

Dot product 1: [ 1, 2, 3 ] · [ -1, -1, 1 ] = -1 - 2 + 3 = 0.

Dot product 2: [ 0, 1, 1 ] · [ -1, -1, 1 ] = 0 - 1 + 1 = 0.

All orthogonality and dimensional invariants hold verified.

Problem 4 (Change of Basis Matrix and Coordinate Transformation):

Consider two bases for R^2:

• Basis B = { b1, b2 } with b1 = [ 2, 1 ]^T and b2 = [ 1, 1 ]^T
• Basis C = { c1, c2 } with c1 = [ 1, 3 ]^T and c2 = [ 1, 2 ]^T

Given a vector v whose coordinates relative to basis B are [v]_B = [ 3, -2 ]^T:

1. Find the change of basis matrix P_(C <- B).
2. Compute the coordinates of v relative to basis C, [v]_C.
3. Verify that both coordinate representations reconstruct the identical Euclidean vector v in R^2.

Solution:

Step 1: Construct the Change of Basis Matrix P_(C <- B).

Recall the relation: P_(C <- B) = (P_(E <- C))^(-1) · P_(E <- B).

The standard transition matrices are composed of basis vectors as columns:

P_(E <- B) = [

[ 2, 1 ],

[ 1, 1 ]

]

P_(E <- C) = [

[ 1, 1 ],

[ 3, 2 ]

]

Calculate (P_(E <- C))^(-1):

det(P_(E <- C)) = (1)(2) - (1)(3) = 2 - 3 = -1.

The inverse of a 2×2 matrix [ [a, b], [c, d] ] is (1/det) · [ [d, -b], [-c, a] ]:

(P_(E <- C))^(-1) = (1 / -1) · [

[ 2, -1 ],

[ -3, 1 ]

] = [

[ -2, 1 ],

[ 3, -1 ]

]

Multiply (P_(E <- C))^(-1) by P_(E <- B):

P_(C <- B) = [

[ -2, 1 ],

[ 3, -1 ]

] · [

[ 2, 1 ],

[ 1, 1 ]

]

Compute entries:

• Entry (1, 1): (-2)(2) + (1)(1) = -4 + 1 = -3
• Entry (1, 2): (-2)(1) + (1)(1) = -2 + 1 = -1
• Entry (2, 1): (3)(2) + (-1)(1) = 6 - 1 = 5
• Entry (2, 2): (3)(1) + (-1)(1) = 3 - 1 = 2

Thus:

P_(C <- B) = [

[ -3, -1 ],

[ 5, 2 ]

]

Step 2: Compute [v]_C.

[v]_C = P_(C <- B) · [v]_B

[v]_C = [

[ -3, -1 ],

[ 5, 2 ]

] · [

3,

-2

]

Row 1: (-3)(3) + (-1)(-2) = -9 + 2 = -7

Row 2: (5)(3) + (2)(-2) = 15 - 4 = 11

So: [v]_C = [ -7, 11 ]^T.

Step 3: Verification of Vector Reconstruction in R^2.

• Reconstruction via Basis B:

v = 3 · b1 + (-2) · b2 = 3[ 2, 1 ]^T - 2[ 1, 1 ]^T = [ 6 - 2, 3 - 2 ]^T = [ 4, 1 ]^T.

• Reconstruction via Basis C:

v = -7 · c1 + 11 · c2 = -7[ 1, 3 ]^T + 11[ 1, 2 ]^T = [ -7 + 11, -21 + 22 ]^T = [ 4, 1 ]^T.

Both basis expansions resolve to the exact identical vector [ 4, 1 ]^T in R^2.