Chapter 3 • Theory & Derivations

Unit 3: Linear Combinations, Span, Linear Independence & Basis Dimension

Linear span, linear independence, the Linear Dependence Lemma, Wronskian determinant test, basis as minimal spanning set and maximal independent set, coordinate isomorphism, Steinitz Exchange Lemma, dimension invariance, and the Dimension Formula for Subspace Sums.

§3.1 Linear Combinations, Span & the Linear Dependence Lemma

1. Linear Combinations and Linear Span

Let $V$ be a vector space over a field $F$. A vector $\vec{v} \in V$ is a linear combination of a non-empty set of vectors $S = \{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_k\} \subseteq V$ if there exist scalars $c_1, c_2, \dots, c_k \in F$ such that:

\vec{v} = c_1 \vec{v}_1 + c_2 \vec{v}_2 + \cdots + c_k \vec{v}_k = \sum_{i=1}^k c_i \vec{v}_i

The Linear Span of $S$, denoted by $\text{span}(S)$ or $\text{span}\{\vec{v}_1, \dots, \vec{v}_k\}$, is the set of all linear combinations of vectors in $S$:

\text{span}(S) = \left\{ \sum_{i=1}^k c_i \vec{v}_i : c_i \in F, \; k \in \mathbb{N}, \; \vec{v}_i \in S \right\}

By convention, the span of the empty set is the zero subspace: $\text{span}(\emptyset) = \{\vec{0}\}$.

Theorem 3.1 (Span is the Minimal Subspace):

Let $S$ be a non-empty subset of $V$. Then:

$\text{span}(S)$ is a subspace of $V$.
$\text{span}(S)$ is the smallest subspace of $V$ containing $S$, in the sense that if $W$ is any subspace of $V$ containing $S$, then $\text{span}(S) \subseteq W$.

Proof:

$\vec{0} = 0\vec{v}_1 \in \text{span}(S)$. If $\vec{u} = \sum a_i \vec{v}_i$ and $\vec{w} = \sum b_i \vec{v}_i$ are in $\text{span}(S)$, then for any $\alpha, \beta \in F$, $\alpha\vec{u} + \beta\vec{w} = \sum (\alpha a_i + \beta b_i)\vec{v}_i \in \text{span}(S)$. Thus $\text{span}(S) \le V$.
If $W \le V$ and $S \subseteq W$, then by closure of $W$ under addition and scalar multiplication, every linear combination $\sum c_i \vec{v}_i$ must belong to $W$. Hence $\text{span}(S) \subseteq W$. $\blacksquare$

2. Linear Independence and Dependence

A set of vectors $S = \{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_k\} \subseteq V$ is Linearly Independent if the vector equation:

c_1 \vec{v}_1 + c_2 \vec{v}_2 + \cdots + c_k \vec{v}_k = \vec{0}

has only the trivial solution $c_1 = c_2 = \cdots = c_k = 0$.

If there exist scalars $c_1, \dots, c_k \in F$, not all zero, such that $\sum_{i=1}^k c_i \vec{v}_i = \vec{0}$, then the set $S$ is Linearly Dependent.

Theorem 3.2 (The Linear Dependence Lemma):

A set $S = \{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_k\}$ ($k \ge 2$) with $\vec{v}_1 \ne \vec{0}$ is linearly dependent if and only if there exists an index $j \in \{2, 3, \dots, k\}$ such that $\vec{v}_j$ is a linear combination of the preceding vectors:

\vec{v}_j \in \text{span}\{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_{j-1}\}

Furthermore, removing $\vec{v}_j$ does not alter the span: $\text{span}(S \setminus \{\vec{v}_j\}) = \text{span}(S)$.

3. The Wronskian Determinant in Function Spaces

In the function space $C^{n-1}[a, b]$, let $f_1(x), f_2(x), \dots, f_n(x)$ be $(n-1)$-times continuously differentiable functions. The Wronskian is defined by the functional determinant:

W(f_1, \dots, f_n)(x) = \det \begin{pmatrix} f_1(x) & f_2(x) & \cdots & f_n(x) \\ f_1'(x) & f_2'(x) & \cdots & f_n'(x) \\ \vdots & \vdots & \ddots & \vdots \\ f_1^{(n-1)}(x) & f_2^{(n-1)}(x) & \cdots & f_n^{(n-1)}(x) \end{pmatrix}

Theorem 3.3 (Wronskian Test for Linear Independence):

If there exists at least one point $x_0 \in [a, b]$ such that $W(f_1, \dots, f_n)(x_0) \ne 0$, then the functions $\{f_1, f_2, \dots, f_n\}$ are linearly independent on $[a, b]$.

§3.2 Bases of Vector Spaces & Unique Coordinate Representation

1. Definition and Characterization of a Basis

A subset $\mathcal{B} = \{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_n\} \subseteq V$ is a Basis of vector space $V$ if:

$\mathcal{B}$ is linearly independent.
$\mathcal{B}$ spans $V$: $\text{span}(\mathcal{B}) = V$.

Equivalent Formulations:

$\mathcal{B}$ is a minimal spanning set of $V$: Spans $V$, but removing any vector reduces the span.
$\mathcal{B}$ is a maximal linearly independent set in $V$: Linearly independent, but adding any vector produces a linearly dependent set.

2. The Unique Representation Theorem

Theorem 3.4 (Unique Representation Theorem):

Let $\mathcal{B} = \{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_n\}$ be a basis for $V$. Then for every vector $\vec{v} \in V$, there exists one and only one ordered $n$-tuple of scalars $(c_1, c_2, \dots, c_n) \in F^n$ such that:

\vec{v} = c_1 \vec{v}_1 + c_2 \vec{v}_2 + \cdots + c_n \vec{v}_n

Proof: Since $\mathcal{B}$ spans $V$, at least one representation exists. To prove uniqueness, suppose there exist two representations:

\vec{v} = \sum_{i=1}^n c_i \vec{v}_i = \sum_{i=1}^n d_i \vec{v}_i

Subtracting the two equations gives:

\sum_{i=1}^n (c_i - d_i)\vec{v}_i = \vec{0}

Since $\mathcal{B}$ is linearly independent, the coefficients must all be zero:

c_i - d_i = 0 \implies c_i = d_i \quad \forall i = 1, 2, \dots, n

Thus the representation is strictly unique. $\blacksquare$

Coordinate Vectors and Isomorphism:

The unique scalars $c_1, \dots, c_n$ are called the coordinates of $\vec{v}$ relative to the ordered basis $\mathcal{B}$, denoted by:

[\vec{v}]_{\mathcal{B}} = \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} \in F^n

The mapping $[\cdot]_{\mathcal{B}} : V \to F^n$ is a vector space isomorphism (bijective linear map), preserving all vector space operations:

$[\vec{u} + \vec{v}]_{\mathcal{B}} = [\vec{u}]_{\mathcal{B}} + [\vec{v}]_{\mathcal{B}}$
$[c\vec{v}]_{\mathcal{B}} = c[\vec{v}]_{\mathcal{B}}$

§3.3 Dimension Theory, Steinitz Exchange & Subspace Sums

1. The Steinitz Exchange Lemma

Theorem 3.5 (Steinitz Exchange Lemma):

Let $V$ be a vector space. Suppose $S = \{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_m\}$ is a linearly independent subset of $V$, and $G = \{\vec{w}_1, \vec{w}_2, \dots, \vec{w}_n\}$ spans $V$. Then:

$m \le n$ (no linearly independent set can have more elements than a spanning set).
There exists a subset of $n - m$ vectors from $G$ which, together with $S$, spans $V$.

Proof by Induction on $m$:

For $m = 1$: Since $\vec{v}_1 \ne \vec{0}$ and $G$ spans $V$, $\vec{v}_1 = \sum_{i=1}^n c_i \vec{w}_i$ with at least one $c_k \ne 0$. We can replace $\vec{w}_k$ with $\vec{v}_1$, and the new set $\{\vec{v}_1\} \cup (G \setminus \{\vec{w}_k\})$ still spans $V$.
By induction, if $k$ vectors $\vec{v}_1, \dots, \vec{v}_k$ have replaced $k$ vectors in $G$, the remaining set spans $V$. If $m > n$, we could replace all $n$ vectors in $G$, leaving $\{\vec{v}_1, \dots, \vec{v}_n\}$ spanning $V$. Then $\vec{v}_{n+1}$ would be a linear combination of $\{\vec{v}_1, \dots, \vec{v}_n\}$, violating the linear independence of $S$. Therefore, $m \le n$. $\blacksquare$

2. Invariance of Basis Cardinality and Definition of Dimension

Theorem 3.6 (Dimension Invariance Theorem):

If $\mathcal{B}_1$ and $\mathcal{B}_2$ are two bases of a finite-dimensional vector space $V$, then they contain the exact same number of elements:

|\mathcal{B}_1| = |\mathcal{B}_2|

Proof: Since $\mathcal{B}_1$ is linearly independent and $\mathcal{B}_2$ spans $V$, by Steinitz Exchange Lemma, $|\mathcal{B}_1| \le |\mathcal{B}_2|$. Reversing roles, since $\mathcal{B}_2$ is linearly independent and $\mathcal{B}_1$ spans $V$, $|\mathcal{B}_2| \le |\mathcal{B}_1|$. Therefore, $|\mathcal{B}_1| = |\mathcal{B}_2|$. $\blacksquare$

Definition of Dimension: The Dimension of a finite-dimensional vector space $V$, denoted by $\dim(V)$ or $\dim_F(V)$, is the number of vectors in any basis of $V$. If $V = \{\vec{0}\}$, $\dim(V) = 0$.

3. The Dimension Theorem for Subspace Sums

Theorem 3.7 (Grassmann's Subspace Sum Dimension Formula):

Let $W_1$ and $W_2$ be finite-dimensional subspaces of a vector space $V$. Then:

\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2)

Proof: Let $k = \dim(W_1 \cap W_2)$. Choose a basis $\mathcal{B}_0 = \{\vec{u}_1, \vec{u}_2, \dots, \vec{u}_k\}$ for $W_1 \cap W_2$.

Since $W_1 \cap W_2 \le W_1$, extend $\mathcal{B}_0$ to a basis of $W_1$:

\mathcal{B}_1 = \{\vec{u}_1, \dots, \vec{u}_k, \; \vec{v}_1, \dots, \vec{v}_r\} \implies \dim(W_1) = k + r

Since $W_1 \cap W_2 \le W_2$, extend $\mathcal{B}_0$ to a basis of $W_2$:

\mathcal{B}_2 = \{\vec{u}_1, \dots, \vec{u}_k, \; \vec{w}_1, \dots, \vec{w}_s\} \implies \dim(W_2) = k + s

We claim that $\mathcal{B} = \{\vec{u}_1, \dots, \vec{u}_k, \; \vec{v}_1, \dots, \vec{v}_r, \; \vec{w}_1, \dots, \vec{w}_s\}$ is a basis for $W_1 + W_2$.

1. Spans $W_1 + W_2$: Any vector $\vec{x} \in W_1 + W_2$ is $\vec{x} = \vec{x}_1 + \vec{x}_2$ with $\vec{x}_1 \in W_1$ and $\vec{x}_2 \in W_2$. Since $\vec{x}_1 \in \text{span}(\mathcal{B}_1)$ and $\vec{x}_2 \in \text{span}(\mathcal{B}_2)$, their sum is in $\text{span}(\mathcal{B})$.

2. Linear Independence: Suppose:

\sum_{i=1}^k a_i \vec{u}_i + \sum_{j=1}^r b_j \vec{v}_j + \sum_{l=1}^s c_l \vec{w}_l = \vec{0}

Rearrange to isolate the $\vec{w}$ terms:

\sum_{l=1}^s c_l \vec{w}_l = -\sum_{i=1}^k a_i \vec{u}_i - \sum_{j=1}^r b_j \vec{v}_j

The left-hand side is in $W_2$, while the right-hand side is in $W_1$. Therefore, the vector $\vec{y} = \sum_{l=1}^s c_l \vec{w}_l$ lies in $W_1 \cap W_2$. Since $\mathcal{B}_0$ is a basis for $W_1 \cap W_2$, $\vec{y}$ can be written in terms of $\vec{u}_1, \dots, \vec{u}_k$:

\sum_{l=1}^s c_l \vec{w}_l = \sum_{i=1}^k d_i \vec{u}_i \implies \sum_{i=1}^k d_i \vec{u}_i - \sum_{l=1}^s c_l \vec{w}_l = \vec{0}

Because $\mathcal{B}_2 = \{\vec{u}_1, \dots, \vec{u}_k, \vec{w}_1, \dots, \vec{w}_s\}$ is a basis of $W_2$, it is linearly independent, so all $c_l = 0$ and $d_i = 0$. Substitute $c_l = 0$ back into the original equation:

\sum_{i=1}^k a_i \vec{u}_i + \sum_{j=1}^r b_j \vec{v}_j = \vec{0}

Since $\mathcal{B}_1$ is a basis for $W_1$, it is linearly independent, so all $a_i = 0$ and $b_j = 0$. Therefore, all coefficients are zero, proving $\mathcal{B}$ is linearly independent!

Thus $\dim(W_1 + W_2) = |\mathcal{B}| = k + r + s$. Computing the right side:

\dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2) = (k + r) + (k + s) - k = k + r + s = \dim(W_1 + W_2)

The proof is complete. $\blacksquare$

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Example 3.1: Linear Independence & Basis Verification in R^4

Consider the following set of vectors in $\mathbb{R}^4$:

\vec{v}_1 = \begin{pmatrix} 1 \\ 2 \\ -1 \\ 3 \end{pmatrix}, \quad \vec{v}_2 = \begin{pmatrix} 2 \\ 5 \\ -1 \\ 8 \end{pmatrix}, \quad \vec{v}_3 = \begin{pmatrix} 1 \\ 1 \\ -2 \\ 1 \end{pmatrix}, \quad \vec{v}_4 = \begin{pmatrix} 3 \\ 7 \\ -2 \\ 11 \end{pmatrix}

(a) Determine whether the set $S = \{\vec{v}_1, \vec{v}_2, \vec{v}_3, \vec{v}_4\}$ is linearly independent. (b) Find the dimension of $W = \text{span}(S)$ and extract a basis for $W$ from the set $S$. (c) Express any redundant vectors as explicit linear combinations of the chosen basis vectors.

Step 1: Form the matrix $A$ with vectors as columns and row reduce:

A = \begin{pmatrix} 1 & 2 & 1 & 3 \\ 2 & 5 & 1 & 7 \\ -1 & -1 & -2 & -2 \\ 3 & 8 & 1 & 11 \end{pmatrix}

Apply elementary row operations:

$R_2 \to R_2 - 2R_1$: $(0, \; 5 - 4 = 1, \; 1 - 2 = -1, \; 7 - 6 = 1)$
$R_3 \to R_3 + R_1$: $(0, \; -1 + 2 = 1, \; -2 + 1 = -1, \; -2 + 3 = 1)$
$R_4 \to R_4 - 3R_1$: $(0, \; 8 - 6 = 2, \; 1 - 3 = -2, \; 11 - 9 = 2)$

Matrix after Column 1 clearance:

\begin{pmatrix} 1 & 2 & 1 & 3 \\ 0 & 1 & -1 & 1 \\ 0 & 1 & -1 & 1 \\ 0 & 2 & -2 & 2 \end{pmatrix}

Now clear Column 2:

$R_1 \to R_1 - 2R_2$: $(1, 0, \; 1 - 2(-1) = 3, \; 3 - 2(1) = 1)$
$R_3 \to R_3 - R_2$: $(0, 0, 0, 0)$
$R_4 \to R_4 - 2R_2$: $(0, 0, 0, 0)$

The Reduced Row Echelon Form is:

\text{rref}(A) = \begin{pmatrix} 1 & 0 & 3 & 1 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}

Step 2: Linear Independence and Basis Determination:

Pivot columns are Columns 1 and 2.
Non-pivot (free) columns are Columns 3 and 4.

Since non-pivot columns exist, $S$ is linearly dependent.

The dimension of $W = \text{span}(S)$ is $\text{rank}(A) = 2$.
A basis for $W$ consists of the original vectors corresponding to the pivot columns:

\mathcal{B} = \{\vec{v}_1, \vec{v}_2\}

Step 3: Linear Combinations for Redundant Vectors: From the RREF:

Column 3: $\begin{pmatrix} 3 \\ -1 \\ 0 \\ 0 \end{pmatrix} \implies \vec{v}_3 = 3\vec{v}_1 - \vec{v}_2$

Check: $3(1, 2, -1, 3)^T - (2, 5, -1, 8)^T = (1, 1, -2, 1)^T = \vec{v}_3$. Match!

Column 4: $\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} \implies \vec{v}_4 = \vec{v}_1 + \vec{v}_2$

Check: $(1, 2, -1, 3)^T + (2, 5, -1, 8)^T = (3, 7, -2, 11)^T = \vec{v}_4$. Match!

Final Answer & Physical Insight

$S$ is linearly dependent. $\dim(\text{span}(S)) = 2$. A basis for $W$ is $\{\vec{v}_1, \vec{v}_2\}$. The redundant vectors satisfy $\vec{v}_3 = 3\vec{v}_1 - \vec{v}_2$ and $\vec{v}_4 = \vec{v}_1 + \vec{v}_2$.

Tier 2: Intermediate Exam Example 3.2: Basis Construction & Dimension of Constrained Polynomial Subspaces

Let $\mathbb{P}_3(\mathbb{R}) = \{a_0 + a_1 t + a_2 t^2 + a_3 t^3 : a_i \in \mathbb{R}\}$ be the 4-dimensional vector space of polynomials of degree at most 3 over $\mathbb{R}$. Consider the subset:

W = \{p(t) \in \mathbb{P}_3(\mathbb{R}) : p(1) = 0 \quad \text{and} \quad p'(0) = 0\}

(a) Prove that $W$ is a subspace of $\mathbb{P}_3(\mathbb{R})$. (b) Find the dimension of $W$. (c) Construct an explicit basis $\mathcal{B}_W$ for $W$. (d) Extend the basis $\mathcal{B}_W$ to a full basis of $\mathbb{P}_3(\mathbb{R})$.

Step 1: Subspace Verification: Let $p(t), q(t) \in W$ and $c, d \in \mathbb{R}$.

$(c p + d q)(1) = c p(1) + d q(1) = c(0) + d(0) = 0$.
$(c p + d q)'(0) = c p'(0) + d q'(0) = c(0) + d(0) = 0$.
The zero polynomial satisfies $\mathbf{0}(1) = 0$ and $\mathbf{0}'(0) = 0$.

By the Subspace Criterion, $W$ is a subspace of $\mathbb{P}_3(\mathbb{R})$.

Step 2: Translate constraints into a linear system: Let $p(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3$. Then $p'(t) = a_1 + 2 a_2 t + 3 a_3 t^2$.

Constraint 1: $p'(0) = a_1 = 0$.
Constraint 2: $p(1) = a_0 + a_1 + a_2 + a_3 = 0$.

Substitute $a_1 = 0$:

a_0 + a_2 + a_3 = 0 \implies a_0 = -a_2 - a_3

Step 3: Parametric representation and basis construction: The coefficients $a_2$ and $a_3$ are free variables:

$$p(t) = (-a_2 - a_3) + 0 t + a_2 t^2 + a_3 t^3 = a_2(t^2 - 1) + a_3(t^3 - 1)$$

Define:

p_1(t) = t^2 - 1, \qquad p_2(t) = t^3 - 1

Every $p \in W$ is a linear combination of $p_1(t)$ and $p_2(t)$. Furthermore, neither polynomial is a scalar multiple of the other, so $\{p_1, p_2\}$ is linearly independent. Thus:

\mathcal{B}_W = \{t^2 - 1, \; t^3 - 1\}, \qquad \dim(W) = 2

Step 4: Extension to full basis of $\mathbb{P}_3(\mathbb{R})$: Since $\dim(\mathbb{P}_3) = 4$, we need to append 2 vectors from the standard basis $\{1, t, t^2, t^3\}$ that are linearly independent of $\mathcal{B}_W$. Consider $q_1(t) = 1$ and $q_2(t) = t$. Evaluate coordinate vectors with respect to the ordered basis $\{1, t, t^2, t^3\}$:

[p_1] = \begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \quad [p_2] = \begin{pmatrix} -1 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \quad [q_1] = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \quad [q_2] = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}

Form the matrix:

M = \begin{pmatrix} -1 & -1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}

Expanding determinant: $\det(M) = -1 \cdot (1 \cdot 1) = -1 \ne 0$. Thus the set $\{t^2 - 1, \; t^3 - 1, \; 1, \; t\}$ is linearly independent and forms a complete basis for $\mathbb{P}_3(\mathbb{R})$.

Final Answer & Physical Insight

$\dim(W) = 2$. An explicit basis for $W$ is $\mathcal{B}_W = \{t^2 - 1, \; t^3 - 1\}$. Extended basis for $\mathbb{P}_3(\mathbb{R})$: $\{t^2 - 1, \; t^3 - 1, \; 1, \; t\}$.

Tier 3: Honors / Proof Challenge Example 3.3: Complete Proof of the Dimension Theorem for Subspace Sums

Let $W_1$ and $W_2$ be two finite-dimensional subspaces of an arbitrary vector space $V$ over field $F$.

(a) Prove Grassmann's formula: $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2)$. (b) Deduce the direct sum dimension formula: If $V = W_1 \oplus W_2$, then $\dim(V) = \dim(W_1) + \dim(W_2)$. (c) In $\mathbb{R}^6$, let $W_1$ and $W_2$ be subspaces with $\dim(W_1) = 4$ and $\dim(W_2) = 5$. Determine all possible values for $\dim(W_1 \cap W_2)$ and $\dim(W_1 + W_2)$.

Part (a): Formal Proof of Grassmann's Formula: Let $k = \dim(W_1 \cap W_2)$, $d_1 = \dim(W_1)$, $d_2 = \dim(W_2)$. Let $\mathcal{U} = \{\vec{u}_1, \dots, \vec{u}_k\}$ be a basis for the subspace $W_1 \cap W_2$. By the Basis Extension Theorem:

Extend $\mathcal{U}$ to a basis of $W_1$: $\mathcal{B}_1 = \{\vec{u}_1, \dots, \vec{u}_k, \; \vec{v}_1, \dots, \vec{v}_{d_1 - k}\}$.
Extend $\mathcal{U}$ to a basis of $W_2$: $\mathcal{B}_2 = \{\vec{u}_1, \dots, \vec{u}_k, \; \vec{w}_1, \dots, \vec{w}_{d_2 - k}\}$.

Let $\mathcal{B} = \{\vec{u}_1, \dots, \vec{u}_k, \; \vec{v}_1, \dots, \vec{v}_{d_1 - k}, \; \vec{w}_1, \dots, \vec{w}_{d_2 - k}\}$.

1. $\text{span}(\mathcal{B}) = W_1 + W_2$:

Every $\vec{x} \in W_1 + W_2$ is $\vec{x}_1 + \vec{x}_2$ with $\vec{x}_1 \in W_1, \vec{x}_2 \in W_2$. Since $\vec{x}_1 \in \text{span}(\mathcal{B}_1) \subseteq \text{span}(\mathcal{B})$ and $\vec{x}_2 \in \text{span}(\mathcal{B}_2) \subseteq \text{span}(\mathcal{B})$, $\vec{x} \in \text{span}(\mathcal{B})$.

2. Linear Independence of $\mathcal{B}$:

Suppose:

\sum_{i=1}^k a_i \vec{u}_i + \sum_{j=1}^{d_1 - k} b_j \vec{v}_j + \sum_{l=1}^{d_2 - k} c_l \vec{w}_l = \vec{0}

Isolate the $\vec{w}_l$ terms:

\sum_{l=1}^{d_2 - k} c_l \vec{w}_l = -\sum_{i=1}^k a_i \vec{u}_i - \sum_{j=1}^{d_1 - k} b_j \vec{v}_j

The left-hand side belongs to $W_2$, while the right-hand side belongs to $W_1$. Therefore, $\vec{y} = \sum_{l=1}^{d_2 - k} c_l \vec{w}_l \in W_1 \cap W_2$. Since $\mathcal{U}$ is a basis for $W_1 \cap W_2$, there exist scalars $\lambda_1, \dots, \lambda_k$ such that:

\sum_{l=1}^{d_2 - k} c_l \vec{w}_l = \sum_{i=1}^k \lambda_i \vec{u}_i \implies \sum_{i=1}^k \lambda_i \vec{u}_i - \sum_{l=1}^{d_2 - k} c_l \vec{w}_l = \vec{0}

Since $\mathcal{B}_2 = \{\vec{u}_1, \dots, \vec{u}_k, \vec{w}_1, \dots, \vec{w}_{d_2 - k}\}$ is a basis for $W_2$, it is linearly independent. Hence all $c_l = 0$ (and $\lambda_i = 0$). Substituting $c_l = 0$ into the original dependence relation yields:

\sum_{i=1}^k a_i \vec{u}_i + \sum_{j=1}^{d_1 - k} b_j \vec{v}_j = \vec{0}

Since $\mathcal{B}_1$ is a basis for $W_1$, it is linearly independent, so all $a_i = 0$ and $b_j = 0$. Therefore, all coefficients are zero, proving $\mathcal{B}$ is linearly independent.

Thus $\mathcal{B}$ is a basis for $W_1 + W_2$, with cardinality:

|\mathcal{B}| = k + (d_1 - k) + (d_2 - k) = d_1 + d_2 - k = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2)

Hence $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2)$. $\blacksquare$

Part (b): Direct Sum Formula: If $V = W_1 \oplus W_2$, then $V = W_1 + W_2$ and $W_1 \cap W_2 = \{\vec{0}\}$. Thus $\dim(W_1 \cap W_2) = 0$, giving:

\dim(V) = \dim(W_1) + \dim(W_2) - 0 = \dim(W_1) + \dim(W_2)

Part (c): Dimension Bounds in $\mathbb{R}^6$: $W_1, W_2 \le \mathbb{R}^6$ with $\dim(W_1) = 4, \dim(W_2) = 5$. Since $W_1 + W_2 \le \mathbb{R}^6$, $\dim(W_1 + W_2) \le 6$. Also, $W_2 \subseteq W_1 + W_2 \implies \dim(W_1 + W_2) \ge \dim(W_2) = 5$. So $\dim(W_1 + W_2) \in \{5, 6\}$.

By Grassmann's formula:

\dim(W_1 \cap W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 + W_2) = 4 + 5 - \dim(W_1 + W_2) = 9 - \dim(W_1 + W_2)

If $\dim(W_1 + W_2) = 5$: $\dim(W_1 \cap W_2) = 9 - 5 = 4$ (here $W_1 \subseteq W_2$).
If $\dim(W_1 + W_2) = 6$: $\dim(W_1 \cap W_2) = 9 - 6 = 3$ (here $W_1 + W_2 = \mathbb{R}^6$).

Possible values: $(\dim(W_1 + W_2), \dim(W_1 \cap W_2)) \in \{(5, 4), (6, 3)\}$.

Final Answer & Physical Insight

Grassmann's formula: $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2)$. For direct sum $W_1 \oplus W_2$, $\dim(W_1 \oplus W_2) = \dim(W_1) + \dim(W_2)$. In $\mathbb{R}^6$ with dimensions 4 and 5: either $\dim(W_1 + W_2) = 6$ with $\dim(W_1 \cap W_2) = 3$, or $\dim(W_1 + W_2) = 5$ with $\dim(W_1 \cap W_2) = 4$.