Linear Independence, Span, Basis, and Dimension

A set of vectors $\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\} \subseteq V$ is linearly Independent if the equation

$\alpha_1 \mathbf{v}_1 + \alpha_2 \mathbf{v}_2 + \cdots + \alpha_k \mathbf{v}_k = \mathbf{0}$

Implies $\alpha_1 = \alpha_2 = \cdots = \alpha_k = 0$. Otherwise the set is linearly dependent.

Proposition 2.1 (Equivalent formulations). The following are equivalent for vectors $\mathbf{v}_1, \ldots, \mathbf{v}_k \in V$:

1. $\{\mathbf{v}_1, \ldots, \mathbf{v}_k\}$ is linearly independent.
2. No $\mathbf{v}_j$ can be written as a linear combination of the remaining vectors.
3. If $\sum_{i=1}^k \alpha_i \mathbf{v}_i = \sum_{i=1}^k \beta_i \mathbf{v}_i$Then $\alpha_i = \beta_i$ for all $i$.

Proof. (1 $\Rightarrow$ 2): If $\mathbf{v}_j = \sum_{i \neq j} \alpha_i \mathbf{v}_i$Then $\sum_{i \neq j} \alpha_i \mathbf{v}_i - \mathbf{v}_j = \mathbf{0}$ gives a non-trivial linear Dependence, contradicting (1).

(2 $\Rightarrow$ 3): If $\sum (\alpha_i - \beta_i)\mathbf{v}_i = \mathbf{0}$Then by linear Independence (which follows from (2)), $\alpha_i = \beta_i$ for all $i$.

(3 $\Rightarrow$ 1): If $\sum \alpha_i \mathbf{v}_i = \mathbf{0} = \sum 0 \cdot \mathbf{v}_i$ Then by (3), $\alpha_i = 0$ for all $i$. $\blacksquare$

The span of a set $S \subseteq V$Denoted $\mathrm{span}(S)$Is the set of all finite linear Combinations of elements of $S$:

$\mathrm{span}(S) = \left\{ \sum_{i=1}^k \alpha_i \mathbf{v}_i : k \in \mathbb{N},\, \alpha_i \in F,\, \mathbf{v}_i \in S \right\}$

Proposition 2.2. $\mathrm{span}(S)$ is always a subspace of $V$. In fact, $\mathrm{span}(S)$ is The smallest subspace containing $S$: if $W$ is any subspace with $S \subseteq W$Then $\mathrm{span}(S) \subseteq W$.

Proof. $\mathrm{span}(S)$ is non-empty since $\mathbf{0} = 0 \cdot \mathbf{v}$ for any $\mathbf{v} \in S$. Closure under addition and scalar multiplication follows directly from the Definition of linear combinations. For minimality, any subspace $W$ containing $S$ must contain all Finite linear combinations of elements of $S$ by Proposition 1.2, so $\mathrm{span}(S) \subseteq W$. $\blacksquare$

A set $B \subseteq V$ is a basis for $V$ if:

1. $B$ is linearly independent, and
2. $\mathrm{span}(B) = V$.

Theorem 2.1. Every vector space has a basis. All bases of a finite-dimensional vector space have The same number of elements.

The dimension of $V$Denoted $\dim(V)$Is the cardinality of any basis for $V$.

Lemma 2.3 (Steinitz Exchange Lemma). Let $\{\mathbf{u}_1, \ldots, \mathbf{u}_k\}$ be a linearly Independent set in $V$And let $\{\mathbf{w}_1, \ldots, \mathbf{w}_m\}$ be a spanning set for $V$. Then $k \leq m$And after relabelling the $\mathbf{w}_j$The set

$\{\mathbf{u}_1, \ldots, \mathbf{u}_k, \mathbf{w}_{k+1}, \ldots, \mathbf{w}_m\}$

Also spans $V$.

Proof. We proceed by induction on $k$. For $k = 0$ there is nothing to prove.

Assume the result holds for $k - 1$. Since $\{\mathbf{u}_1, \ldots, \mathbf{u}_k\}$ is linearly Independent, $\mathbf{u}_k \neq \mathbf{0}$ and $\mathbf{u}_k \in \mathrm{span}\{\mathbf{w}_1, \ldots, \mathbf{w}_m\}$ Since the $\mathbf{w}_j$ span $V$. Therefore $\mathbf{u}_k = \sum_{j=1}^m \alpha_j \mathbf{w}_j$ for some $\alpha_j \in F$And not all $\alpha_j$ are zero.

After relabelling, assume $\alpha_1 \neq 0$. Then $\mathbf{w}_1 = \alpha_1^{-1}(\mathbf{u}_k - \sum_{j=2}^m \alpha_j \mathbf{w}_j)$ So $\mathbf{w}_1 \in \mathrm{span}\{\mathbf{u}_k, \mathbf{w}_2, \ldots, \mathbf{w}_m\}$. It follows that

$\mathrm{span}\{\mathbf{w}_1, \ldots, \mathbf{w}_m\} = \mathrm{span}\{\mathbf{u}_k, \mathbf{w}_2, \ldots, \mathbf{w}_m\} = V$

Now $\{\mathbf{u}_1, \ldots, \mathbf{u}_{k-1}\}$ is linearly independent and $\{\mathbf{u}_k, \mathbf{w}_2, \ldots, \mathbf{w}_m\}$ spans $V$. By the inductive hypothesis, $k - 1 \leq m - 1$ (so $k \leq m$) and after relabelling, $\{\mathbf{u}_1, \ldots, \mathbf{u}_{k-1}, \mathbf{w}_k, \ldots, \mathbf{w}_m\}$ spans $V$. Since $\mathbf{u}_k$ is already in this span, the full set $\{\mathbf{u}_1, \ldots, \mathbf{u}_k, \mathbf{w}_{k+1}, \ldots, \mathbf{w}_m\}$ also spans $V$. $\blacksquare$

Theorem 2.4 (Dimension is Well-Defined). If $V$ is finite-dimensional, then any two bases of $V$ have the same number of elements.

Proof. Let $\mathcal{B}_1$ and $\mathcal{B}_2$ be two bases with $\lvert\mathcal{B}_1\rvert = k$ and $\lvert\mathcal{B}_2\rvert = m$. Applying the Steinitz exchange lemma with $\mathcal{B}_1$ as the Independent set and $\mathcal{B}_2$ as the spanning set gives $k \leq m$. Swapping roles gives $m \leq k$. Hence $k = m$. $\blacksquare$

Theorem 2.5 (Dimension Formula). If $U$ and $W$ are subspaces of a finite-dimensional vector Space $V$Then

$\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W)$

Theorem 2.6 (Rank-Nullity Theorem). Let $A \in \mathcal{M}_{m \times n}(F)$. Then

$\mathrm{rank}(A) + \mathrm{nullity}(A) = n$

Where $\mathrm{rank}(A) = \dim(\mathrm{col}(A))$ and $\mathrm{nullity}(A) = \dim(\mathrm{null}(A))$.

Proof. Let $\{\mathbf{v}_1, \ldots, \mathbf{v}_k\}$ be a basis for $\mathrm{null}(A)$Where $k = \mathrm{nullity}(A)$. Extend this to a basis $\{\mathbf{v}_1, \ldots, \mathbf{v}_k, \mathbf{v}_{k+1}, \ldots, \mathbf{v}_n\}$ for $F^n$.

We claim that $\{A\mathbf{v}_{k+1}, \ldots, A\mathbf{v}_n\}$ is a basis for $\mathrm{col}(A)$.

Spanning: For any $\mathbf{y} \in \mathrm{col}(A)$There exists $\mathbf{x} \in F^n$ With $\mathbf{y} = A\mathbf{x}$. Writing $\mathbf{x} = \sum_{i=1}^n \alpha_i \mathbf{v}_i$

$\mathbf{y} = A\left(\sum_{i=1}^n \alpha_i \mathbf{v}_i\right) = \sum_{i=1}^n \alpha_i A\mathbf{v}_i = \sum_{i=k+1}^n \alpha_i A\mathbf{v}_i$

Since $A\mathbf{v}_i = \mathbf{0}$ for $i \leq k$.

Linear independence: If $\sum_{i=k+1}^n \alpha_i A\mathbf{v}_i = \mathbf{0}$Then $A\left(\sum_{i=k+1}^n \alpha_i \mathbf{v}_i\right) = \mathbf{0}$So $\sum_{i=k+1}^n \alpha_i \mathbf{v}_i \in \mathrm{null}(A)$. Since $\{\mathbf{v}_1, \ldots, \mathbf{v}_k\}$ Is a basis for the null space, $\sum_{i=k+1}^n \alpha_i \mathbf{v}_i = \sum_{i=1}^k \beta_i \mathbf{v}_i$ For some $\beta_i$Giving $\sum_{i=1}^n (-\beta_i)\mathbf{v}_i + \sum_{i=k+1}^n \alpha_i \mathbf{v}_i = \mathbf{0}$. By linear independence of the full basis, $\alpha_i = 0$ for all $i \geq k + 1$.

Therefore $\mathrm{rank}(A) = n - k = n - \mathrm{nullity}(A)$. $\blacksquare$

Problem. Find a basis for and the dimension of the subspace $W = \mathrm{span}\{(1, 2, -1, 0), (3, 1, 0, 2), (-1, 3, -2, -2)\}$ of $\mathbb{R}^4$.

Problem. Find a basis for the null space of

$A = \begin{pmatrix} 1 & 2 & 1 & -1 \\ 2 & 4 & 0 & 1 \\ 0 & 0 & 1 & 3 \end{pmatrix}$

Problem. Determine whether the vectors $\mathbf{v}_1 = (1, 2, 3)$, $\mathbf{v}_2 = (4, 5, 6)$, $\mathbf{v}_3 = (7, 8, 9)$ form a basis For $\mathbb{R}^3$.

:::tip To check if $n$ vectors in $\mathbb{R}^n$ form a basis, compute the determinant of the matrix whose Columns are those vectors. If $\det \neq 0$They form a basis; if $\det = 0$They do not.

Problem. Let $V = \mathcal{P}_3(\mathbb{R})$ (polynomials of degree at most 3). Find the dimension Of the subspace $W = \{p \in \mathcal{P}_3 : p(1) = p(-1) = 0\}$.

• Linear independence of infinitely many vectors. The definition only directly applies to finite subsets. A set $S$ is linearly independent if every finite subset of $S$ is linearly independent.
• Dimension and spanning. A set of $n$ vectors in $\mathbb{R}^n$ that spans $\mathbb{R}^n$ must be linearly independent (and hence a basis). Similarly, $n$ linearly independent vectors in $\mathbb{R}^n$ must span $\mathbb{R}^n$.
• The empty set spans $\{\mathbf{0}\}$. The span of the empty set is the trivial subspace, and the empty set is a basis for $\{\mathbf{0}\}$. The dimension of the zero space is 0.

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