Vector Spaces, Subspaces, and Linear Independence
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You should be comfortable with basic vector arithmetic in \(\mathbb{R}^3\) and the concept of a scalar field. Familiarity with the idea of a "set" and basic logic (implication, negation) is assumed.
Why this matters
Linear independence is the algebraic definition of "new information" in a physical system.
In physics, we constantly decompose complex states into simpler components. Whether we are analyzing the normal modes of a vibrating string, the spin states of an electron, or the basis vectors of a coordinate system, we rely on the guarantee that these components are distinct and non-redundant. If a set of vectors is linearly dependent, one vector can be constructed from the others, meaning it carries no independent physical degree of freedom. This lecture establishes the rigorous criteria for independence and the structure of subspaces, which are the "rooms" within the larger vector space where physical states reside.
The Intuition of Independence
Imagine you are navigating a city using only two directions: North and East. These two directions are independent because you cannot reach a point by going only North if you need to go East. However, if you added a third direction, "North-East," this new direction is not independent; it is a combination of the first two. In vector space terms, the North-East vector is "linearly dependent" on North and East because it can be written as a sum of them. A set of vectors is linearly independent if and only if none of them can be written as a linear combination of the others. This is the fundamental test for whether a set of vectors provides a genuine basis for describing a physical state without redundancy. If your basis vectors are dependent, your description of the system is over-specified and mathematically unstable.
Formal Definition and the Trivial Solution
We define linear independence formally to avoid ambiguity. Let \(V\) be a vector space over a field \(\mathbb{F}\) (typically \(\mathbb{R}\) or \(\mathbb{C}\)). A set of vectors \(\{v_1, v_2, \dots, v_n\}\) is linearly independent if the only solution to the equation \(c_1 v_1 + c_2 v_2 + \dots + c_n v_n = \mathbf{0}\) is the trivial solution \(c_1 = c_2 = \dots = c_n = 0\). If there exists a non-trivial set of coefficients (where at least one \(c_i \neq 0\)) that satisfies this equation, the set is linearly dependent. This definition is crucial because it transforms a geometric question ("do these vectors point in the same direction?") into an algebraic one ("does this homogeneous system have non-zero solutions?"). In quantum mechanics, where states are vectors in a Hilbert space, the same test decides whether a set of states is genuinely independent. Independence is weaker than orthogonality: orthogonal non-zero vectors are always independent, but independent vectors need not be orthogonal.
Subspaces and Closure
A subspace \(W\) of a vector space \(V\) is a subset that is itself a vector space under the same operations. For \(W\) to be a subspace, it must satisfy three conditions: it must contain the zero vector, it must be closed under vector addition, and it must be closed under scalar multiplication. This means that if \(u, v \in W\), then \(u + v \in W\), and if \(c \in \mathbb{F}\) and \(u \in W\), then \(c u \in W\). Physically, subspaces represent constrained systems. For example, the set of all states of a particle confined to a plane through the origin in 3D space forms a 2D subspace of the full 3D configuration space. The linear independence of a set of vectors within a subspace determines the dimension of that subspace. If a subspace is spanned by \(k\) linearly independent vectors, its dimension is \(k\). This connects directly to the degrees of freedom available to the physical system within that constraint.
Dimension as an Invariant
Once we have established what it means for vectors to be independent, we can define the dimension of a vector space. The dimension \(\dim V\) is the maximum number of linearly independent vectors in \(V\). A critical theorem, which we will rely on heavily in later lectures, is the Invariance of Dimension. This theorem states that if a vector space has a finite basis, then all bases of that space have the same number of elements. This ensures that the dimension is a well-defined property of the space itself, not an artifact of how we choose to describe it. For instance, whether we describe a 3D space using Cartesian coordinates \((x, y, z)\) or spherical coordinates \((r, \theta, \phi)\), the dimension remains 3. This invariance is the bedrock of coordinate-free physics, allowing us to make statements about the system that are independent of the observer's frame of reference.
The derivation
Invariance of Dimension of a Vector Space
Let \(V\) be a vector space over a field \(\mathbb{F}\) that is spanned by a finite set. Then any two bases of \(V\) are finite and have exactly the same number of elements. Consequently the integer \(\dim V\), defined as the cardinality of any basis, is a well-defined invariant of \(V\) that does not depend on the choice of basis.
A common pitfall is assuming that if two vectors are orthogonal, they are automatically linearly independent. While this is true for non-zero vectors, the zero vector is orthogonal to every vector, yet any set containing the zero vector is linearly dependent. Always check for the zero vector before assuming independence based on orthogonality. C
Worked example
Problem. Determine whether the set of vectors \(S = \{ \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \}\) in \(\mathbb{R}^3\) is linearly independent, where \(\mathbf{v}_1 = (1, 0, 0)\), \(\mathbf{v}_2 = (0, 1, 0)\), and \(\mathbf{v}_3 = (1, 1, 0)\).
- Set up the linear combination equation equal to the zero vector: \(c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = \mathbf{0}\).
- Substitute the vector components: \(c_1(1, 0, 0) + c_2(0, 1, 0) + c_3(1, 1, 0) = (0, 0, 0)\).
- Write the system of linear equations for each component: \(c_1 + c_3 = 0\), \(c_2 + c_3 = 0\), and \(0 = 0\).
- Observe that we have 3 variables but only 2 independent equations. We can choose \(c_3 = 1\), which implies \(c_1 = -1\) and \(c_2 = -1\).
- Since there exists a non-trivial solution \((-1, -1, 1)\), the vectors are linearly dependent.
Units check. This is a dimensionless algebraic problem, so no physical units are required. The consistency of the vector components (all dimensionless numbers) ensures the arithmetic is valid.
Three quick tests
- Is the set \(\{ (1, 2), (2, 4) \}\) linearly independent?
Answer
No, because \((2, 4) = 2(1, 2)\), so one is a scalar multiple of the other. - If \(V\) is a 3D vector space, can a subspace \(W\) have dimension 4?
Answer
No, the dimension of a subspace cannot exceed the dimension of the parent space. - Does the set containing only the zero vector \(\{\mathbf{0}\}\) have dimension 0 or 1?
Answer
Dimension 0, because the empty set is a basis for the zero vector space, and the zero vector itself is linearly dependent.
Problem set for PU-104 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.