Dimension Formulas for Subspaces

The sum–intersection dimension formula is a workhorse well beyond pure linear algebra courses.

• Linear systems and rank–nullity companions: When solving simultaneous systems of linear equations defined by two different subspaces (e.g., solution sets of two separate homogeneous systems), the formula computes the dimension of the combined solution space and detects when the systems are "independent enough" to force a large intersection, directly informing consistency and uniqueness questions tied to Rank–Nullity Theorem.
• Coding theory and cryptography: Linear error-correcting codes are subspaces of $\mathbb{F}_q^n$; combining two codes via sum and intersection, and tracking dimensions, is used in analyzing code concatenation and in secret-sharing schemes built from linear codes.
• Computer graphics and robotics: Configuration spaces and constraint subspaces (e.g., the space of joint velocities satisfying multiple simultaneous linear constraints) are intersected, and their dimensions computed via this formula, to determine the true degrees of freedom of a mechanical system.
• Statistics: In linear regression and the analysis of variance, model subspaces (column spaces of design matrices) are summed and intersected; the formula underlies dimension counts used in computing degrees of freedom for nested and combined models.
• Quantum mechanics: Subspaces of a Hilbert space corresponding to eigenspaces or symmetry sectors combine according to the same dimension arithmetic (in the finite-dimensional setting), governing degeneracy counts when two conserved quantities' eigenspaces overlap.
• Algebraic geometry and combinatorics: Grassmannians, the parameter spaces of $k$-dimensional subspaces of an $n$-dimensional space, rely on exactly this dimension calculus to define Schubert cell dimensions and intersection conditions.
• Foreshadowing further theory: The dimension formula is also the seed of the idea behind Quotient Spaces: one can prove $\dim(U+W) = \dim U + \dim W - \dim(U \cap W)$ again, more conceptually, via the isomorphism $(U+W)/W \cong U/(U\cap W)$, a fact that becomes available once quotient space dimension theory is developed.

Ask the tutor: Can you show me how to compute $\dim(U \cap W)$ when $U$ and $W$ are each given as solution sets of homogeneous linear systems, rather than as spans of vectors?

Think of $U$ and $W$ as two flat sheets (subspaces) sitting inside a room ($V$). If the sheets only touch at the origin, gluing them together to form $U+W$ costs you nothing: every direction in $U$ is genuinely new relative to $W$, and vice versa, so the dimensions simply add. But if the sheets already share a whole line, or plane, of common directions, then when you list a spanning set for $U$ together with a spanning set for $W$, the shared directions get counted twice. The formula's job is to subtract that double-counted overlap exactly once — precisely as in $|A \cup B| = |A| + |B| - |A\cap B|$ for finite sets. The proof strategy makes this precise: start with a basis for the "shared part" $U \cap W$, then extend it in two independent directions — once using vectors that live only in $U$, once using vectors that live only in $W$. The claim is that combining all three pieces (the shared basis, the $U$-only extension, the $W$-only extension) gives a basis for $U+W$, with no redundancy and nothing missing.

Example 1: Two planes in $\mathbb{R}^4$ meeting in a line. Let $V = \mathbb{R}^4$. Define

Both $U$ and $W$ are $2$-dimensional (each spanning list is clearly linearly independent, since neither spanning vector is a scalar multiple of the other in either list — check the second coordinates $0,1$ for $U$ and the ratio of first coordinates $1,1$ against last coordinates $0,-1$ for $W$). Finding $U \cap W$. A vector in $U$ has the form $a(1,0,1,0) + b(0,1,0,1) = (a, b, a, b)$. A vector in $W$ has the form $c(1,0,1,0) + d(1,1,1,-1) = (c+d,\, d,\, c+d,\, -d)$. Setting these equal:

The second and fourth equations give $b = d$ and $b = -d$, so $d = -d$, forcing $d = 0$ (over $\mathbb{R}$, or any field of characteristic $\ne 2$). Then $b = 0$ and $a = c$. So every common vector has the form $(a, 0, a, 0) = a(1,0,1,0)$. Thus $U \cap W = \operatorname{span}\{(1,0,1,0)\}$, which is $1$-dimensional (the vector $(1,0,1,0)$ is nonzero). Applying the formula.

Direct check. $U + W$ is spanned by $(1,0,1,0), (0,1,0,1), (1,1,1,-1)$. Row-reduce:

which has rank $3$ (three nonzero pivot rows, and no further cancellation is possible). This confirms $\dim(U+W) = 3$, matching the formula, and since $3 < 4$, $U+W$ is a proper subspace of $\mathbb{R}^4$ — in particular $U+W \ne V$ here.

Example 2: Polynomial subspaces (a direct sum case). Let $V = \mathcal{P}_4(\mathbb{R})$, the real polynomials of degree at most $4$, with $\dim V = 5$ (basis $1, x, x^2, x^3, x^4$). Define

Dimension of $U$. An even polynomial of degree $\le 4$ has the form $a + bx^2 + cx^4$, so $U = \operatorname{span}\{1, x^2, x^4\}$, and this list is linearly independent (it consists of monomials of distinct degrees, which are always linearly independent — a linear dependence among monomials of different degrees would force a nonzero polynomial identity of the form $\sum c_i x^{n_i}=0$, impossible unless all $c_i=0$ by comparing coefficients). Hence $\dim U = 3$. Dimension of $W$. An odd polynomial of degree $\le 4$ has the form $bx + cx^3$ (no constant, $x^2$, or $x^4$ term, since those are even-degree). So $W = \operatorname{span}\{x, x^3\}$, linearly independent by the same monomial argument, giving $\dim W = 2$. Intersection. If $p \in U \cap W$ then $p(x) = p(-x)$ and $p(-x) = -p(x)$ simultaneously, so $p(x) = -p(x)$, i.e. $2p(x) = 0$, forcing $p = 0$ (valid since $\mathbb{R}$ has characteristic $0$, so $2 \ne 0$). Hence $U \cap W = \{0\}$, so $\dim(U\cap W) = 0$, and the sum is direct. Applying the formula.

Since $U + W$ is a $5$-dimensional subspace of the $5$-dimensional space $V$, in fact $U+W = V$ (a subspace of the same finite dimension as the whole space must equal the whole space, since a proper subspace has strictly smaller dimension — itself a consequence of the Basis Extension Theorem). So every polynomial of degree $\le 4$ decomposes uniquely as (even part) $+$ (odd part), confirming the familiar even–odd decomposition of polynomials as an instance of $V = U \oplus W$:

A frequent trap: students assume the three-subspace analogue

mimicking set-theoretic inclusion–exclusion. This is false in general for subspaces. A standard counterexample: in $\mathbb{R}^2$, let $U_1 = \operatorname{span}\{(1,0)\}$, $U_2=\operatorname{span}\{(0,1)\}$, $U_3=\operatorname{span}\{(1,1)\}$. Each pairwise intersection is $\{0\}$ and the triple intersection is $\{0\}$, so the right side would give $1+1+1-0-0-0+0 = 3$. But $U_1+U_2+U_3 = \mathbb{R}^2$ has dimension $2$, not $3$. The two-subspace formula is a genuine theorem; it does not extend by naive analogy to three or more subspaces.