Sums of Subspaces

Sums of subspaces are a basic organizing tool across mathematics and its applications. In pure linear algebra, they underlie the sharper notion of a Direct Sum of Subspaces used to decompose vector spaces into independent pieces, essential for diagonalization, Jordan form, and the structure theory of linear operators. In functional analysis, sums (and their topological refinements, closed sums) of subspaces of Hilbert or Banach spaces describe how signals or functions decompose into orthogonal or complementary components, foundational to Fourier analysis and wavelet theory. In coding theory and cryptography, sums of subspaces over finite fields describe how codewords or key spaces combine, and the dimension formula governs redundancy and error-correcting capacity. In statistics, sums of subspaces appear when decomposing the column space of a design matrix into main-effect and interaction subspaces in the analysis of variance (ANOVA). In physics and engineering, sums of eigenspaces or invariant subspaces of an operator describe how a system's state space decomposes into modes, essential to normal mode analysis in mechanical and electrical systems, and to the spectral decomposition used in quantum mechanics.

Ask the tutor: How does the sum of two subspaces change if their intersection is nontrivial, and why does that lead to the dimension formula $\dim(U+W) = \dim U + \dim W - \dim(U \cap W)$?

Picture two lines through the origin in $\mathbb{R}^2$ that are not equal to each other — say $U_1 = \operatorname{span}(1,0)$ and $U_2 = \operatorname{span}(0,1)$. Their union $U_1 \cup U_2$ is the "plus sign" shape consisting of both axes. This union is not a subspace: the vector $(1,0) \in U_1$ and the vector $(0,1) \in U_2$ both lie in the union, but their sum $(1,1)$ lies on neither axis, so it is not in $U_1 \cup U_2$. The union fails to be closed under addition. The sum $U_1 + U_2$ repairs this defect by declaring that every such combination $u_1+u_2$ must be included. In this example, $U_1+U_2 = \mathbb{R}^2$, because every vector $(x,y)$ can be written as $(x,0)+(0,y)$ with $(x,0)\in U_1$ and $(0,y)\in U_2$. The sum is the smallest subspace that "closes up" the union under the vector space operations — and this is exactly what the span of a set does to any set of vectors. The picture to keep in mind: the sum sweeps out every vector reachable by adding one representative from each subspace, filling in a solid region (here, all of the plane) around the "skeleton" formed by the union. A second useful picture is in $\mathbb{R}^3$: two distinct planes through the origin intersect in a line, but their sum is forced to be all of $\mathbb{R}^3$, since two planes already contain enough independent directions to reach every vector by summation.

Example 1: Two non-trivial planes in $\mathbb{R}^3$ with an asymmetric spanning description. Let $V = \mathbb{R}^3$, and define

Here $U$ is a plane through the origin (it is 2-dimensional, since $(1,2,0)$ and $(0,1,1)$ are not parallel), and $W$ is a line through the origin. We compute $U+W$. By Part 2 of the theorem,

To find the dimension and an explicit description, row-reduce the matrix whose rows are these three vectors:

Subtract row 1 from row 3: row 3 becomes $(0,-2,3)$. Add $2\times$ row 2 to the new row 3: $(0,-2,3) + 2(0,1,1) = (0,0,5)$. Since $(0,0,5) \neq 0$, the three original rows are linearly independent (the row-reduced matrix has rank 3, an upper triangular form with nonzero pivots $1,1,5$). Hence $\{(1,2,0),(0,1,1),(1,0,3)\}$ is a basis of $\mathbb{R}^3$, so

This matches intuition: a plane and a line not lying in that plane together must span all of $\mathbb{R}^3$. We can verify directly that $(1,0,3) \notin U$: if $(1,0,3) = a(1,2,0)+b(0,1,1)$ for scalars $a,b$, then comparing first coordinates $a=1$, comparing third coordinates $b=3$, but then the second coordinate gives $2a+b = 2+3 = 5 \neq 0$, a contradiction. So indeed $W \not\subseteq U$, consistent with the sum growing strictly larger than $U$ alone (and reaching the whole space, since $\dim U = 2$ already and adding one more independent direction fills $\mathbb{R}^3$).

Example 2: A sum of subspaces of a polynomial space that is not all of the ambient space. Let $V = \mathcal{P}_3(\mathbb{R})$, the vector space of real polynomials of degree at most $3$, which has dimension $4$ with basis $\{1, x, x^2, x^3\}$. Define subspaces

First, verify $U$ is a subspace: it is the kernel of the linear map $\operatorname{ev}_0 : \mathcal{P}_3(\mathbb{R}) \to \mathbb{R}$ sending $p \mapsto p(0)$, and kernels of linear maps are subspaces. Concretely, $U = \operatorname{span}(x, x^2, x^3)$, since a polynomial $a_0 + a_1x+a_2x^2+a_3x^3$ satisfies $p(0) = a_0 = 0$ exactly when its constant term vanishes, and clearly $\{x,x^2,x^3\}$ is a basis of this $3$-dimensional subspace (they are the images of standard basis vectors under a coordinate projection, hence linearly independent, and by dimension count they span $U$ since $\dim U = 3$ by the rank-nullity count on $\operatorname{ev}_0$). By Part 2 of the theorem,

(We may drop the repeated spanning vector $x^3$ since removing a duplicate from a spanning set does not change its span.) Since $1+x^2 = 1\cdot(1) + 1\cdot(x^2)$, and $x^2$ is already in the list, the presence of $1+x^2$ together with $x^2$ lets us recover the constant polynomial $1 = (1+x^2) - x^2$. So

Let us double check this is correct and not accidental, by exhibiting an arbitrary $a_0+a_1x+a_2x^2+a_3x^3 \in V$ as a sum $u+w$ with $u \in U$, $w \in W$. Take

Indeed $u \in U$ since its constant term is $0$, and $w \in W$ since it is a scalar multiple of $1+x^2$. Their sum is

as required (the $x^2$-coefficients $a_0 + (a_2-a_0) = a_2$ combine correctly). This confirms $U+W = V$ by direct construction, corroborating the spanning-set computation. Note by contrast that if we had instead taken $W' = \operatorname{span}(x^3)$ alone, then $U + W' = \operatorname{span}(x,x^2,x^3) = U$ itself, since $x^3 \in U$ already — a reminder that summing in a subspace that is already contained in $U$ changes nothing, consistent with Part 3 (the sum is the smallest subspace containing both, and if $W' \subseteq U$ then $U$ itself already contains both, so the smallest such subspace is $U$).

Grassmann's original insight was that subspaces of varying dimensions could be combined by an addition-like operation, and that this operation obeyed a precise arithmetic — the germ of what is now proven rigorously as the sum-of-subspaces theorem and its accompanying dimension formula. His work anticipated, by roughly half a century, the axiomatic vector space formalism in which the sum of subspaces is now routinely defined and used.