Direct Sums

A direct sum combines vector spaces as independent summands inside one larger vector space. It is the linear-algebra language of decompositions into subspaces, blocks, sectors, and representation pieces.

Direct sums should not be confused with tensor products. A finite direct sum has additive dimension and represents subspaces or sector labels. Its vectors may have components in several summands at once. A tensor product has multiplicative dimension and represents simultaneous factors.

Helpful background. vector spaces for linear combinations and linear independence; bases and coordinates for expanding a vector in a chosen basis.

External Direct Sum

For vector spaces $V$ and $W$ over the same field, the external direct sum is

$$V\oplus W = \{(v,w):v\in V,\ w\in W\}.$$

Addition and scalar multiplication are componentwise:

$$(v,w)+(v',w') = (v+v',w+w'),$$

and

$$\alpha(v,w) = (\alpha v,\alpha w).$$

If $V$ and $W$ are finite-dimensional, then

$$\dim(V\oplus W) = \dim V+\dim W.$$

For finitely many summands, write

$$\bigoplus_{a=1}^m V_a.$$

An element is a tuple with one component in each summand.

Internal Direct Sum

Let $U$ and $W$ be subspaces of a vector space $V$. We say

$$V=U\oplus W$$

if every vector $v\in V$ can be written uniquely as

$$v=u+w, \qquad u\in U,\quad w\in W.$$

Equivalently,

$$V=U+W, \qquad U\cap W=\{0\}.$$

For finitely many subspaces,

$$V=\bigoplus_a V_a$$

means every $v\in V$ has a unique expansion

$$v=\sum_a v_a, \qquad v_a\in V_a.$$

The uniqueness condition is the essential content. Without it, a sum of subspaces may have redundant overlap.

Orthogonal Direct Sums

In an inner-product space, a direct sum is orthogonal if distinct summands are mutually orthogonal:

$$V_a\perp V_b \qquad \text{for }a\ne b.$$

Then

$$\left\lVert \sum_a v_a \right\rVert^2 = \sum_a \lVert v_a\rVert^2.$$

If $P_a$ is the orthogonal projector onto $V_a$, an orthogonal decomposition of the whole space gives

$$P_aP_b=\delta_{ab}P_a, \qquad \sum_a P_a=I.$$

This is the same algebra that appears in finite-dimensional spectral decompositions and projective measurements.

Block Matrices

A direct-sum decomposition gives block matrix notation. If

$$V=U\oplus W,$$

then a vector can be displayed as

$$\begin{pmatrix} u\\ w \end{pmatrix}, \qquad u\in U,\quad w\in W.$$

An operator preserving both subspaces has block diagonal form:

$$A = \begin{pmatrix} A_U & 0\\ 0 & A_W \end{pmatrix}.$$

An operator with off-diagonal blocks can move vectors between summands:

$$A = \begin{pmatrix} A_{UU} & A_{UW}\\ A_{WU} & A_{WW} \end{pmatrix}.$$

Here $A_{WU}$ maps $U$ into $W$, while $A_{UW}$ maps $W$ into $U$.

Block diagonalization is useful when a Hamiltonian preserves sectors. Each block can be diagonalized separately, reducing both the computation and the conceptual clutter.

Direct Sums Versus Tensor Products

If $\dim V=2$ and $\dim W=3$, then

$$\dim(V\oplus W)=5, \qquad \dim(V\otimes W)=6.$$

The direct sum describes a space with a two-dimensional sector and a three-dimensional sector. The tensor product describes two simultaneous systems or degrees of freedom whose basis labels come in ordered pairs.

For the physics-facing distinction, see Direct Sums versus Tensor Products. The mathematical tensor-product construction is Tensor Products.

Superselection Preview

A superselection decomposition is often written

$$\mathcal H = \bigoplus_q \mathcal H_q,$$

where $q$ labels charge, particle number, or another conserved sector. If allowed observables preserve sectors, they are block diagonal:

$$A = \bigoplus_q A_q.$$

The direct sum alone is not a superselection rule. The rule is an additional physical statement about which observables and operations are allowed. The physical discussion is Particle-Number Superselection Preview.

Representation Decomposition Preview

Direct sums also describe how a representation splits into invariant pieces. For example, the tensor product of two spin-$1/2$ representation spaces decomposes into total-spin sectors:

$$\frac12\otimes\frac12 = 1\oplus0.$$

The left side is a tensor product of two spin systems. The right side is a direct-sum decomposition of the resulting four-dimensional space into a three-dimensional triplet sector and a one-dimensional singlet sector.

The physics version is Singlet and Triplet States, and the representation-theory machinery is Tensor Product Representations together with Angular Momentum Algebra.

Worked Example

Let

$$V=\mathbb C^3, \qquad U=\operatorname{span}\{e_1,e_2\}, \qquad W=\operatorname{span}\{e_3\}.$$

Then

$$V=U\oplus W.$$

Every vector has the unique decomposition

$$\begin{pmatrix} z_1\\ z_2\\ z_3 \end{pmatrix} = \begin{pmatrix} z_1\\ z_2\\ 0 \end{pmatrix} + \begin{pmatrix} 0\\ 0\\ z_3 \end{pmatrix}.$$

An operator preserving this decomposition has the form

$$A = \begin{pmatrix} a & b & 0\\ c & d & 0\\ 0 & 0 & e \end{pmatrix}.$$

The upper-left $2\times2$ block acts within $U$, and the lower-right entry acts within $W$.

Common Mistakes

• Confusing direct sums with tensor products.
• Forgetting that an internal direct sum requires uniqueness of decomposition.
• Treating a direct-sum sector label as an independent subsystem label.
• Inferring superselection merely from the symbol $\oplus$.
• Ignoring off-diagonal blocks that move states between summands.
• Calling a direct-sum superposition entanglement without a tensor-product or algebraic subsystem split.
• Forgetting that block diagonalization depends on the chosen decomposition.

References

• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• P. R. Halmos, Finite-Dimensional Vector Spaces, 2nd ed., Springer, 1974.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Let $U=\operatorname{span}\{(1,0,0)^T\}$ and $W=\operatorname{span}\{(0,1,0)^T,(0,0,1)^T\}$ in $\mathbb R^3$. Show that $\mathbb R^3=U\oplus W$.

Solution

Every vector decomposes as

$$\begin{pmatrix} x\\ y\\ z \end{pmatrix} = \begin{pmatrix} x\\ 0\\ 0 \end{pmatrix} + \begin{pmatrix} 0\\ y\\ z \end{pmatrix}.$$

The first vector lies in $U$, the second lies in $W$, and $U\cap W=\{0\}$. Therefore the decomposition is unique.

2. If $\dim V=4$ and $\dim W=5$, compare $\dim(V\oplus W)$ and $\dim(V\otimes W)$.

Solution

The direct sum has dimension

$$\dim(V\oplus W)=4+5=9.$$

The tensor product has dimension

$$\dim(V\otimes W)=4\cdot5=20.$$

3. Suppose $P_1,P_2,P_3$ are orthogonal projectors satisfying $P_aP_b=\delta_{ab}P_a$ and $P_1+P_2+P_3=I$. What direct-sum decomposition do they define?

Solution

They define the orthogonal direct sum of their ranges:

$$V = \operatorname{im}P_1 \oplus \operatorname{im}P_2 \oplus \operatorname{im}P_3.$$

Every vector decomposes uniquely as

$$v=P_1v+P_2v+P_3v.$$