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Geometry and Algebra

Vector spaces: subspaces, linear independence, bases and dimension

Definition and examples, subspaces, linear combinations and spans, linear dependence, bases and coordinates, dimension and Grassmann's formula.

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So far we have worked with vectors in the plane and in space. Linear algebra takes a step of abstraction: it takes the properties of vector addition and multiplication by a number, and calls “vector space” any set that respects them. That way the same theorems hold for triples of numbers, for polynomials, for matrices and for signals. In this lesson we cover spaces and subspaces, linear combinations, linear dependence, bases and dimension.

Vector spaces

A real vector space is a set V with an addition \mathbf u + \mathbf v and a multiplication by scalars \lambda\mathbf v that satisfy the properties we are used to: addition is associative and commutative, there is a zero vector \mathbf 0, every vector has an opposite, and \lambda(\mathbf u + \mathbf v) = \lambda\mathbf u + \lambda\mathbf v, (\lambda + \mu)\mathbf v = \lambda\mathbf v + \mu\mathbf v, \lambda(\mu\mathbf v) = (\lambda\mu)\mathbf v, 1\,\mathbf v = \mathbf v.

Examples:

  • \mathbb R^n, the n-tuples of real numbers;
  • \mathbb R[x]_{\le d}, polynomials of degree at most d;
  • M_{m,n}(\mathbb R), the m\times n matrices;
  • continuous functions on an interval, with pointwise addition. The signals in a circuit live in spaces of this kind.

Subspaces

A subset W of V is a subspace if it is itself a vector space with the same operations. Three checks are enough: \mathbf 0\in W; if \mathbf u, \mathbf v\in W then \mathbf u + \mathbf v\in W; if \mathbf v\in W then \lambda\mathbf v\in W.

Subspaces of R3: a plane and a line through the origin, and a shifted plane that is not a subspace
In \mathbb R^3 the subspaces are the origin, the lines and planes through the origin (L and W) and the whole space. The shifted plane W + \mathbf v_0 is not a subspace: it does not contain \mathbf 0.

Examples. The solutions of a homogeneous system A\mathbf x = \mathbf 0 form a subspace of \mathbb R^n: if A\mathbf u = \mathbf 0 and A\mathbf v = \mathbf 0 then A(\mathbf u + \mathbf v) = \mathbf 0 and A(\lambda\mathbf u) = \mathbf 0. The solutions of a non-homogeneous system are not (they do not contain \mathbf 0): they form a shifted subspace, like the grey plane in the figure. The plane x + y + z = 0 is a subspace, the plane x + y + z = 1 is not.

Linear combinations and spanning sets

A linear combination of the vectors \mathbf v_1, \dots, \mathbf v_k is a vector of the form \lambda_1\mathbf v_1 + \dots + \lambda_k\mathbf v_k. The set of all linear combinations is called the span and is written \langle\mathbf v_1, \dots, \mathbf v_k\rangle. It is the smallest subspace containing those vectors. Two non-parallel vectors of \mathbb R^3 span a plane through the origin; three non-coplanar vectors span the whole of \mathbb R^3.

Linear dependence and independence

The vectors \mathbf v_1, \dots, \mathbf v_k are linearly independent if the only linear combination giving the zero vector is the one with all coefficients zero:

    \[ \lambda_1\mathbf v_1 + \dots + \lambda_k\mathbf v_k = \mathbf 0 \quad\Rightarrow\quad \lambda_1 = \dots = \lambda_k = 0 \]

Otherwise they are dependent, and at least one of them can be written as a combination of the others: it is “redundant”, it adds no new direction.

How to check. Put the vectors as columns of a matrix and solve the homogeneous system: they are independent if the only solution is zero, that is if the echelon form has a pivot in every column. In \mathbb R^n, n vectors are independent if and only if the determinant of the matrix with them as columns is non-zero.

Example. (1, 2, 3), (0, 1, 1), (2, 5, 7): the determinant is 1\cdot(7 - 5) - 0 + 2\cdot(2 - 3) = 2 - 2 = 0. They are dependent, and indeed (2, 5, 7) = 2\,(1, 2, 3) + (0, 1, 1).

Bases and coordinates

A basis of V is a set of linearly independent vectors that span V. Every vector can then be written as a combination of the basis vectors in exactly one way: the coefficients are its coordinates.

Proof: coordinates are unique

If \mathbf v = \sum\lambda_i\mathbf b_i = \sum\mu_i\mathbf b_i, subtracting gives \sum(\lambda_i - \mu_i)\mathbf b_i = \mathbf 0. By independence all coefficients are zero, so \lambda_i = \mu_i for every i. This is where independence is needed: with dependent vectors the same vector would have infinitely many expressions.

Standard bases: in \mathbb R^n the vectors \mathbf e_1 = (1, 0, \dots, 0), …, \mathbf e_n; in polynomials of degree at most d the monomials 1, x, x^2, \dots, x^d. The coordinates of the polynomial 3 - x + 2x^2 are (3, -1, 2): with a basis, a polynomial becomes a triple of numbers and can be handled like a vector of \mathbb R^3.

Dimension

Theorem. All bases of a vector space have the same number of elements. This number is the dimension of V, \dim V.

The idea of the proof (Steinitz exchange lemma): if V is spanned by m vectors, then any set of more than m vectors is dependent. Applying this both ways to two bases, they have the same number of elements.

\dim\mathbb R^n = n, \dim\mathbb R[x]_{\le d} = d + 1, \dim M_{m,n} = mn. The space of all continuous functions, on the other hand, is infinite-dimensional.

Practical consequences, in a space of dimension n:

  • more than n vectors are always dependent;
  • fewer than n vectors cannot span the whole space;
  • n independent vectors automatically form a basis (and so do n spanning vectors).

Sum and intersection of subspaces

If U and W are subspaces, so are U\cap W and U + W = \{\mathbf u + \mathbf w\}. Grassmann’s formula links their dimensions:

    \[ \dim(U + W) = \dim U + \dim W - \dim(U\cap W) \]

Two distinct planes through the origin in \mathbb R^3: \dim(U + W) = 3, so \dim(U\cap W) = 2 + 2 - 3 = 1. They meet in a line, as expected. If U\cap W = \{\mathbf 0\} the sum is called direct, U\oplus W, and every vector of the sum splits in a unique way.

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