[ALGEBRA] Coordinates and bases

This page builds bases from given vectors and then uses a basis to turn each vector into a column of coordinates. Coordinates let calculations in any finite-dimensional space be done with columns in $F^n$ .

Constructing a basis from data

[DEFINITION]

Basis

A set $B$ in a vector space $V$ is a basis for $V$ if it is linearly independent and spans $V$ .

[1]:Span
[2]:Linear independence
[3]:Eigenbasis
[4]:Diagonalisation
[5]:A system with zero initial derivative
[6]:Values on a basis determine a linear map
22 more

[DEFINITION]

Dimension

If a vector space $V$ has a finite basis, then $V$ is finite-dimensional and its dimension, denoted $\dim(V)$ , is the common number of vectors in all bases of $V$ .

[1]:Basis
[2]:All bases have the same size
[3]:A system with zero initial derivative
[4]:Nullity
[5]:Rank
[6]:Rank-nullity
7 more

A basis can be built from given vectors in two directions, both using pivot columns. A spanning set is cut down to a maximal independent subset and an independent set is extended by standard basis vectors until it spans.

If $S$ is a finite non-empty subset of a vector space, then $S$ contains a subset which is a basis for $\operatorname{span}(S)$ . In particular, a non-zero vector space spanned by a finite set has a basis.

[METHOD]

Reduce a spanning set to a basis

For a finite set $S=\{u_1,\ldots,u_n\}\subseteq\mathbb R^m$ :

Form the matrix $A=[\,u_1\ \cdots\ u_n\,]$ .
Row reduce $A$ to echelon form.
Keep the original vectors whose columns correspond to pivot columns of the echelon form.

The remaining original vectors form a basis for $\operatorname{span}(S)$ .

Row operations identify which columns are redundant, but the basis must use the corresponding columns of the original matrix. Removing the redundant vectors preserves the span.

[EXAMPLE]

Eight vectors in $\mathbb R^5$

Suppose the row-reduced matrix has columns corresponding to eight original vectors:

\begin{pmatrix} 1&2&3&4&5&7&1&5\\ 0&0&0&1&6&8&2&6\\ 0&0&0&0&1&9&3&7\\ 0&0&0&0&0&0&1&8\\ 0&0&0&0&0&0&0&0 \end{pmatrix}.

The leading columns are $1,4,5,7$ . Retaining the four original vectors in those positions gives a basis for the span of the eight vectors, so that span has dimension $4$ .

[routine]

[1]:Reduce a spanning set to a basis

If $V$ is spanned by a finite set and $S$ is an independent subset of $V$ , then there exists a basis for $V$ containing $S$ . In other words, every independent subset can be extended to a basis. To build it, keep adjoining a vector outside the current span until the set spans $V$ . Each addition preserves independence and the dimension of $V$ bounds the size of any independent set, so the process stops after finitely many steps.

[METHOD]

Extend an independent set to a basis

For an independent set $S=\{u_1,\ldots,u_n\}\subseteq\mathbb R^m$ :

Form the matrix with columns

[\,u_1\ \cdots\ u_n\ e_1\ \cdots\ e_m\,],

where $e_1,\ldots,e_m$ are the standard basis vectors. 2. Row reduce this matrix to echelon form. 3. Keep the original columns corresponding to all pivot columns.

The resulting vectors form a basis for $\mathbb R^m$ containing $S$ .

The standard basis supplies candidate vectors outside the current span. The pivot columns keep the given independent vectors and add only the candidates needed to reach dimension $m$ . As before, take the columns from the original matrix: row reduction changes the vectors but not the pivot positions.

[EXAMPLE]

Extending a set in $\mathbb R^3$

Let

S=\left\{ \begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}1\\2\\4\end{pmatrix}, \begin{pmatrix}1\\2\\5\end{pmatrix} \right\}.

Writing $u_1,u_2,u_3$ for these vectors, the first two are independent and

u_3=2u_2-u_1,

so $\{u_1,u_2\}$ is a basis for $\operatorname{span}(S)$ . Following the construction, adjoin the standard basis columns:

\left[ u_1\ u_2\ e_1\ e_2\ e_3 \right] = \begin{pmatrix} 1&1&1&0&0\\ 2&2&0&1&0\\ 3&4&0&0&1 \end{pmatrix} \longrightarrow \begin{pmatrix} 1&0&0&2&-1\\ 0&1&0&-\frac32&1\\ 0&0&1&-\frac12&0 \end{pmatrix}.

The leading columns are $1,2,3$ . Keeping the corresponding original columns gives the basis

\left\{ \begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}1\\2\\4\end{pmatrix}, \begin{pmatrix}1\\0\\0\end{pmatrix} \right\}

for $\mathbb R^3$ .

[routine]

[1]:Extend an independent set to a basis

Coordinates record the unique coefficients

Once an ordered basis is fixed, the unique coefficients of a vector form a column of scalars.

[DEFINITION]

Coordinate vector

Let $B=(u_1,\ldots,u_n)$ be an ordered basis for a vector space $V$ . If

u=x_1u_1+\cdots+x_nu_n,

then the coordinate vector of $u$ with respect to $B$ is the column

[u]_B= \begin{pmatrix} x_1\\ \vdots\\ x_n \end{pmatrix}.

The order of $B$ fixes the order of the entries. Changing the order of the basis changes the coordinate vector even though it does not change $u$ .

[1]:Basis
[2]:Independence and uniqueness
[3]:Matrix representation of a linear map
[4]:Construct a matrix representation
[5]:Matrix representations
[6]:Compute a coordinate vector
5 more

[METHOD]

Compute a coordinate vector

To find $[u]_B$ for an ordered basis $B=(u_1,\ldots,u_n)$ :

Solve

u=x_1u_1+\cdots+x_nu_n

for the coefficients. 2. Place the coefficients in the order determined by $B$ :

[u]_B= \begin{pmatrix} x_1\\ \vdots\\ x_n \end{pmatrix}.

Step 1 is a span membership problem and its solution is unique because $B$ is independent.

[EXAMPLE]

Coordinates in the standard polynomial basis

For the ordered basis

B=(1,x,x^2,\ldots,x^n)

of $P_n(F)$ over the chosen scalar field $F$ , the polynomial

p=a_0+a_1x+\cdots+a_nx^n, \qquad a_0,\ldots,a_n\in F,

has coordinate vector

[p]_B= \begin{pmatrix} a_0\\ a_1\\ \vdots\\ a_n \end{pmatrix}.

[routine]

[1]:Coordinate vector
[2]:Compute a coordinate vector

[EXAMPLE]

Coordinates in a non-standard polynomial basis

Let

B=(1+x,\ x+x^2,\ 1+x^2).

First check that these polynomials form a basis. If $\lambda_1(1+x)+\lambda_2(x+x^2)+\lambda_3(1+x^2)=0$ , equating the constant, linear and quadratic coefficients gives

\lambda_1+\lambda_3=0,\qquad \lambda_1+\lambda_2=0,\qquad \lambda_2+\lambda_3=0.

These equations force all three coefficients to be zero. Thus $B$ is an independent set of three vectors in the three-dimensional space $P_2(\mathbb R)$ , so it is a basis. To find the coordinates of $x^2+x+1$ , solve

x^2+x+1 =\lambda_1(1+x)+\lambda_2(x+x^2)+\lambda_3(1+x^2).

Equating coefficients gives

\lambda_1+\lambda_3=1,\qquad \lambda_1+\lambda_2=1,\qquad \lambda_2+\lambda_3=1.

Thus

\lambda_1=\lambda_2=\lambda_3=\frac12,

and

[x^2+x+1]_B= \begin{pmatrix} \frac12\\[2pt] \frac12\\[2pt] \frac12 \end{pmatrix}.

[routine]

[1]:Compute a coordinate vector

Orthonormal coordinates

When the basis vectors in $\mathbb R^m$ are orthonormal, dot products give the coordinates directly, with no system to solve.

[DEFINITION]

Orthonormal set

A set $S=\{u_1,\ldots,u_n\}$ in $\mathbb R^m$ is orthonormal if

u_i\cdot u_j= \begin{cases} 1,&i=j,\\ 0,&i\ne j. \end{cases}

[1]:Orthonormal sets are independent
[2]:Orthonormal basis

[THEOREM]

Orthonormal sets are independent

Every orthonormal set is linearly independent.

If $\lambda_1u_1+\cdots+\lambda_nu_n=0$ , take the dot product with $u_k$ . Orthonormality leaves $\lambda_k=0$ . This holds for every $k$ .

[1]:Orthonormal set
[2]:Linear independence

[DEFINITION]

Orthonormal basis

An orthonormal basis of a subspace of $\mathbb R^m$ is a basis that is also an orthonormal set.

[THEOREM]

Coordinates in an orthonormal basis

If $B=(u_1,\ldots,u_n)$ is an ordered orthonormal basis of a subspace $V\subseteq\mathbb R^m$ and $v\in V$ , then

[v]_B= \begin{pmatrix} u_1\cdot v\\ \vdots\\ u_n\cdot v \end{pmatrix}.

Each dot product extracts one coefficient because the other basis vectors are perpendicular to it.

[1]:Orthonormal basis
[2]:Coordinate vector
[3]:Proof
[4]:An orthonormal basis of $\mathbb R^3$

[PROOF]

Write $v=x_1u_1+\cdots+x_nu_n$ . Taking the dot product with $u_k$ gives

u_k\cdot v =\sum_{j=1}^n x_j(u_k\cdot u_j) =x_k.

These are exactly the entries of $[v]_B$ .

[1]:Coordinates in an orthonormal basis

[EXAMPLE]

An orthonormal basis of $\mathbb R^3$

The set

B=\left( \frac1{\sqrt3}\begin{pmatrix}1\\1\\1\end{pmatrix}, \frac1{\sqrt2}\begin{pmatrix}1\\-1\\0\end{pmatrix}, \frac1{\sqrt6}\begin{pmatrix}1\\1\\-2\end{pmatrix} \right)

is orthonormal, hence independent. It has three vectors in $\mathbb R^3$ , so it is a basis. For

v=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix},

the orthonormal formula gives

[v]_B= \begin{pmatrix} (a_1+a_2+a_3)/\sqrt3\\ (a_1-a_2)/\sqrt2\\ (a_1+a_2-2a_3)/\sqrt6 \end{pmatrix}.

[routine]

Coordinate arithmetic and matrix tests

[THEOREM]

Coordinates preserve vector-space operations

If $B$ is an ordered basis for $V$ , $u,v\in V$ and $\lambda\in F$ , then

u=v\iff [u]_B=[v]_B, \qquad [u+v]_B=[u]_B+[v]_B, \qquad [\lambda u]_B=\lambda[u]_B.

These identities follow by adding or scaling the coefficients in the unique representations.

[1]:Coordinate vector
[2]:Basis

[COROLLARY]

Square-matrix basis criteria

For a square matrix $A$ with columns $u_1,\ldots,u_n$ , the following are equivalent:

the columns are linearly independent;
the columns span $\mathbb R^n$ ;
the columns form a basis for $\mathbb R^n$ .

For a square matrix, these basis criteria are equivalent to the familiar tests for invertibility.

Data fitting uses coordinates in $P_n(\mathbb R)$ to express interpolation conditions. Later, coordinates in the domain and codomain turn a linear transformation into a matrix.

Coordinates and bases

Basis

Dimension

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