Type a command or search the notes. Tab cycles completions. Up and down recall history when blank. Escape dismisses, then leaves the prompt.
MATH
[1251]

Examples from Basis and dimension

All examples from Basis and dimension.

Back to Basis and dimension

A basis of R3\mathbb R^3

Consider

B={(102),(0−11),(−210)}.B=\left\{ \begin{pmatrix}1\\0\\2\end{pmatrix}, \begin{pmatrix}0\\-1\\1\end{pmatrix}, \begin{pmatrix}-2\\1\\0\end{pmatrix} \right\}.

The matrix with these vectors as columns row reduces to I3I_3, so its columns are linearly independent and span R3\mathbb R^3. Thus BB is a basis for R3\mathbb R^3.

A basis for a generated subspace

Let

S={(12),(−21),(11),(−1−2)}.S=\left\{ \begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}-2\\1\end{pmatrix}, \begin{pmatrix}1\\1\end{pmatrix}, \begin{pmatrix}-1\\-2\end{pmatrix} \right\}.

The first two columns are leading columns when the matrix formed from SS is row reduced. They are independent and the remaining vectors are redundant, so

{(12),(−21)}\left\{ \begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}-2\\1\end{pmatrix} \right\}

is a basis for span⁡(S)\operatorname{span}(S).

A basis of P1P_1

For p=a0+a1xp=a_0+a_1x, write

p=λ1(1+x)+λ2(x+2).p=\lambda_1(1+x)+\lambda_2(x+2).

Equating coefficients gives

λ1+2λ2=a0,λ1+λ2=a1.\lambda_1+2\lambda_2=a_0, \qquad \lambda_1+\lambda_2=a_1.

The coefficient matrix has determinant −1-1, so this system has a unique solution for every a0,a1a_0,a_1. Therefore {1+x,x+2}\{1+x,x+2\} spans P1P_1 and is linearly independent, so it is a basis.

[routine]

  1. [1]:Basis
  2. [2]:Span
  3. [3]:Linear independence

Standard bases and dimensions

The standard basis of R2\mathbb R^2 is

{(10),(01)}.\left\{ \begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix} \right\}.

The standard polynomial basis is {1,x,…,xn}\{1,x,\ldots,x^n\} for Pn(R)P_n(\mathbb R). For M2×2(R)M_{2\times2}(\mathbb R), the standard matrix basis is

{(1000),(0100),(0010),(0001)}.\left\{ \begin{pmatrix}1&0\\0&0\end{pmatrix}, \begin{pmatrix}0&1\\0&0\end{pmatrix}, \begin{pmatrix}0&0\\1&0\end{pmatrix}, \begin{pmatrix}0&0\\0&1\end{pmatrix} \right\}.

More generally, the standard bases give

dim⁡(Rn)=n,dim⁡(Pn(R))=n+1,dim⁡(Mm×n(R))=mn.\dim(\mathbb R^n)=n, \qquad \dim(P_n(\mathbb R))=n+1, \qquad \dim(M_{m\times n}(\mathbb R))=mn.

Each count is the number of free coefficients: a polynomial in Pn(R)P_n(\mathbb R) has n+1n+1 coefficients and an m×nm\times n matrix has one for each of its mnmn entries.

[routine]

  1. [1]:Dimension

Counting prevents spanning

Two polynomials cannot span P2(R)P_2(\mathbb R), because

2<dim⁡(P2(R))=3.2<\dim(P_2(\mathbb R))=3.

Likewise, three polynomials in P1(R)P_1(\mathbb R) must be linearly dependent, since

3>dim⁡(P1(R))=2.3>\dim(P_1(\mathbb R))=2.

On this page