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MATH
[1251]

Examples from Coordinates and bases

All examples from Coordinates and bases.

Eight vectors in R5\mathbb R^5

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

(1234571500016826000019370000001800000000).\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 1,4,5,71,4,5,7. Retaining the four original vectors in those positions gives a basis for the span of the eight vectors, so that span has dimension 44.

Extending a set in R3\mathbb R^3

Let

S={(123),(124),(125)}.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 u1,u2,u3u_1,u_2,u_3 for these vectors, the first two are independent and

u3=2u2−u1,u_3=2u_2-u_1,

so {u1,u2}\{u_1,u_2\} is a basis for span⁡(S)\operatorname{span}(S). Following the construction, adjoin the standard basis columns:

[u1 u2 e1 e2 e3]=(111002201034001)⟶(1002−1010−321001−120).\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 1,2,31,2,3. Keeping the corresponding original columns gives the basis

{(123),(124),(100)}\left\{ \begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}1\\2\\4\end{pmatrix}, \begin{pmatrix}1\\0\\0\end{pmatrix} \right\}

for R3\mathbb R^3.

Coordinates in the standard polynomial basis

For the ordered basis

B=(1,x,x2,…,xn)B=(1,x,x^2,\ldots,x^n)

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

p=a0+a1x+⋯+anxn,a0,…,an∈F,p=a_0+a_1x+\cdots+a_nx^n, \qquad a_0,\ldots,a_n\in F,

has coordinate vector

[p]B=(a0a1⋮an).[p]_B= \begin{pmatrix} a_0\\ a_1\\ \vdots\\ a_n \end{pmatrix}.

Coordinates in a non-standard polynomial basis

Let

B=(1+x, x+x2, 1+x2).B=(1+x,\ x+x^2,\ 1+x^2).

First check that these polynomials form a basis. If λ1(1+x)+λ2(x+x2)+λ3(1+x2)=0\lambda_1(1+x)+\lambda_2(x+x^2)+\lambda_3(1+x^2)=0, equating the constant, linear and quadratic coefficients gives

λ1+λ3=0,λ1+λ2=0,λ2+λ3=0.\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 BB is an independent set of three vectors in the three-dimensional space P2(R)P_2(\mathbb R), so it is a basis. To find the coordinates of x2+x+1x^2+x+1, solve

x2+x+1=λ1(1+x)+λ2(x+x2)+λ3(1+x2).x^2+x+1 =\lambda_1(1+x)+\lambda_2(x+x^2)+\lambda_3(1+x^2).

Equating coefficients gives

λ1+λ3=1,λ1+λ2=1,λ2+λ3=1.\lambda_1+\lambda_3=1,\qquad \lambda_1+\lambda_2=1,\qquad \lambda_2+\lambda_3=1.

Thus

λ1=λ2=λ3=12,\lambda_1=\lambda_2=\lambda_3=\frac12,

and

[x2+x+1]B=(121212).[x^2+x+1]_B= \begin{pmatrix} \frac12\\[2pt] \frac12\\[2pt] \frac12 \end{pmatrix}.

An orthonormal basis of R3\mathbb R^3

The set

B=(13(111),12(1−10),16(11−2))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 R3\mathbb R^3, so it is a basis. For

v=(a1a2a3),v=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix},

the orthonormal formula gives

[v]B=((a1+a2+a3)/3(a1−a2)/2(a1+a2−2a3)/6).[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}.

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