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4.5 Coordinates and Basis: Class Notes

Definition

If $S=\{{𝐯}_1,{𝐯}_2,\dots ,{𝐯}_n\}$ is a set of vectors in a finite-dimensional vector space $V$, then $S$ is called a basis for $V$ if:
a) $S$ spans $V$
b) $S$ is linearly independent.

$\{{\overrightarrow{e}}_1,{\overrightarrow{e}}_2,{\overrightarrow{e}}_3\}$ are the standard basis vectors for $R^3$. These relate to mutually perpendicular axes. However, any 3 vectors in $R^3$ that do not lie in the same plane can be used to represent any other vector in $R^3$. These 3 vectors are called basis vectors.

#1

Use the determinant of a coefficient matrix to show that the following set of vectors forms a basis for $R^2$: $\{(2,1),(3,0)\}$.

Solution

Let $A=\left[\begin{array}{cc}2& 3\\ 1& 0\end{array}\right]$ (the vectors are its columns).

If $A\overrightarrow{x}=\overrightarrow{b}$ always has a solution, then the set spans $R^2$. If $A\overrightarrow{x}=\overrightarrow{0}$ has only the trivial solution, then the set is independent.

$det(A)=-3\ne 0$. Thus both conditions are met, and the set is a basis for $R^2$.

#8

Show that the following vectors do not form a basis for $P_2$.
$1-3x+2x^2$, $1+x+4x^2$, $1-7x$

Solution

Put the coefficients of each polynomial (constant, $x$, $x^2$) in a column and find the determinant. Use the diagonal method: copy the first two columns to the right, add the products down the diagonals to the right (blue), and subtract the products up the diagonals to the right (red).

$$\left|\begin{array}{ccc}1& 1& 1\\ -3& 1& -7\\ 2& 4& 0\end{array}\right|\begin{array}{cc}1& 1\\ -3& 1\\ 2& 4\end{array}$$

$$0-14-12-2+28+0=0$$

So the vectors do not form a basis for $P_2$.

Note that $2(1-3x+2x^2)-(1+x+4x^2)=1-7x$, so the set is linearly dependent.

Definition

A basis in which the listed order of the vectors matters is called an ordered basis.

Some Standard bases

The basis for the zero vector space $V=\{\overrightarrow{0}\}$ is $\varnothing$ (the empty set).

Theorem 4.5.1 Uniqueness of Basis Representation

If $S=\{{𝐯}_1,{𝐯}_2,\dots ,{𝐯}_n\}$ is a basis for a vector space $V$, then every vector $𝐯$ in $V$ can be expressed in the form $𝐯=c_1{𝐯}_1+c_2{𝐯}_2+\dots +c_n{𝐯}_n$ in exactly one way.

Proof

Let $\overrightarrow{v}\in V$. Since $S$ spans $V$, there are scalars $c_1,c_2,\dots ,c_n$ such that

$$\overrightarrow{v}=c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2+\cdots +c_n{\overrightarrow{v}}_n.$$

Suppose also that

$$\overrightarrow{v}=k_1{\overrightarrow{v}}_1+k_2{\overrightarrow{v}}_2+\cdots +k_n{\overrightarrow{v}}_n.$$

Subtract the two expressions:

$$\overrightarrow{0}=\overrightarrow{v}-\overrightarrow{v}=(c_1-k_1){\overrightarrow{v}}_1+(c_2-k_2){\overrightarrow{v}}_2+\cdots +(c_n-k_n){\overrightarrow{v}}_n$$

Since $S$ is linearly independent, $c_i-k_i=0$ for all $i$, so $c_i=k_i$ for all $i$.

Definition

If $S=\{{𝐯}_1,{𝐯}_2,\dots ,{𝐯}_n\}$ is an ordered basis for a vector space $V$, and $𝐯=c_1{𝐯}_1+c_2{𝐯}_2+\dots +c_n{𝐯}_n$ is the expression for a vector $𝐯$ in terms of the basis $S$, then the scalars $c_1,c_2,\dots ,c_n$ are called the coordinates of $𝐯$ relative to the basis $𝐒$. The vector $(c_1,c_2,\dots ,c_n)$ in $R^n$ constructed from these coordinates is called the coordinate vector of $𝐯$ relative to the basis $𝐒$; it is denoted by $(𝐯{)}_S=(c_1,c_2,\dots ,c_n)$.

Example

Consider $𝐯=(-1,7,2)\in R^3$. The set $B=\{{𝐯}_1,{𝐯}_2,{𝐯}_3\}$, where ${𝐯}_1=(2,-1,-1)$, ${𝐯}_2=(-2,1,2)$, and ${𝐯}_3=(3,5,4)$ forms a basis for $R^3$ (verify). Find $(𝐯{)}_S$ (where $S$ is the standard basis for $R^3$) and $(𝐯{)}_B$.

Solution

First verify that $B$ is a basis. The system

$$c_1(2,-1,-1)+c_2(-2,1,2)+c_3(3,5,4)=(b_1,b_2,b_3)$$

should always have a solution, and

$$c_1(2,-1,-1)+c_2(-2,1,2)+c_3(3,5,4)=(0,0,0)$$

should have only the trivial solution. Both hold because the determinant of the coefficient matrix is not zero:

$$\left|\begin{array}{ccc}2& -2& 3\\ -1& 1& 5\\ -1& 2& 4\end{array}\right|=-13\ne 0.$$

Now consider $c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2+c_3{\overrightarrow{v}}_3=(-1,7,2)$. Find $c_1,c_2,c_3$.

$$\left[\begin{array}{ccc}2& -2& 3\\ -1& 1& 5\\ -1& 2& 4\end{array}\right]\overrightarrow{c}=\left[\begin{array}{c}-1\\ 7\\ 2\end{array}\right]⟶\left[\begin{array}{cccc}\hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle -1}\hfill \\ \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 5}\hfill & \hfill {\displaystyle 7}\hfill \\ \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 2}\hfill \end{array}\right]$$

becomes

$$\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -6}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -4}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

So $(c_1,c_2,c_3)=(-6,-4,1)$. That means

$$-6(2,-1,-1)-4(-2,1,2)+1(3,5,4)=(-1,7,2).$$

Since $(c_1,c_2,c_3)=(-6,-4,1)$ are the coefficients used with $B=\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3\}$ that yield $\overrightarrow{v}=(-1,7,2)$, the vector $(\overrightarrow{v}{)}_B=(-6,-4,1)$ is the coordinate vector of $(-1,7,2)$ relative to the basis $B=\{(2,-1,-1),(-2,1,2),(3,5,4)\}$.

Note: $\overrightarrow{v}$ itself is the coordinate vector relative to the standard basis $\{{\overrightarrow{e}}_1,{\overrightarrow{e}}_2,{\overrightarrow{e}}_3\}$, so $(\overrightarrow{v}{)}_S=(-1,7,2)$.

#16

First show that the set $S=\{A_1,A_2,A_3,A_4\}$ is a basis for $M_{22}$, then express $A$ as a linear combination of the vectors in $S$, and then find the coordinate vector of $A$ relative to $S$.

$A_1=\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]$, $A_2=\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]$, $A_3=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, $A_4=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$; $A=\left[\begin{array}{cc}6& 2\\ 5& 3\end{array}\right]$

Solution

Write $c_1{A}_1+c_2{A}_2+c_3{A}_3+c_4{A}_4=A$ and match entries: each row of the coefficient matrix below comes from one position, in the order upper left (magenta), upper right (blue), lower left (orange), lower right (green).

$$\left[\begin{array}{cccc}1& 1& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 1\\ 0& 0& 1& 0\end{array}\right]⟶\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]✓$$

The matrix reduces to the identity, so $S$ is a basis for $M_{22}$.

By inspection, $c_2=2$ (only $A_2$ has an upper-right entry) and $c_3=3$ (only $A_3$ has a lower-right entry). Then $c_1=1$ (from $c_1+c_2+c_3=6$) and $c_4=4$ (from $c_1+c_4=5$).

$$A=1A_1+2A_2+3A_3+4A_4$$