MAT223: 8_Change of Basis

Coordinates & Change of Basis §

• Basis for a subspace $\mathcal{V}$ is a linearly independent spanning subspace
• Dimension of a subspace $\mathcal{V}$ is the size of a basis for $\mathcal{V}$ (number of basis vectors)

Find a basis and the dimension of $W=\text{span}\left\{\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}1\\ -2\\ -1\end{array}\right],\left[\begin{array}{c}2\\ -4\\ -2\end{array}\right],\left[\begin{array}{c}-2\\ 5\\ 3\end{array}\right],\left[\begin{array}{c}3\\ -6\\ -2\end{array}\right]\right\}$ §

Find a maximal linearly independent subset using matrix

$$\left[\begin{array}{ccccc}0& 1& 2& -2& 3\\ 0& -2& -4& 5& -6\\ 0& -1& -2& 3& -2\end{array}\right]\text{RREF}\left[\begin{array}{ccccc}0& 1& 2& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]$$

Column 2, 4, 5 are linearly independent

$$\mathcal{B}=\left\{\left[\begin{array}{c}1\\ -2\\ -1\end{array}\right],\left[\begin{array}{c}-2\\ 5\\ 3\end{array}\right],\left[\begin{array}{c}3\\ -6\\ -2\end{array}\right]\right\}$$

$\mathcal{B}$ is a basis of $W$ and dim$(W)=3$, then $W={\mathbb{R}}^3$

Representation in Basis §

• Let $\mathcal{B}=\{{\vec{b}}_1,\dots ,{\vec{b}}_n\}$ be a basis for a subspace $V$ and let $\vec{v}\in V$. The representation of $\vec{v}$ in the $\mathcal{B}$ basis, notated $[\vec{v}{]}_{\mathcal{B}}$ is the column matrix

  $$[\vec{v}{]}_{\mathcal{B}}=\left[\begin{array}{c}{\alpha }_a\\ ⋮\\ {\alpha }_n\end{array}\right]$$

  Where ${\alpha }_1,\dots ,{\alpha }_n$ uniquely satisfy $\vec{v}={\alpha }_1{\vec{v}}_1+\cdots +{\alpha }_n{\vec{v}}_n$, conversely,

  $${\left[\begin{array}{c}{\alpha }_a\\ ⋮\\ {\alpha }_n\end{array}\right]}_{\mathcal{B}}={\alpha }_1{\vec{v}}_1+\cdots +{\alpha }_n{\vec{v}}_n$$

  Is notation for the linear combination of ${\vec{b}}_1,\dots ,{\vec{b}}_n$ with coefficients ${\alpha }_1,\dots ,{\alpha }_n$.

  Let $\mathcal{E}=\{{\vec{e}}_1,{\vec{e}}_2\}$ be the standard basis for ${\mathbb{R}}^2$ and let $\mathcal{C}=\{{\vec{c}}_1,{\vec{c}}_2\}$ where ${\vec{c}}_1={\vec{e}}_1+{\vec{e}}_2$, and ${\vec{e}}_2=3{\vec{e}}_2$ be nother basis for ${\mathbb{R}}^2$. Given that $\vec{v}=2{\vec{e}}_1-{\vec{e}}_2$, find $[\vec{v}{]}_{\mathcal{E}}$ and $[\vec{v}{]}_{\mathcal{C}}$ §

  Since $\vec{v}=2{\vec{e}}_1-{\vec{e}}_2$, get that $[\vec{v}{]}_{\mathcal{E}}=\left[\begin{array}{c}2\\ -1\end{array}\right]$

  To find $[\vec{v}{]}_{\mathcal{C}}$, need to write $\vec{v}$ as a linear combination of ${\vec{c}}_1,{\vec{c}}_2$: let $\vec{v}=x{\vec{c}}_1+y{\vec{c}}_2$ then

  $$\vec{v}=x{\vec{c}}_1+y{\vec{c}}_2=x({\vec{e}}_1+{\vec{e}}_2)+3y{\vec{e}}_2=x{\vec{e}}_1+(x+3y){\vec{e}}_2$$

  Using $\vec{v}=2{\vec{e}}_1-{\vec{e}}_2$, and the equation above, get

  $$2{\vec{e}}_1+{\vec{e}}_2=x{\vec{e}}_1+(x+3y){\vec{e}}_2$$

  So than

  $$(x-2){\vec{e}}_1+(x+3y+1){\vec{e}}_2=\vec{0}$$

  Since ${\vec{e}}_1$ and ${\vec{e}}_2$ are linearly independent, the only way for the above equation to be satisfied:

  $$\{\begin{array}{ccc}x& & =2\\ x& +3y& =-1\end{array}$$

  Solving, and get $\vec{v}=2{\vec{c}}_1-{\vec{c}}_2$ so $[\vec{v}{]}_{\mathcal{C}}=\left[\begin{array}{c}2\\ -1\end{array}\right]$

Procedure to find a basis for a set of vectors §

• From standard to other basis

  $$\begin{gathered} \text{from }[\vec{x}{]}_{\mathcal{E}}=M[\vec{x}{]}_{\mathcal{B}}\text{solve for}(x,y,z):\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{ccc}\dots & \dots & \dots \\ \dots & \dots & \dots \\ \dots & \dots & \dots \end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right] \\ \text{or }{M}^{-1}[\vec{x}{]}_{\mathcal{E}}=[\vec{x}{]}_{\mathcal{B}}\text{solve for}(x,y,z):\left[\begin{array}{ccc}\dots & \dots & \dots \\ \dots & \dots & \dots \\ \dots & \dots & \dots \end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}x\\ y\\ z\end{array}\right] \end{gathered}$$

• From basis to standard basis

  $$\text{from }[\vec{x}{]}_{\mathcal{E}}=M[\vec{x}{]}_{\mathcal{B}}\text{solve for}(x,y,z):\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}\dots & \dots & \dots \\ \dots & \dots & \dots \\ \dots & \dots & \dots \end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]$$

• Let $X,Y$ be matrices such that $X[\vec{x}{]}_{\mathcal{E}}=[\vec{x}{]}_{\mathcal{B}}$ and $Y[\vec{x}{]}_{\mathcal{B}}=[\vec{x}{]}_{\mathcal{E}}$ then for all $\vec{x}$, $XY=YX$

Consider the basis: $\mathcal{B}=\left\{\left[\begin{array}{c}2\\ 2\\ 4\end{array}\right],\left[\begin{array}{c}-1\\ 0\\ 3\end{array}\right],\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]\right\}$ and $[\vec{x}{]}_{\mathcal{B}}=\left[\begin{array}{c}2\\ 1\\ -3\end{array}\right]$ Find $\vec{x}$ in standard basis §

$$\vec{x}=2\left[\begin{array}{c}2\\ 2\\ 4\end{array}\right]+1\left[\begin{array}{c}-1\\ 0\\ 3\end{array}\right]-3\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}4\\ 4\\ 4\end{array}\right]+\left[\begin{array}{c}-1\\ 0\\ 3\end{array}\right]+\left[\begin{array}{c}-3\\ -3\\ -3\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ 8\end{array}\right]$$

Therefore $\vec{x}=(0,1,8)$ in standard basis

Consider the basis: $\mathcal{B}=\left\{\left[\begin{array}{c}2\\ 2\\ 4\end{array}\right],\left[\begin{array}{c}-1\\ 0\\ 3\end{array}\right],\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]\right\}$ and $\vec{x}=\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]$ Find $[\vec{x}{]}_{\mathcal{B}}$ §

$[\vec{x}{]}_{\mathcal{B}}$ has the coordinates $(c_1,c_2,c_3)$

$$\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]=c_1\left[\begin{array}{c}2\\ 2\\ 4\end{array}\right]+c_2\left[\begin{array}{c}-1\\ 0\\ 3\end{array}\right]+c_3\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]⟹\left[\begin{array}{cccc}2& -1& 1& 2\\ 2& 0& 1& 3\\ 4& 3& 1& 4\end{array}\right]\text{RREF}\left[\begin{array}{cccc}1& 0& 0& 2\\ 0& 1& 0& 1\\ 0& 0& 1& 5\end{array}\right]$$

Then $\vec{x}=(-1,1,5)$ in standard basis

Notation Conventions §

• If a problem involves only one basis, then write $\left[\begin{array}{c}x\\ y\end{array}\right]$ means ${\left[\begin{array}{c}x\\ y\end{array}\right]}_{\mathcal{E}}$ where $\mathcal{E}$ is the standard basis

• If there are multiple basis, then always write ${\left[\begin{array}{c}x\\ y\end{array}\right]}_{\mathcal{X}}$ to specify a vector relative to basis $\mathcal{X}$

• $\vec{x}\ne [\vec{x}{]}_{\mathcal{E}}=\left[\begin{array}{c}a\\ b\end{array}\right],{\left[\begin{array}{c}a\\ b\end{array}\right]}_{\mathcal{E}}=\vec{x}\ne [\vec{x}{]}_{\mathcal{E}}$

  • Left sides of the two non-equal signs is a true vector, where as the right side is a meaningless list of numbers. (Unit it has been assigned a basis)
    $$[\text{true vector}{]}_{\mathcal{X}}=\text{list of numbers},[\text{list of numbers}{]}_{\mathcal{X}}=\text{true vector}$$

Orientation of a Basis §

• Orthonormal bases - bases consisting of unit vectors that are orthogonal to each other

  • Not all bases are orthonormal

• The ordered basis $\mathcal{B}$ is Right-handed (positively oriented) if it can be continuously transformed to the standard basis (${\vec{b}}_i\mapsto {\vec{e}}_i$) while remaining linearly independent throughout the transformation. Otherwise, $\mathcal{B}$ would be left-handed (negatively oriented)

  • By convention, the standard basis is right-handed

  For the two basis here: §

  The $\mathcal{A}$ basis can be rotated to get to $\mathcal{E}$ while remaining the order (${\vec{a}}_1\mapsto {\vec{e}}_1$ and ${\vec{a}}_2\mapsto {\vec{e}}_2$), then it has the same orientation as the standard basis (positive)

  The $\mathcal{B}$ basis cannot be rotated to get to $\mathcal{E}$ while remaining the order, then it has a different orientation as the standard basis (negative)

  For $\mathcal{X}$ §

  Keep ${\vec{x}}_1$ in place and transform ${\vec{x}}_2$ smoothly along the dotted line. During the entire transformation, ${\vec{x}}_1,{\vec{x}}_2$ is linearly independent, therefore, $\mathcal{X}$ is positively oriented

  For $\mathcal{Y}$ §

  This time, along the route of ${\vec{y}}_2$’s transformation, the set ${\vec{y}}_1$ and ${\vec{y}}_2$ would become linearly dependent at some point. Therefore, $\mathcal{Y}$ is negatively oriented

Reversing Orientation §

1. Reflect $\mathcal{E}=\{{\vec{e}}_1,{\vec{e}}_2\}$ with the line $y=x$, this sends $\{{\vec{e}}_1,{\vec{e}}_2\}\mapsto \{{\vec{e}}_2,{\vec{e}}_1\}$

2. Reflect $\mathcal{E}=\{{\vec{e}}_1,{\vec{e}}_2\}$ with he line $x=0$, this sends $\{{\vec{e}}_1,{\vec{e}}_2\}\mapsto \{-{\vec{e}}_1,{\vec{e}}_2\}$

   • This has the opposite orientation as $\mathcal{E}$, it is negatively oriented

$$\left[\begin{array}{cc}2& 0\\ 1& 1\end{array}\right]{\left[\begin{array}{c}2\\ 3\end{array}\right]}_{\mathcal{E}}=\left[\begin{array}{c}2\\ 3\end{array}\right]$$