MATH 323 Lecture 18

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Announcements

• Test 2 will be November 13
• Math Club meeting Monday, November 5, 19:00 BLOC 117 — Dr. Frank Sotile on hyperbolic soccerballs.
• Office hours this week are 14:20–15:20

Similarity

$L:V\to V$

Let $A_L$ be the transformation matrix for $L$.

1. if $B$ is the matrix representing $L$ w.r.t. $[{\overrightarrow{u}}_1,{\overrightarrow{u}}_2]$
2. if $A$ is the matrix representing $L$ w.r.t. $[{\overrightarrow{e}}_1,{\overrightarrow{e}}_2]$
3. $U$ is the transition matrix corresponding to the change of basis from $[{\overrightarrow{u}}_1,{\overrightarrow{u}}_2]$ to $[{\overrightarrow{e}}_1,{\overrightarrow{e}}_2]$

Then $B=U^{-1}AU$.

$A$ is similar to $B$ iff the above equation holds for some nonsingular $U$

Example

Consider $L(\overrightarrow{x})=⟨2x_1,x_1+x_2⟩$

$A_{L,[{\overrightarrow{e}}_1,{\overrightarrow{e}}_2]}=\left(\begin{array}{cc}2& 0\\ 1& 1\end{array}\right)$

Let $[{\overrightarrow{u}}_1=⟨1,1⟩,{\overrightarrow{u}}_2=⟨-1,1⟩]$ be another basis

$\begin{array}{rl}L({\overrightarrow{u}}_1)& =A{\overrightarrow{u}}_1=\left(\begin{array}{cc}2& 0\\ 1& 1\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)=\left(\begin{array}{c}2\\ 2\end{array}\right)\\ L({\overrightarrow{u}}_2)& =A{\overrightarrow{u}}_2=\left(\begin{array}{cc}2& 0\\ 1& 1\end{array}\right)\left(\begin{array}{c}-1\\ 1\end{array}\right)=\left(\begin{array}{c}-2\\ 0\end{array}\right)\end{array}$

Transition matrix from $[{\overrightarrow{e}}_1,{\overrightarrow{e}}_2]$ to $[{\overrightarrow{u}}_1,{\overrightarrow{u}}_2]$ is $U=({\overrightarrow{u}}_1,{\overrightarrow{u}}_2)=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$

Transition matrix from $[{\overrightarrow{u}}_1,{\overrightarrow{u}}_2]$ to $[{\overrightarrow{e}}_1,{\overrightarrow{e}}_2]$ is $U^{-1}=\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{1}{2}\end{array}\right)$

$\begin{array}{rlrl}{U}^{-1}L({\overrightarrow{u}}_1)& =\left(\begin{array}{c}2\\ 0\end{array}\right)& U^{-1}L({\overrightarrow{u}}_2)& =\left(\begin{array}{c}-1\\ 1\end{array}\right)\end{array}$

The matrix $B=\left(\begin{array}{cc}2& -1\\ 0& 1\end{array}\right)=U^{-1}AU$ represents $L$ w.r.t. $[{\overrightarrow{u}}_1,{\overrightarrow{u}}_2]$

Theorem 4.3.1

Let $E=[v_1,\dots ,v_n]$ and $F=[w_1,\dots ,w_n]$ be two ordered bases for a vector space $V$, and let $L$ be a linear operator on $V$.

Let $S$ be the transition matrix representing the change from $F$ to $E$.

If $A$ is the matrix representing $L$ w.r.t. $E$ and $B$ is the matrix representing $L$ w.r.t. $F$, then $B=S^{-1}AS$.

Proof

$\overrightarrow{x}\in {\mathbb{ℝ}}^n$, $v=x_1{w}_1+\dots +x_n{w}_n$, and $\hat{x}=⟨x_1,\dots ,x_n⟩=[v{]}_F$.

Let $\overrightarrow{y}=S\overrightarrow{x}=[v{]}_E$, $\overrightarrow{t}=A\overrightarrow{y}=[L(v){]}_E$, $\overrightarrow{z}=B\overrightarrow{x}=[L(v){]}_F$, where $S$ is transition matrix from $F$ to $E$ and $S^{-1}$ is transition matrix from $E$ to $F$.

$S^{-1}\overrightarrow{t}=\overrightarrow{z}$, $S^{-1}AS\overrightarrow{x}=\overrightarrow{z}=B\overrightarrow{x}$

Finding new Bases

$A$ represents $L$ w.r.t $E=[v_1,\dots ,v_n]$

Suppose we haeve $w_1=S_{11}{v}_1+\dots +S_{n1}{v}_n$ through $w_n=S1n{v}_1+\dots +S_{nn}{v}_n$. Then $F=[w_1,\dots ,w_n]$ gives us a new basis.

Example

$D:P_3\to P_2=\frac{\mathrm{d}}{\mathrm{d}x}$ find $B$ representing $D$ w.r.t. $[1,x,x^2]$ and $A$ representing $D$ w.r.t. $[1,2x,4x^2-2]$.

$\begin{array}{rlrlrl}D(1)& =0& D(x)& =1& D(x^2)& =2x\\ B& =\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 2\\ 0& 0& 0\end{array}\right]\\ D(1)& =0& D(2x)& =2& D(4x^2-2)& =8x\\ A& =\left[\begin{array}{ccc}0& 2& 0\\ 0& 0& 4\\ 0& 0& 0\end{array}\right]\end{array}$

Write $[1,2x,4x^2-2]$ as a linear combination of $[1,x,x^2]$ to find $S$:

$\begin{array}{rl}1& =1\cdot 1+0\cdot x+0\cdot x^2\\ 2x& =0\cdot 1+2\cdot x+0\cdot x^2\\ 4x^2-2& =-2\cdot 1+0\cdot x+4\cdot x^2\\ S& =\left[\begin{array}{ccc}1& 0& -2\\ 0& 2& 0\\ 0& 0& 4\end{array}\right]\\ S^{-1}& =\left[\begin{array}{ccc}1& 0& \frac{1}{2}\\ 0& \frac{1}{2}& 0\\ 0& 0& \frac{1}{4}\end{array}\right]\end{array}$

Thus it holds that $B=S^{-1}AS$.

Chapter 5: Orthogonality

Scalar Product

$\overrightarrow{x},\overrightarrow{y}\in {\mathbb{ℝ}}^n$

Scalar product (also called dot product) is defined as

$\begin{array}{rl}\overrightarrow{x}\cdot \overrightarrow{y}& =\sum _{i=1}^n{x}_i{y}_i=x_1{y}_1+\dots +x_n{y}_n\\ & ={\overrightarrow{x}}^T\overrightarrow{y}=(x_1,\dots ,x_n)\left(\begin{array}{c}{y}_1\\ ⋮\\ y_n\end{array}\right)\end{array}$

Length

The length of an $n$-dimensional vector is given by

$\Vert \overrightarrow{x}\Vert =\sqrt{\sum _{x=1}^n{x}_i^2}$

The distance between two vectors $\overrightarrow{x}$ and $\overrightarrow{y}$ can be found from $\Vert \overrightarrow{x}-\overrightarrow{y}\Vert$

Theorem 5.1.1

If $\overrightarrow{x},\overrightarrow{y}\in {\mathbb{ℝ}}^n$ and $\theta$ is the angle between them, then

$\overrightarrow{x}\cdot \overrightarrow{y}=\Vert \overrightarrow{x}\Vert \Vert \overrightarrow{y}\Vert \cos \theta$