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

Then $B=U^{-1}\,A\,U$.

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

Example

Consider $L({\vec {x}})=\left\langle 2x_{1},x_{1}+x_{2}\right\rangle$

$A_{L,[{\vec {e}}_{1},{\vec {e}}_{2}]}={\begin{pmatrix}2&0\\1&1\end{pmatrix}}$

Let $[{\vec {u}}_{1}=\left\langle 1,1\right\rangle ,{\vec {u}}_{2}=\left\langle -1,1\right\rangle ]$ be another basis

${\begin{aligned}L({\vec {u}}_{1})&=A{\vec {u}}_{1}={\begin{pmatrix}2&0\\1&1\end{pmatrix}}\,{\begin{pmatrix}1\\1\end{pmatrix}}={\begin{pmatrix}2\\2\end{pmatrix}}\\L({\vec {u}}_{2})&=A{\vec {u}}_{2}={\begin{pmatrix}2&0\\1&1\end{pmatrix}}\,{\begin{pmatrix}-1\\1\end{pmatrix}}={\begin{pmatrix}-2\\0\end{pmatrix}}\end{aligned}}$

Transition matrix from $[{\vec {e}}_{1},{\vec {e}}_{2}]$ to $[{\vec {u}}_{1},{\vec {u}}_{2}]$ is $U=({\vec {u}}_{1},{\vec {u}}_{2})={\begin{pmatrix}1&-1\\1&1\end{pmatrix}}$

Transition matrix from $[{\vec {u}}_{1},{\vec {u}}_{2}]$ to $[{\vec {e}}_{1},{\vec {e}}_{2}]$ is $U^{-1}={\begin{pmatrix}{\frac {1}{2}}&{\frac {1}{2}}\\-{\frac {1}{2}}&{\frac {1}{2}}\end{pmatrix}}$

${\begin{aligned}U^{-1}L({\vec {u}}_{1})&={\begin{pmatrix}2\\0\end{pmatrix}}&U^{-1}L({\vec {u}}_{2})&={\begin{pmatrix}-1\\1\end{pmatrix}}\end{aligned}}$

The matrix $B={\begin{pmatrix}2&-1\\0&1\end{pmatrix}}=U^{-1}\,A\,U$ represents $L$ w.r.t. $[{\vec {u}}_{1},{\vec {u}}_{2}]$

Theorem 4.3.1

Let $E=[v_{1},\ldots ,v_{n}]$ and $F=[w_{1},\ldots ,w_{n}]$ 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 $B=S^{-1}\,A\,S$.

Proof

${\vec {x}}\in \mathbb {R} ^{n}$, $v=x_{1}\,w_{1}+\dots +x_{n}\,w_{n}$, and ${\hat {x}}=\left\langle x_{1},\ldots ,x_{n}\right\rangle =[v]_{F}$.

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

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

Finding new Bases

$A$ represents $L$ w.r.t $E=[v_{1},\ldots ,v_{n}]$

Suppose we haeve $w_{1}=S_{11}\,v_{1}+\dots +S_{n1}\,v_{n}$ through $w_{n}=S{1n}\,v_{1}+\dots +S_{nn}\,v_{n}$. Then $F=[w_{1},\ldots ,w_{n}]$ 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. $[1,2x,4x^{2}-2]$.

${\begin{aligned}D(1)&=0&D(x)&=1&D(x^{2})&=2x\\B&={\begin{bmatrix}0&1&0\\0&0&2\\0&0&0\end{bmatrix}}\\D(1)&=0&D(2x)&=2&D(4x^{2}-2)&=8x\\A&={\begin{bmatrix}0&2&0\\0&0&4\\0&0&0\end{bmatrix}}\end{aligned}}$

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

${\begin{aligned}1&=1\cdot 1+0\cdot x+0\cdot x^{2}\\2x&=0\cdot 1+2\cdot x+0\cdot x^{2}\\4x^{2}-2&=-2\cdot 1+0\cdot x+4\cdot x^{2}\\S&={\begin{bmatrix}1&0&-2\\0&2&0\\0&0&4\end{bmatrix}}\\S^{-1}&={\begin{bmatrix}1&0&{\frac {1}{2}}\\0&{\frac {1}{2}}&0\\0&0&{\frac {1}{4}}\end{bmatrix}}\end{aligned}}$

Thus it holds that $B=S^{-1}\,A\,S$.

Chapter 5: Orthogonality

Scalar Product

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

Scalar product (also called dot product) is defined as

${\begin{aligned}{\vec {x}}\cdot {\vec {y}}&=\sum _{i=1}^{n}x_{i}\,y_{i}=x_{1}\,y_{1}+\dots +x_{n}\,y_{n}\\&={\vec {x}}^{T}{\vec {y}}=(x_{1},\ldots ,x_{n}){\begin{pmatrix}y_{1}\\\vdots \\y_{n}\end{pmatrix}}\end{aligned}}$

Length

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

$\left\|{\vec {x}}\right\|={\sqrt {\sum _{x=1}^{n}x_{i}^{2}}}$

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

Theorem 5.1.1

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

${\vec {x}}\cdot {\vec {y}}=\|{\vec {x}}\|\|{\vec {y}}\|\cos {\theta }$