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

${\displaystyle L:V\to V}$

Let ${\displaystyle A_{L}}$ be the transformation matrix for ${\displaystyle L}$.

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

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

${\displaystyle A}$ is similar to ${\displaystyle B}$ iff the above equation holds for some nonsingular ${\displaystyle U}$

Example

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

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

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

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

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

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

Theorem 4.3.1

Let ${\displaystyle E=[v_{1},\ldots ,v_{n}]}$ and ${\displaystyle F=[w_{1},\ldots ,w_{n}]}$ be two ordered bases for a vector space ${\displaystyle V}$, and let ${\displaystyle L}$ be a linear operator on ${\displaystyle V}$.

Let ${\displaystyle S}$ be the transition matrix representing the change from ${\displaystyle F}$ to ${\displaystyle E}$.

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

Proof

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

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

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

Finding new Bases

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

Suppose we haeve ${\displaystyle w_{1}=S_{11}\,v_{1}+\dots +S_{n1}\,v_{n}}$ through ${\displaystyle w_{n}=S{1n}\,v_{1}+\dots +S_{nn}\,v_{n}}$. Then ${\displaystyle F=[w_{1},\ldots ,w_{n}]}$ gives us a new basis.

Example

${\displaystyle D:P_{3}\to P_{2}={\frac {\mathrm {d} }{\mathrm {d} x}}}$ find ${\displaystyle B}$ representing ${\displaystyle D}$ w.r.t. ${\displaystyle [1,x,x^{2}]}$ and ${\displaystyle A}$ representing ${\displaystyle D}$ w.r.t. ${\displaystyle [1,2x,4x^{2}-2]}$.

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

{\displaystyle {\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 ${\displaystyle B=S^{-1}\,A\,S}$.

Chapter 5: Orthogonality

Scalar Product

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

Scalar product (also called dot product) is defined as

{\displaystyle {\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 ${\displaystyle n}$-dimensional vector is given by

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

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

Theorem 5.1.1

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

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

Footnotes

1. For ${\displaystyle B=U^{-1}\,A\,U}$, we say that ${\displaystyle A}$ and ${\displaystyle B}$ are conjugate by ${\displaystyle U}$.