459 lines
17 KiB
TeX
459 lines
17 KiB
TeX
|
|
\subsection{Euclidian n-Spaces}
|
|
|
|
\begin{defn}[Vector Definitions]\ \\
|
|
\begin{itemize}
|
|
\item $n$-vector : $v = (v_1, v_2, \dots, v_n)$
|
|
\item $\vec{PQ} // \vec{P'Q'} \implies \vec{PQ} = \vec{P'Q'}$
|
|
\item $|| \vec{PQ} || = \sqrt{(a_2 - a_1)^2 + (b_2 - b_1)^2}$
|
|
\item $u + v = (u_1 + v_1, u_2 + v_2), u = (u_1, u_2), v = (v_1, v_2)$
|
|
\item $n$-vector can be viewed as a row matrix / column matrix
|
|
\item $\mathbb{R}^n = \left\{ (v_1, v_2, \dots, v_n) | v_1, v_2, \dots, v_n \right\} \in \mathbb{R}$, Euclidean $n$-space
|
|
\end{itemize}
|
|
\end{defn}
|
|
|
|
A linear system is given in implicit form.
|
|
|
|
$\begin{cases}
|
|
a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1 \\
|
|
a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2 \\
|
|
\vdots \\
|
|
a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n = b_m \\
|
|
\end{cases}$
|
|
|
|
and its general solution is in the explicit form
|
|
|
|
\begin{defn} Straight lines in $\mathbb{R}^2$
|
|
\begin{itemize}
|
|
\item Implicit: $\left\{(x,y)|ax+by=c\right\}$
|
|
\item Explicit: (Equation Form)
|
|
\begin{itemize}
|
|
\item If $a \neq 0$, then $\left\{\left(\dfrac{c-bt}{a},t\right)| t \in \mathbb{R}\right\}$
|
|
\item If $b \neq 0$, then $\left\{\left(s,\dfrac{c-as}{b}\right)| s \in \mathbb{R}\right\}$
|
|
\end{itemize}
|
|
\item Explicit: (Vector form)
|
|
\begin{itemize}
|
|
\item A point on the line $(x_0, y_0)$ and its direction vector $(a,b)\neq 0$
|
|
\item $(x_0, y_0) + t(a,b)$
|
|
\item $\left\{(x_0 + ta, y_0 + tb|t \in \mathbb{R}\right\}$
|
|
\end{itemize}
|
|
\end{itemize}
|
|
\end{defn}
|
|
|
|
\begin{defn} Planes in $\mathbb{R}^3$
|
|
\begin{itemize}
|
|
\item Implicit: $\left\{(x,y,z)|ax+by+cz=d\right\}$
|
|
\item Explicit: (Equation Form)
|
|
\begin{itemize}
|
|
\item If $a \neq 0$, then $\left\{\left(\dfrac{c-bs-ct}{a},s,t\right)| s,t \in \mathbb{R}\right\}$
|
|
\item If $b \neq 0$, then $\left\{\left(s,\dfrac{d-as-ct}{b},t\right)| s,t \in \mathbb{R}\right\}$
|
|
\item If $c \neq 0$, then $\left\{\left(s,t,\dfrac{d-as-bt}{c}\right)| s,t \in \mathbb{R}\right\}$
|
|
\end{itemize}
|
|
\item Explicit: (Vector Form)
|
|
\begin{itemize}
|
|
\item $\left\{(x_0, y_0, z_0) + s(a_1, b_1,c_1) + t(a_2,b_2,c_2)|s,t \in \mathbb{R}\right\}$
|
|
\item $(a_1,b_1,c_1)$ and $(a_2, b_2, c_2)$ are non-parallel vectors, parallel to the plane
|
|
\end{itemize}
|
|
\end{itemize}
|
|
\end{defn}
|
|
|
|
Example: Plane is given by $\left\{(1+s-t, 2+s-2t,4-s-3t)|s,t\in\mathbb{R}\right\}$
|
|
\begin{itemize}
|
|
\item Let $x=1+s-t,y=2+s-2t,z=4-s-3t$
|
|
\item $\begin{amatrix}{2}
|
|
1&-1&x-1\\
|
|
1&-2&y-2\\
|
|
-1&-3&z-4
|
|
\end{amatrix} \to \begin{amatrix}{2}
|
|
1&-1&x-1\\
|
|
0&-1&-x + y-1\\
|
|
0&0&5x-4y+z-1
|
|
\end{amatrix}$
|
|
\item For system to be consistent, $5x-4y+z = 1$
|
|
\item Implicit: $\left\{(x,y,z)|5x-4y+z=1\right\}$
|
|
\end{itemize}
|
|
|
|
\begin{defn} Lines in $\mathbb{R}^3$ is the intersection of 2 non-parallel planes
|
|
\begin{itemize}
|
|
\item Implicit: $\left\{(x,y,z) | a_1x + b_1y + c_1z = d_1 \text{ and } a_2x+b_2y+c_2z=d_2\right\}$
|
|
\item Explicit $\left\{(x_0 + ta, y_0 + tb, z_0 + tc) | t \in \mathbb{R}\right\}$
|
|
\end{itemize}
|
|
\end{defn}
|
|
|
|
It is easy to go from implicit to explicit form, by just solving the linear equation. To have an implicit form of line, we need to find 2 non parallel planes $a_ix + b_iy+c_iz = d_i (i=1,2)$ containing the line
|
|
|
|
Example: Line is $\{(t-2, -2t+3,t+1) | t \in \mathbb{R}\}$.
|
|
|
|
\begin{itemize}
|
|
\item $t=x+2, -2t = y-3, t=z-1$
|
|
\item $\begin{amatrix}{1}
|
|
1&x+2\\
|
|
-2&y-3\\
|
|
1&z-1
|
|
\end{amatrix} \to \begin{amatrix}{1}
|
|
1&x+2\\
|
|
0&2x + y + 1\\
|
|
0&-x + z - 3
|
|
\end{amatrix}$
|
|
\item Implicit Form: $\{(x,y,z) | 2x+y+1 = 0 \text{ and } -x+z-3 = 0$
|
|
|
|
\end{itemize}
|
|
|
|
|
|
\subsection{Linear Combinations and Linear Spans}
|
|
|
|
\begin{defn} Linear Combination
|
|
\begin{itemize}
|
|
\item Linear combination of $v_1, v_2, \dots, v_k$ has the form
|
|
\item $c_1v_1 + c_2v_2 + \dots + c_kv_k, c_1,c_2,\dots,c_k \in \mathbb{R}$
|
|
\item $0$ is always a linear combination of $v_1, v_2, \dots, v_k$
|
|
\item to check if $v$ is a linear combination of $v_1, v_2, v_3$, solve for $(v_1, v_2, v_3 | v)$ and check if the REF is consistent
|
|
\end{itemize}
|
|
\end{defn}
|
|
|
|
\begin{defn} Linear Span
|
|
\begin{itemize}
|
|
\item Let $S = \{v_1, v_2, \dots, v_k\}$ be a subset of $\mathbb{R}^n$
|
|
\item Set of all linear combinations of $v_1, v_2,\dots,v_k$
|
|
\item $\{c_1v_1+ c_2v_2+\dots+c_kv_k | c_1,c_2,\dots,c_k \in \mathbb{R}$
|
|
\item is called the Span of $S$, $\text{Span}(S)$
|
|
\end{itemize}
|
|
\end{defn}
|
|
|
|
Example:
|
|
|
|
\begin{itemize}
|
|
\item Let $S = \{(2,1,3), (1,-1,2),(3,0,5)\}$
|
|
\subitem $(3,3,4) \in \text{Span}(S), (1,2,4) \not\in \text{Span}(S)$
|
|
\item Let $S = \{(1,0,0), (0,1,0),(0,0,1)\}$
|
|
\subitem for any $(x, y, z) \in \mathbb{R}^3, (x,y,z) = x(1,0,0) + y(0,1,0)+z(0,0,1)$
|
|
\subitem Therefore, $\text{Span}(S) = \mathbb{R}^3$
|
|
\end{itemize}
|
|
|
|
More Examples:
|
|
|
|
|
|
\begin{itemize}
|
|
\item Let $S = \{(1,0,0,-1),(0,1,1,0)\}$ be subset of $\mathbb{R}^4$
|
|
\begin{itemize}
|
|
\item $a(1,0,0-1) + b(0,1,1,0) = (a,b,b,-a), (a,b \in \mathbb{R})$
|
|
\item span(S) = $\{(a,b,b,-a) | a,b \in \mathbb{R}$
|
|
\end{itemize}
|
|
\item Let $V = \{(2a+b,a,3b-a) | a,b \in \mathbb{R} \} \subseteq \mathbb{R}^3$
|
|
\begin{itemize}
|
|
\item $a(2a+b,a,3b-a) = a(2,1,-1) + b(1,0,3), (a,b \in \mathbb{R})$
|
|
\item span(V) = $\{(a,b,b,-a) | a,b \in \mathbb{R}$
|
|
\item $V = \text{span}\{(2,1,-1),(1,0,3)\}$
|
|
\end{itemize}
|
|
\item Prove that span$\{(1,0,1),(1,1,0),(0,1,1)\} = \mathbb{R}^3$
|
|
\begin{itemize}
|
|
\item It is clear span$\{(1,0,1),(1,1,0),(0,1,1)\} \subseteq \mathbb{R}^3$
|
|
\item let $(x,y,z) \in \mathbb{R}^3$. Show that there exists $a,b,c \in \mathbb{R}$ s.t.
|
|
\begin{itemize}
|
|
\item $(x,y,z) = a(1,0,1) + b(1,1,0) + c(0,1,1)$
|
|
\item Do gaussian Elimination on $(1,0,1),(1,1,0),(0,1,1) | (x,y,z)$
|
|
\item if the system is always consistent then span$\{\dots\} = \mathbb{R}$
|
|
\item IF the system is consistent $\iff$ condition, then $\not \subseteq \mathbb{R}^3$
|
|
\end{itemize}
|
|
\end{itemize}
|
|
\end{itemize}
|
|
|
|
\begin{defn} Criterion for Span$(S) = \mathbb{R}^n$
|
|
\begin{itemize}
|
|
\item Let $S = \{v_1,v_2,\dots,\v_k\} \subseteq \mathbb{R}^n$
|
|
\item for an arbitrary $v \in \mathbb{R}^n$, we shall check the consistency of the equation $c_1v_1 + v_2v_2 + \dots + c_kv_k = v$
|
|
\item View $v_j$ as column vectors, $A = \left(v_1\ v_2\ \dots\ v_k\right)$
|
|
\subitem The equation is $Ax=v$
|
|
\item Let $R$ be a REF of $A$
|
|
\subitem $(A | v) \to (R | v')$
|
|
\subitem Since $v \in \mathbb{R}^n$ is arbitrary, $v' \in \mathbb{R}^n$ is also arbitrary
|
|
\subitem span$(S) = \mathbb{R}^n \iff Ax = v$ is consistent for every $v \in \mathbb{R}^n$
|
|
\subitem span$(S) = \mathbb{R}^n \iff Rx = v'$ is consistent for every $v' \in \mathbb{R}^n$
|
|
\subitem span$(S) = \mathbb{R}^n \iff$ rightmost column of $(R | v')$ is non pivot for any $v' \in \mathbb{R}^n$
|
|
\subitem span$(S) = \mathbb{R}^n \iff$ All rows in $R$ are nonzero
|
|
\end{itemize}
|
|
|
|
TLDR:
|
|
|
|
\begin{enumerate}
|
|
\item Let $S = \{v_1,v_2,\dots,\v_k\} \subseteq \mathbb{R}^n$
|
|
\item View $v_j$ as column vectors, $A = \left(v_1\ v_2\ \dots\ v_k\right)$
|
|
\item Find REF $R$ of $A$
|
|
\subitem If $R$ has zero row, then span$(S) \neq \mathbb{R}^n$
|
|
\subitem If $R$ has no zero row, then span$(S) = \mathbb{R}^n$
|
|
\end{enumerate}
|
|
|
|
Other rules
|
|
|
|
\begin{itemize}
|
|
\item $\mathbb{R}^n$ cannot be spanned by $n-1$ vectors
|
|
\subitem $\mathbb{R}^3$ cannot be spanned by 2 vectors
|
|
\end{itemize}
|
|
\end{defn}
|
|
|
|
\begin{defn} Properties of Linear Spans
|
|
\begin{itemize}
|
|
\item $0 \in $span(S), span(S) $\neq \emptyset$
|
|
\item $v \in $span(S) and $c \in \mathbb{R} \to cv \in$ span(S).
|
|
\item $u \in$ span(S) and $v \in$ span(S) $\to u+v \in$ span(S).
|
|
\end{itemize}
|
|
|
|
Check if $\text{span}(S_1) \subseteq \text{span}S_2$
|
|
\begin{itemize}
|
|
\item Let $S = \{v_1,v_2,\dots,\v_k\} \subseteq \mathbb{R}^n$
|
|
\item View $v_j$ as column vectors, $A = \left(v_1\ v_2\ \dots\ v_k\right)$
|
|
\item Check whether $Ax = u$, where $u$ is one of the vectors in $S_1$
|
|
\subitem If $Ax=u$ is consistent, u $\subseteq$ span(S)
|
|
\subitem If $Ax=u$ is inconsistent, u $\not\subseteq$ span(S)
|
|
\end{itemize}
|
|
\end{defn}
|
|
|
|
\subsection{Subspaces}
|
|
|
|
|
|
\begin{defn} Subspaces
|
|
\begin{itemize}
|
|
\item Let $V \subseteq \mathbb{R}^n$. Then $V$ is the subspace of $\mathbb{R}^n$
|
|
\item If there exists $v_1, \dots, v_k \in \mathbb{R}^n$, then V is the subspace spanned by $S = \{v_1, \dots, v_k\}$.
|
|
\end{itemize}
|
|
|
|
To validate if $V$ is subspace of $\mathbb{R}^n$
|
|
\begin{itemize}
|
|
\item $0 \in V$
|
|
\item $c \in \mathbb{R}$ and $v \in V \to cv \in V$
|
|
\item $u \in V$ and $v \in V \to u+v \in V$
|
|
\end{itemize}
|
|
\end{defn}
|
|
|
|
\begin{defn} Subspaces of $\mathbb{R}^1,\mathbb{R}^2,\mathbb{R}^3$
|
|
\end{defn}
|
|
|
|
\begin{defn} Solution Space
|
|
|
|
\end{defn}
|
|
|
|
|
|
\subsection{Linear Independence}
|
|
|
|
\begin{defn} Linear Independence
|
|
\begin{itemize}
|
|
\item Let $S=\{v_1,\dots,v_k\}$ be a subset of $\mathbb{R}^n$
|
|
\item Equation $c_1v_1 + \dots + c_kv_k = 0$ has trivial solution $c_1 = \dots = c_k = 0$
|
|
\item If equation has non-trivial solution, then
|
|
\begin{itemize}
|
|
\item $S$ is a linearly dependent set
|
|
\item $v_1, \dots, v_k$ is a linearly dependent set
|
|
\item Exists $c_1,\dots,c_k \in \mathbb{R}$ not all zero s.t. $c_1v_1 + \dots + c_kv_k = 0$
|
|
\end{itemize}
|
|
\item If equation has only the trivial solution, then
|
|
\begin{itemize}
|
|
\item S is linearly independent set
|
|
\item $v_1,\dots,v_k$ are linearly independent
|
|
\end{itemize}
|
|
\end{itemize}
|
|
|
|
How do you calculate whether trivial or non trivial? Solve for $Ax = 0$, perform gaussian elimination and identify if non-pivot columns exist. If there are non-pivot columns, then there are infintely many solutions, and thus, linearly dependent. If all columns are pivot, then system has only trivial solution, and thus, linearly independent set.
|
|
\end{defn}
|
|
|
|
\begin{defn} Properties of Linear Independence \\
|
|
Let $S_1$ and $S_2$ be finite subsets of $\mathbb{R}^n$ s.t. $S_1 \subseteq S_2$
|
|
\begin{itemize}
|
|
\item $S_1$ linearly dependent $\to$ $S_2$ linearly dependent
|
|
\item $S_2$ linearly independent $\to$ $S_1$ linearly independent
|
|
\end{itemize}
|
|
|
|
Let $S = \{v_1, v_2, \dots, v_k \} \subseteq \mathbb{R}^n, k \geq 2$
|
|
\begin{itemize}
|
|
\item $S$ is linearly dependent $\iff v_i$ is a linear combination of other vectors in $S$
|
|
\item S is linearly independent $\iff$ no vector in $S$ can be written as a linear combination of other vectors
|
|
\end{itemize}
|
|
|
|
Suppose $S = \{v_1, v_2, \dots, v_k\}$ is linearly dependent
|
|
\begin{itemize}
|
|
\item Let $V = \text{span}(S)$
|
|
\item If $v_i \in S$ is a linear combination of other vectors, remove $v_i$ from $S$.
|
|
\item Repeat until we obtain linearly independent set $S'$.
|
|
\item span$(S') = V$ and $S'$ has no redundant vector to span $V$.
|
|
\end{itemize}
|
|
|
|
Let $S = \{v_1, v_2, \dots, v_k \} \subseteq \mathbb{R}^n$ be linearly independent
|
|
|
|
\begin{enumerate}
|
|
\item Suppose span$(S) \neq \mathbb{R}^n$
|
|
\item pick $v_{k+1} \in \mathbb{R}^n$ but $v_{k+1} \not\in$ span$(S) \neq \mathbb{R}^n$
|
|
\item $\{v_1, ..., v_k, v_{k+1}\}$ is linearly independent
|
|
\item Repeat until $\{v_1, ..., v_k, ..., v_m\}$ is linearly independent and span$(S') = \mathbb{R}^n$
|
|
\end{enumerate}
|
|
\begin{itemize}
|
|
\item If $m > n$ then $S$ is linearly dependent
|
|
\item If $m < n$, then $S$ cannot span $\mathbb{R}^n$
|
|
\item If $m = n$, then $S$ is linearly independent and spans $\mathbb{R}^n$
|
|
\end{itemize}
|
|
\end{defn}
|
|
|
|
\begin{defn} Vector Spaces
|
|
\begin{itemize}
|
|
\item $V$ is vector space if $V$ is subspace of $\mathbb{R}^n$
|
|
\item $W$ and $V$ are vector space such that $W \subseteq V$, $W$ is a subspace of $V$
|
|
\end{itemize}
|
|
\end{defn}
|
|
|
|
\subsection{Bases}
|
|
\subsubsection{Definition}
|
|
$S$ is basis for $V$ if $S$ is
|
|
\begin{enumerate}
|
|
\item Linearly Independent
|
|
\item Span$(S) = V$
|
|
\end{enumerate}
|
|
|
|
\begin{note}
|
|
To show that Vector $S$ is a basis vector for $\mathbb{R}^n$, show that $S$ is linearly independent.
|
|
|
|
$S \xrightarrow[\text{Elimination}]{\text{Gaussian}} R$
|
|
\begin{enumerate}
|
|
\item Linear Independence
|
|
\begin{enumerate}
|
|
\item Show All columns are pivot. $\therefore$ system has only trivial solution
|
|
\item $S$ is linearly independent
|
|
\end{enumerate}
|
|
\item Span$(S) = \mathbb{R}^n$
|
|
\begin{enumerate}
|
|
\item REF has no zero row
|
|
\item span$(S) = \mathbb{R}^n$
|
|
\end{enumerate}
|
|
\item We can conclude $S$ is basis for $\mathbb{R}^n$
|
|
\end{enumerate}
|
|
|
|
\end{note}
|
|
|
|
|
|
Basis for Vector space $V$ contains
|
|
\begin{itemize}
|
|
\item Smallest possible number of vectors that spans $V$
|
|
\item largest possible number of vectors that is linearly independent $V$
|
|
\end{itemize}
|
|
|
|
\subsubsection{Coordinate Vector}
|
|
\begin{theorem} Coordinate Vectors
|
|
|
|
\begin{itemize}
|
|
\item Let $S = \{v_1, \dots, v_k \}$ be a subset of vector space $V$
|
|
|
|
$S$ is basis for $V$ $\iff$ every vector in $V$ can be written as $v = c_1v_1 + \dots + c_kv_k$
|
|
\item Let $S = \{v_1, \dots, v_k \}$ be a basis for vector space $V$
|
|
|
|
For every $v \in V$, there exists a unique $c_1, \dots, c_k \in \mathbb{R}$ such that $v = c_1v_1 + \dots + c_kv_k$
|
|
\end{itemize}
|
|
|
|
$(v)_S = (c_1, \dots, c_k)$ is the coordinate vector of $v$ relative to $S$.
|
|
|
|
Column vector $[v]_S = \begin{pmatrix}c_1\\c_2\\\vdots\\c_k\end{pmatrix}$ is also coordinate vector
|
|
|
|
Let $a = (v_1 \; \dots \; v_k)$. Then $[v]_S$ is the unique solution to $Ax = v$. We can write $A[v]_S = v$
|
|
\end{theorem}
|
|
|
|
\noindent To calculate Coordinate vector for $v$ relative to $S$, then view each vector in $S$ as a column vector, and let $A = (v_1 \: \dots \: v_k)$ and solve for $Ax=v$
|
|
|
|
\begin{note} Criterion for bases
|
|
|
|
Let $T = \{v_1, \dots, v_k\}$ be subset of $\mathbb{R}^n$
|
|
\begin{itemize}
|
|
\item $k > n$, then $T$ is linearly dependent
|
|
\item $k < n$, then span$(T) \neq \mathbb{R}^n$
|
|
\end{itemize}
|
|
If $T$ is basis for $\mathbb{R}^n$, then $k=n$
|
|
|
|
Let $V$ be a vector space having basis $S$ with $|S| = N$
|
|
\begin{itemize}
|
|
\item Let $T = \{v_1, \dots, v_k\}$be subset of $V$
|
|
\item if $k > n$, then $\{(v_1)_S, \dots, (v_k)_S\}$ is linearly dependent on $\mathbb{R}^n \therefore T$ is linearly dependent on $V$
|
|
\item if $k < n$, then span$(\{(v_1)_S, \dots, (v_k)_S\}) \neq \mathbb{R}^n \therefore$ span$(T) \neq V$
|
|
\end{itemize}
|
|
If $T$ is a basis for $V$, then $|T| = n = |S|$. If $S$ and $T$ are bases for vector space $V$ then $|S| = |T|$
|
|
\end{note}
|
|
|
|
\subsection{Dimensions}
|
|
Let $V$ be a vector space and $S$ be basis for $V$. $\dim(V) = |S|$
|
|
|
|
\subsubsection{Examples}
|
|
\begin{itemize}
|
|
\item $\varnothing$ is basis for $\{0\}$, $\dim(\{0\}) = |\varnothing| = 0$
|
|
\item $\mathbb{R}^n$ has standard basis $E = \{e_1,\dots,e_n\}, \dim(\mathbb{R}^n) = n$
|
|
\end{itemize}
|
|
|
|
\subsubsection{Dimension of Solution Space}
|
|
|
|
Let $Ax=0$ be a homogeneous linear system.
|
|
|
|
Solution set of $Ax = 0$ is a vector space $V$.
|
|
|
|
Let $R$ be REF of $A$. The \# non pivot columns = \# arbitrary params = dimension of $V$.
|
|
|
|
\subsubsection{Properties of Dimensions}
|
|
|
|
\begin{theorem} Dimensions
|
|
|
|
Let $S$ be a subset of vector space $V$, the following are equivalent
|
|
\begin{itemize}
|
|
\item $S$ is basis for $V$
|
|
\item $S$ is linearly independent and $|S| = \dim(V)$
|
|
\item $S$ spans $V$ and $|S| = \dim(V)$
|
|
\end{itemize}
|
|
|
|
Let $U$ be subspace of $V$. Then $\dim(U) \leq \dim(V)$
|
|
\begin{itemize}
|
|
\item $U = V \iff \dim(U) = \dim(V)$
|
|
\item $U \neq V \iff \dim(U) < \dim(V)$
|
|
\end{itemize}
|
|
|
|
Let $A$ be a square matrix of order $n$
|
|
\begin{itemize}
|
|
\item $A$ is invertible
|
|
\item $Ax=b$ has unique solution
|
|
\item $Ax=0$ has only trivial solution
|
|
\item RREF of $A$ is $I_n$
|
|
\item $\det(A) \neq 0$
|
|
\item rows of $A$ form basis for $\mathbb{R}^n$
|
|
\item columns of $A$ form basis for $\mathbb{R}^n$
|
|
\end{itemize}
|
|
\end{theorem}
|
|
|
|
\subsection{Transition Matrices}
|
|
\begin{defn} Let $V$ be vector space and
|
|
$S = \{u_1, \dots, u_k\}$ and $T$ be bases for $V$.
|
|
\begin{itemize}
|
|
\item $P = ([u_1]_T \dots [u_k]_T)$ is the transition matrix from $S$ to $T$
|
|
\item $P[w]_S = [w]_T,$ $\forall w \in V$
|
|
\end{itemize}
|
|
|
|
Let $S_1, S_2, S_3$ be bases for vector space $V$
|
|
\begin{itemize}
|
|
\item $P$ be transition matrix from $S_1$ to $S_2$
|
|
\item $Q$ be transition matrix from $S_2$ to $S_3$
|
|
\item $[v]_{S_1} \xrightarrow{P} [v]_{S_2} \xrightarrow{Q} [v]_{S_3}$
|
|
\item $[v]_{S_3} = Q[v]_{S_2} = QP[v]_{S_1}$
|
|
\item $QP$ is transition matrix from $S_1$ to $S_3$
|
|
\end{itemize}
|
|
|
|
Let $S, T$ be bases for vector space $V$
|
|
\begin{itemize}
|
|
\item $P$ be transition matrix from $S$ to $T$
|
|
\item $P$ is invertible
|
|
\item $P^{-1}$ is transition matrix from $T$ to $S$
|
|
\end{itemize}
|
|
\end{defn}
|
|
|
|
To calc transition matrices for $S$ and $T$, given that $S = \{(1, 1), (1, -1)\} = \{u_1, u_2\}, T = \{(1, 0), (1, 1) \} = \{v_1, v_2\}$
|
|
|
|
$ (v_1 v_2 | u_1 u_2) =
|
|
\left(\begin{array}{@{}*{2}{c}|c|c@{}}
|
|
1 & 1 & 1 & 1 \\
|
|
0 & 1 & 1 & 0 \\
|
|
\end{array}\right) \xrightarrow{R_1 - R_2}
|
|
\left(\begin{array}{@{}*{2}{c}|c|c@{}}
|
|
1 & 0 & 0 & 2 \\
|
|
0 & 1 & 1 & -1 \\
|
|
\end{array}\right) $
|
|
|
|
Transition matrix from $S$ to $T$: $P = \begin{pmatrix}0 & 2 \\ 1 & -1\end{pmatrix}$
|