Improve parantheses in some instances
This commit is contained in:
parent
2dfb8db4a8
commit
6fb48c603f
GPG Key ID:
9303E1C32E3A14A0
52
LinAlg2.tex
52
LinAlg2.tex
|
|
@ -395,11 +395,11 @@ $\varphi$ alternierend und $a_i = a_j$ für $i\neq j \implies \varphi(a_1, \dots
|
|||
$b_i = \sum\lambda_{ij}a_j$, dass
|
||||
\[
|
||||
\varphi(b_1, \dots, b_n) =
|
||||
\varphi(a_1, \dots, a_n)(\sum_{\pi \in S_n}\sgn(\pi)\lambda_{1\pi(1)}\cdots\lambda_{n\pi(n)})
|
||||
\varphi(a_1, \dots, a_n)\left(\sum_{\pi \in S_n}\sgn(\pi)\lambda_{1\pi(1)}\cdots\lambda_{n\pi(n)}\right)
|
||||
\]
|
||||
\item Sei $c\in\K\setminus\{0\}$. Dann ist die Abbildung
|
||||
\[
|
||||
\varphi(b_1, \dots, b_n) = c(\sum_{\pi \in S_n}\sgn(\pi)\lambda_{1\pi(1)}\cdots\lambda_{n\pi(n)})
|
||||
\varphi(b_1, \dots, b_n) = c\left(\sum_{\pi \in S_n}\sgn(\pi)\lambda_{1\pi(1)}\cdots\lambda_{n\pi(n)}\right)
|
||||
\]
|
||||
eine alternierende nicht ausgeartete n-Linearform.
|
||||
\end{enumerate}
|
||||
|
|
@ -410,7 +410,7 @@ $\varphi$ alternierend und $a_i = a_j$ für $i\neq j \implies \varphi(a_1, \dots
|
|||
\item Man verifiziert leicht, dass $\varphi$ n-linear ist. Weiters ist $\varphi$ nicht ausgeartet, da
|
||||
\[
|
||||
\varphi(a_1, \ldots, a_n) =
|
||||
c(\sum_{\pi\in S_n}\sgn(\pi)\delta_{1\pi(1)}, \cdots, \delta_{n\pi(n)}) = c \cdot 1 \neq 0
|
||||
c\left(\sum_{\pi\in S_n}\sgn(\pi)\delta_{1\pi(1)}, \cdots, \delta_{n\pi(n)}\right) = c \cdot 1 \neq 0
|
||||
\]
|
||||
z.Z.: $\varphi$ alternierend. Seien $b_1, \dots, b_n$ linear abhängig.\\
|
||||
O.B.d.A. $b_1=\mu_2b_2+\cdots+\mu_nb_n$. Dann gilt
|
||||
|
|
@ -419,10 +419,10 @@ $\varphi$ alternierend und $a_i = a_j$ für $i\neq j \implies \varphi(a_1, \dots
|
|||
$\varphi(b_1, \dots, b_n) = 0$ falls $b_1 = b_i, i\in\{2, \dots, n\}$.
|
||||
Dann gilt aber $\lambda_{1j}=\lambda_{ij} \forall j$.
|
||||
\begin{align*}
|
||||
\varphi(b_i, \dots, b_i, \dots, b_n) & = c\cdot\sum_{\pi\in S_n} \sgn(\pi) \lambda_{i\pi(1)}\cdots\lambda_{i\pi(i)}\cdots\lambda_{n\pi(n)} \\
|
||||
& =c\cdot \Bigg(\sum_{\pi\in A_n}\sgn(\pi)\lambda_{i\pi(i)}\cdots\lambda_{i\pi(i)}\cdots\lambda_{n\pi(n)} \\
|
||||
& +\sum_{\pi\in A_n}\underbrace{\sgn(\pi\circ(1i))}_{=-\sgn(\pi)}\lambda_{i\pi(i)}\cdots\lambda_{i\pi(i)}\cdots\lambda_{n\pi(n)}\Bigg) \\
|
||||
& =c\cdot\sum_{\pi\in A_n}(\sgn(\pi)-\sgn(\pi)) \cdot \cdots \\
|
||||
\varphi(b_i, \dots, b_i, \dots, b_n) & = c\cdot\sum_{\pi\in S_n} \sgn(\pi) \lambda_{i\pi(1)}\cdots\lambda_{i\pi(i)}\cdots\lambda_{n\pi(n)} \\
|
||||
& \begin{multlined}=c\cdot \Bigg(\sum_{\pi\in A_n}\sgn(\pi)\lambda_{i\pi(i)}\cdots\lambda_{i\pi(i)}\cdots\lambda_{n\pi(n)} \\
|
||||
+\sum_{\pi\in A_n}\underbrace{\sgn(\pi\circ(1i))}_{=-\sgn(\pi)}\lambda_{i\pi(i)}\cdots\lambda_{i\pi(i)}\cdots\lambda_{n\pi(n)}\Bigg)\end{multlined} \\
|
||||
& =c\cdot\sum_{\pi\in A_n}(\sgn(\pi)-\sgn(\pi)) \cdot \cdots \\
|
||||
& \cdot \cdots \lambda_{i\pi(i)}\cdots\lambda_{i\pi(i)}\cdots\lambda_{n\pi(n)}=0
|
||||
\end{align*}
|
||||
\end{enumerate}
|
||||
|
|
@ -1354,16 +1354,16 @@ $A = \begin{pmatrix}
|
|||
\end{flalign}
|
||||
Wegen $b_{ij}(A) = a_{ji} I_N - \delta_{ij}A$ gilt weiters
|
||||
\begin{equation}
|
||||
\forall i \in [n]: \sum_{j=1}^{n} b_{ij}(A) e_j = (\sum_{j=1}^{n} a_{ji} e_j) - A e_i = 0
|
||||
\forall i \in [n]: \sum_{j=1}^{n} b_{ij}(A) e_j = \left(\sum_{j=1}^{n} a_{ji} e_j\right) - A e_i = 0
|
||||
\label{eq:2.2.18.3}
|
||||
\end{equation}
|
||||
Es folgt $\forall k \in [n]$
|
||||
\begin{align*}
|
||||
\chi_A (A) e_k & = \sum_{j=1}^{n} \delta_{jk} \chi(A) e_j & \\
|
||||
\chi_A (A) e_k & = \sum_{j=1}^{n} \delta_{jk} \chi(A) e_j & \\
|
||||
& \underbrace{=}_{\mathclap{\text{\ref{eq:2.2.18.2}}}}
|
||||
\sum_{j=1}^{n} \sum_{i=1}^{n} c_{ki}(A) b_{ij}(A) e_j & \\
|
||||
& = \sum_{i=1}^{n} c_{ki}(A) (\sum_{j=1}^{n} b_{ij(A) e_j}) & \\
|
||||
& \underbrace{=}_{\mathclap{\text{\ref{eq:2.2.18.3}}}} 0 & \\
|
||||
\sum_{j=1}^{n} \sum_{i=1}^{n} c_{ki}(A) b_{ij}(A) e_j & \\
|
||||
& = \sum_{i=1}^{n} c_{ki}(A) \left(\sum_{j=1}^{n} b_{ij(A) e_j}\right) & \\
|
||||
& \underbrace{=}_{\mathclap{\text{\ref{eq:2.2.18.3}}}} 0 & \\
|
||||
& \implies \chi_A(A) = 0
|
||||
\end{align*}
|
||||
\end{proof}
|
||||
|
|
@ -1527,11 +1527,11 @@ $\underset{\mathrlap{\text{\dq fast alle Matrizen sind diagonalisierbar\dq}}}
|
|||
\item [$i>1$:]
|
||||
\begin{align*}
|
||||
& \alpha(c_i) = \alpha(\tilde{b}_i) =
|
||||
\alpha(\sum_{j=2}^n \mu_{ij} b_j)
|
||||
\alpha\left(\sum_{j=2}^n \mu_{ij} b_j\right)
|
||||
= \sum_{j=2}^n \mu_{ij} \alpha(b_j) \\
|
||||
& = \sum_{j=2}^n\mu_{ij}(a_{1j} b_1 + \beta(b_j))
|
||||
= (\underbrace{\sum_{j=2}^n \mu_{ij} a_{1j}}_
|
||||
{\displaystyle\sigma_i})
|
||||
= \underbrace{\left(\sum_{j=2}^n \mu_{ij} a_{1j}\right)}_
|
||||
{\displaystyle\sigma_i}
|
||||
+ \sum_{j=2}^n \mu_{ij}\beta(b_j) \\
|
||||
& = \sigma_i b_1+ \beta(\sum_{j=2}^n \mu_{ij} b_j)
|
||||
= \sigma_i b_1 + \beta(\tilde{b}_i) \\
|
||||
|
|
@ -1649,11 +1649,11 @@ Wir wollen zeigen, dass $\alpha/A$ genau dann eine Jordan-Normalform besitzt, we
|
|||
\item[1)] folgt aus $\alpha(V_{i+1}) \subseteq \alpha(V_i)$
|
||||
\item[3)] Sei $\sum\limits_{i}\mu_i \alpha(c_i^m) \in V_{m-2}$
|
||||
\begin{align*}
|
||||
& \implies \alpha^{m-2}(\sum_{i}\mu_i \alpha(c_i^m)) = 0 \\
|
||||
& \implies \alpha^{m-1} (\sum_{i} \mu_i \alpha(c_i^m)) = 0 \\
|
||||
& \implies \sum \mu_i c_i^m \in V_{m-1} \\
|
||||
& \underbrace{\implies}_{\mathclap{\substack{(c_i^m) \text{ liegen} \\
|
||||
\text{im Komplement} \\
|
||||
& \implies \alpha^{m-2} \left(\sum_{i}\mu_i \alpha(c_i^m)\right) = 0 \\
|
||||
& \implies \alpha^{m-1} \left(\sum_{i} \mu_i \alpha(c_i^m)\right) = 0 \\
|
||||
& \implies \sum \mu_i c_i^m \in V_{m-1} \\
|
||||
& \underbrace{\implies}_{\mathclap{\substack{(c_i^m) \text{ liegen} \\
|
||||
\text{im Komplement} \\
|
||||
\text{von } V_{m-1}}}}
|
||||
\mu_i = 0, \forall i \implies \sum_{i} \mu_i \alpha(c_i^m) = 0
|
||||
\end{align*}
|
||||
|
|
@ -3622,17 +3622,17 @@ Sei $ B= (b_1, \dots, b_n)$ Basis, so ist $M_B(\sigma) := (\sigma(b_i, b_j))_{i,
|
|||
\item Analog wie b)
|
||||
\item $u = \sum \lambda_i b_i, v=\sum \mu_j b_j, A = M_B(\sigma)$
|
||||
\begin{align*}
|
||||
\sigma(u, v) & = \sigma(\sum \lambda_i b_i, \sum \mu_j b_j) = \sum_{i=1}^n \lambda_i
|
||||
\sigma(u, v) & = \sigma\left(\sum \lambda_i b_i, \sum \mu_j b_j\right) = \sum_{i=1}^n \lambda_i
|
||||
\underbrace{\sum_{j=1}^n \underbrace{\sigma(b_i, b_j)}_{a_{ij}} \overline \mu_j}
|
||||
_{A {}_B\overline{\Phi(v)}} \\
|
||||
_{A {}_B\overline{\Phi(v)}} \\
|
||||
& = {}_B \Phi(u)^T M_B(\sigma) {}_B \overline{\Phi(v)}
|
||||
\end{align*}
|
||||
\item Analog wie d)
|
||||
\item $b'_i = \sum_k a_{ki} b_k, M_{B'} = (\sigma(b'_i, b'_j))_{i,j}$
|
||||
\begin{align*}
|
||||
\sigma(b'_i, b'_j) & = \sigma(\sum_k a_{ki} b_k, \sum_l a_{lj} b_l) = \sum_k a_{ki} \sum_l
|
||||
\underbrace{\sigma(b_i,b_j)}_{M_B(\sigma)} \overline{a_{lj}} \\
|
||||
& = (A^T M_B(\sigma)\overline A)_{ij}
|
||||
\sigma(b'_i, b'_j) & = \sigma\left(\sum_k a_{ki} b_k, \sum_l a_{lj} b_l\right)
|
||||
= \sum_k a_{ki} \sum_l \underbrace{\sigma(b_i,b_j)}_{M_B(\sigma)} \overline{a_{lj}} \\
|
||||
& = \left(A^T M_B(\sigma)\overline A\right)_{ij}
|
||||
\end{align*}
|
||||
\item \begin{align*}
|
||||
\underset{\rotatebox{70}{$=$}}{\sigma(v, w)}
|
||||
|
|
@ -4196,7 +4196,7 @@ $Q = \{(x,y) \in \R^2: \psi(x, y) = 0\}$ \\
|
|||
$\psi(x, y):= a_1 x^2 + a_2 xy + a_3 y^2 a_4 x + a_5 y + a_6$ \\
|
||||
Gegeben: $(x_i,y_i)^m_{i=1}$ Suche $x = (a_1, \dots, a_6)^T$ mit $\norm x = 1$ sodass
|
||||
\[
|
||||
\sum_{i=1}^m (a_1 x_i^2 + a_2 x_i y_i + a_3 y_i^2 + a_4 x_i + a_5 y_i + a_6)^2=\norm{Ax}^2
|
||||
\sum_{i=1}^m \left(a_1 x_i^2 + a_2 x_i y_i + a_3 y_i^2 + a_4 x_i + a_5 y_i + a_6\right)^2=\norm{Ax}^2
|
||||
\]
|
||||
minimal.
|
||||
$A = \begin{pmatrix}
|
||||
|
|
|
|||
Loading…
Reference in New Issue