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| \documentclass{../../../lecture} | |||
| \documentclass[uebung]{../../../lecture} | |||
| \title{Lineare Algebra I: Übungsblatt 8} | |||
| \author{Christian Merten, Mert Biyikli} | |||
| \begin{document} | |||
| \punkte | |||
| \begin{aufgabe} | |||
| \begin{enumerate}[(a)] | |||
| \item Beh.: $\underline{w}$ ist eine Basis von $W = K[X]_{\le 3}$ | |||
| \begin{proof} | |||
| Zz.: $\underline{w}$ ist linear unabhängig | |||
| Seien $a, b, c, d \in K$ mit | |||
| \begin{align*} | |||
| a X^{0} + b (X^{0} + X^{1}) + c (X^{1} - X^{2} + X^{3} + d (X^{3} + X^{0}) &= 0 \\ | |||
| \implies X^{0}(a + b + d) + X^{1} (b + c) + X^{2} (-c) + X^{3}(c + d) = 0 | |||
| .\end{align*} Wegen $\underline{v}$ linear unabhängig, folgt: | |||
| \begin{align*} | |||
| c = 0 \implies d = 0 \implies b = 0 \implies a = 0 | |||
| .\end{align*} | |||
| Zz.: $\underline{w}$ ist Erzeugendensystem | |||
| Sei $v \in K[X]_{\le 3}$ beliebig, dann ex. $a, b, c, d \in K$ wegen $\underline{v}$ Basis | |||
| s.d. $v = a X^{0} + b X^{1} + c X^{2} + d X^{3} $. | |||
| Wähle nun $\alpha := a - b - 2c - d, \beta := b + c, \gamma := -c, \delta := c+d$. | |||
| Damit folgt direkt: | |||
| \begin{align*} | |||
| v &= \alpha X^{0} + \beta (X^{0} + x^{1}) + \gamma (X^{1} - X^{2} + X^{3}) + | |||
| \delta (X^{3} + X^{0}) \\ | |||
| &= a X^{0} + b X^{1} + c X^{2} + d X^{3} | |||
| .\end{align*} | |||
| \end{proof} | |||
| \item Seien $\phi_{\underline{v}}\colon K^{4} \to K[X]_{\le 3}$ und | |||
| $\phi_{\underline{w}}\colon K^{4} \to K[X]_{\le 3}$ die kanonischen Isomorphismen. | |||
| \begin{enumerate}[(i)] | |||
| \item | |||
| \[ | |||
| M_{\underline{v}}^{\underline{v}}(\partial) = A := | |||
| \begin{pmatrix} | |||
| 0 & 1 & 0 & 0 \\ | |||
| 0 & 0 & 2 & 0 \\ | |||
| 0 & 0 & 0 & 3 \\ | |||
| 0 & 0 & 0 & 0 \\ | |||
| \end{pmatrix} | |||
| .\] | |||
| \begin{proof} | |||
| Zu zeigen.: $F_{\underline{v}}^{\underline{v}}(A) = \partial$. | |||
| Zu überprüfen für die vier Basisvektoren von $K[X]_{\le 3}$ aus $\underline{v}$. | |||
| \[ | |||
| F_{\underline{v}}^{\underline{v}}(A) = | |||
| \phi_{\underline{v}} \circ F_{4,4}(A) \circ \phi_{\underline{v}}^{-1} | |||
| .\] | |||
| \begin{enumerate} | |||
| \item $v_1 = X_0$, $\phi_{\underline{v}}^{-1}(X^{0}) = (1, 0, 0, 0)$ | |||
| \[ | |||
| \begin{pmatrix} | |||
| 0 & 1 & 0 & 0 \\ | |||
| 0 & 0 & 2 & 0 \\ | |||
| 0 & 0 & 0 & 3 \\ | |||
| 0 & 0 & 0 & 0 \\ | |||
| \end{pmatrix} \cdot | |||
| \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} | |||
| .\] | |||
| $\implies \phi_{\underline{v}}(0, 0,0,0) = 0 = \partial(X_0)$ | |||
| \item $v_2 = X_1$, $\phi_{\underline{v}}^{-1}(X^{1}) = (0, 1, 0, 0)$ | |||
| \[ | |||
| \begin{pmatrix} | |||
| 0 & 1 & 0 & 0 \\ | |||
| 0 & 0 & 2 & 0 \\ | |||
| 0 & 0 & 0 & 3 \\ | |||
| 0 & 0 & 0 & 0 \\ | |||
| \end{pmatrix} \cdot | |||
| \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0\end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} | |||
| .\] | |||
| $\implies \phi_{\underline{v}}(1, 0, 0, 0) = X^{0} = \partial(X_1)$ | |||
| \item $v_3 = X_2$, $\phi_{\underline{v}}^{-1}(X^{2}) = (0, 0, 1, 0)$ | |||
| \[ | |||
| \begin{pmatrix} | |||
| 0 & 1 & 0 & 0 \\ | |||
| 0 & 0 & 2 & 0 \\ | |||
| 0 & 0 & 0 & 3 \\ | |||
| 0 & 0 & 0 & 0 \\ | |||
| \end{pmatrix} \cdot | |||
| \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0\end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 0 \\ 0 \end{pmatrix} | |||
| .\] | |||
| $\implies \phi_{\underline{v}}(0, 2, 0, 0) = 2X^{1} = \partial(X_2)$ | |||
| \item $v_4 = X_3$, $\phi_{\underline{v}}^{-1}(X^{3}) = (0, 0, 0, 1)$ | |||
| \[ | |||
| \begin{pmatrix} | |||
| 0 & 1 & 0 & 0 \\ | |||
| 0 & 0 & 2 & 0 \\ | |||
| 0 & 0 & 0 & 3 \\ | |||
| 0 & 0 & 0 & 0 \\ | |||
| \end{pmatrix} \cdot | |||
| \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 3 \\ 0 \end{pmatrix} | |||
| .\] | |||
| $\implies \phi_{\underline{v}}(0, 0, 3, 0) = 3X^{2} = \partial(X_3)$ | |||
| \end{enumerate} | |||
| $\implies F_{\underline{v}}^{\underline{v}}(A) = \partial$ | |||
| \end{proof} | |||
| \item | |||
| \[ | |||
| M_{\underline{w}}^{\underline{v}}(id_W) = A := | |||
| \begin{pmatrix} | |||
| 1 & -1 & -2 & -1 \\ | |||
| 0 & 1 & 1 & 0 \\ | |||
| 0 & 0 & -1 & 0 \\ | |||
| 0 & 0 & 1 & 1 \\ | |||
| \end{pmatrix} | |||
| .\] | |||
| \begin{proof} | |||
| Zu zeigen.: $F_{\underline{w}}^{\underline{v}}(A) = id$ | |||
| Zu überprüfen für die vier Basisvektoren von $K[X]_{\le 3}$ aus $\underline{v}$. | |||
| \[ | |||
| F_{\underline{w}}^{\underline{v}}(A) = | |||
| \phi_{\underline{w}} \circ F_{4,4}(A) \circ \phi_{\underline{v}}^{-1} | |||
| .\] | |||
| \begin{enumerate} | |||
| \item $v_1 = X_0$, $\phi_{\underline{v}}^{-1}(X^{0}) = (1, 0, 0, 0)$ | |||
| \[ | |||
| \begin{pmatrix} | |||
| 1 & -1 & -2 & -1 \\ | |||
| 0 & 1 & 1 & 0 \\ | |||
| 0 & 0 & -1 & 0 \\ | |||
| 0 & 0 & 1 & 1 \\ | |||
| \end{pmatrix} \cdot | |||
| \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0\end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} | |||
| .\] | |||
| $\implies \phi_{\underline{w}}(1, 0,0,0) = X^{0} = id_W(X^{0})$ | |||
| \item $v_2 = X_1$, $\phi_{\underline{v}}^{-1}(X^{1}) = (0, 1, 0, 0)$ | |||
| \[ | |||
| \begin{pmatrix} | |||
| 1 & -1 & -2 & -1 \\ | |||
| 0 & 1 & 1 & 0 \\ | |||
| 0 & 0 & -1 & 0 \\ | |||
| 0 & 0 & 1 & 1 \\ | |||
| \end{pmatrix} \cdot | |||
| \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0\end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 0 \\ 0 \end{pmatrix} | |||
| .\] | |||
| $\implies \phi_{\underline{w}}(-1, 1, 0, 0) = -X^{0} + X^{0} + X^{1} = X^{1} = id_W(X^{1})$ | |||
| \item $v_3 = X_2$, $\phi_{\underline{v}}^{-1}(X^{2}) = (0, 0, 1, 0)$ | |||
| \[ | |||
| \begin{pmatrix} | |||
| 1 & -1 & -2 & -1 \\ | |||
| 0 & 1 & 1 & 0 \\ | |||
| 0 & 0 & -1 & 0 \\ | |||
| 0 & 0 & 1 & 1 \\ | |||
| \end{pmatrix} \cdot | |||
| \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0\end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ -1 \\ 1 \end{pmatrix} | |||
| .\] | |||
| $\implies \phi_{\underline{v}}(-2, 1, -1, 1) = -2X^{0} + X^{0} + X^{1} - X^{1} + X^{2} - X^{3} + X^{3} + X^{0} = X^{2} = id_W(X^{2})$ | |||
| \item $v_4 = X_3$, $\phi_{\underline{v}}^{-1}(X^{3}) = (0, 0, 0, 1)$ | |||
| \[ | |||
| \begin{pmatrix} | |||
| 1 & -1 & -2 & -1 \\ | |||
| 0 & 1 & 1 & 0 \\ | |||
| 0 & 0 & -1 & 0 \\ | |||
| 0 & 0 & 1 & 1 \\ | |||
| \end{pmatrix} \cdot | |||
| \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1\end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 0 \\ 1 \end{pmatrix} | |||
| .\] | |||
| $\implies \phi_{\underline{v}}(-1, 0, 0, 1) = -X^{0} + X^{3} + X^{0} = X^{3} = id_W(X^{3})$ | |||
| \end{enumerate} | |||
| $\implies F_{\underline{w}}^{\underline{v}}(A) = id_W$ | |||
| \end{proof} | |||
| \item | |||
| \[ | |||
| M_{\underline{v}}^{\underline{w}}(id_W) = A := | |||
| \begin{pmatrix} | |||
| 1 & 1 & 0 & 1 \\ | |||
| 0 & 1 & 1 & 0 \\ | |||
| 0 & 0 & -1 & 0 \\ | |||
| 0 & 0 & 1 & 1 \\ | |||
| \end{pmatrix} | |||
| .\] | |||
| \begin{proof} | |||
| Erfolgt analog zu (ii). | |||
| \end{proof} | |||
| \item | |||
| \[ | |||
| M_{\underline{w}}^{\underline{w}}(\partial) = M_{\underline{w}}^{\underline{v}}(id_W) \cdot | |||
| M_{\underline{v}}^{\underline{v}}(\partial) \cdot M_{\underline{v}}^{\underline{w}}(id_W)= | |||
| \begin{pmatrix} | |||
| 0 & 1 & -3 & -6 \\ | |||
| 0 & 0 & 1 & 3 \\ | |||
| 0 & 0 & -3 & -3 \\ | |||
| 0 & 0 & 3 & 3 \\ | |||
| \end{pmatrix} | |||
| .\] | |||
| \item | |||
| \[ | |||
| M_{\underline{w}}^{\underline{v}}(\partial) = M_{\underline{w}}^{\underline{v}}(id_W) \cdot | |||
| M_{\underline{v}}^{\underline{v}}(\partial) = | |||
| \begin{pmatrix} | |||
| 0 & 1 & -2 & -6 \\ | |||
| 0 & 0 & 2 & 3 \\ | |||
| 0 & 0 & 0 & -3 \\ | |||
| 0 & 0 & 0 & 3 \\ | |||
| \end{pmatrix} | |||
| .\] | |||
| \item | |||
| \[ | |||
| M_{\underline{v}}^{\underline{w}}(\partial) = M_{\underline{v}}^{\underline{v}}(\partial) \cdot | |||
| M_{\underline{v}}^{\underline{w}}(id_W) = | |||
| \begin{pmatrix} | |||
| 0 & 1 & 1 & 0 \\ | |||
| 0 & 0 & -2 & 0 \\ | |||
| 0 & 0 & 3 & 3 \\ | |||
| 0 & 0 & 0 & 0 \\ | |||
| \end{pmatrix} | |||
| .\] | |||
| \end{enumerate} | |||
| \end{enumerate} | |||
| \end{aufgabe} | |||
| \begin{aufgabe} | |||
| \begin{aufgabe} Sei $f\colon U \to V$ und $g\colon V \to W$ lineare Abbildungen | |||
| zwischen endlich dimensionalen Vektorräumen. | |||
| \begin{enumerate}[(a)] | |||
| \item Beh.: $\text{dim } \text{ker}(g\circ f) \le \text{dim } \text{ker} g + \text{dim } \text{ker }f$ | |||
| \begin{proof} | |||
| Schränke $g$ auf $\text{Bild}(f)$ ein durch $g'\colon \text{Bild}(f) \to W$ mit | |||
| $v \mapsto g(v)$. | |||
| \begin{align*} | |||
| \text{ker }(g \circ f) &= \text{dim } U - \text{dim}(\text{Bild}(g \circ f)) \\ | |||
| &= \text{dim }U - \text{Rg}(g')\\ | |||
| &= \text{Rg}(f) - \text{Rg}(g') + \text{dim }U - \text{Rg}(f) \\ | |||
| &= \text{ker }g' + \text{ker }f \\ | |||
| &\le \text{ker }g + \text{ker }f | |||
| .\end{align*} | |||
| \end{proof} | |||
| \item Beh.: $\text{Rg}(f) - \text{Rg}(g \circ f) \le \text{dim } V - \text{Rg}(g)$ | |||
| \begin{proof} | |||
| Aus (a) folgt: | |||
| \begin{align*} | |||
| \text{dim } \text{ker}(g \circ f) &\le \text{dim } \text{ker}(g) + \text{dim } \text{ker}(f) \\ | |||
| \implies \text{dim } U - \text{Rg}(g \circ f) &\le \text{dim } V - \text{Rg}(g) + \text{dim } U - \text{Rg}(f) \\ | |||
| \implies \text{Rg}(f) - \text{Rg}(g \circ f) &\le \text{dim } V - \text{Rg}(g) | |||
| .\end{align*} | |||
| \end{proof} | |||
| \item Beh.: Für $A \in M_{n, m}(K)$ und $B \in M_{l,n}(K)$ gilt | |||
| $S\text{Rg}(A) - S\text{Rg}(B \cdot A) \le n - S\text{Rg}(B) $. | |||
| \begin{proof} | |||
| Seien $A \in M_{n,m}(K)$ und $B \in M_{l,n}(K)$ beliebig, dann definiere | |||
| $f := F_{n, m}(A)$ und $g := F_{l, n}(B)$. Damit folgt: $F_{m, l}(B \cdot A) = g \circ f$. | |||
| Dann folgt aus (b) direkt: | |||
| \[ | |||
| \text{Rg}(f) - \text{Rg}(g \circ f) \le \text{dim } K^{n} - \text{Rg}(g) | |||
| .\] Mit $\text{Rg}(f) = S\text{Rg}(A) $, $\text{Rg}(g) = S\text{Rg}(B) $ und | |||
| $\text{Rg}(g \circ f) = S\text{Rg}(A\cdot B)$ ergibt sich | |||
| \[ | |||
| S\text{Rg}(A) - S\text{Rg}(B \cdot A) \le n - S\text{Rg}(B) | |||
| .\] | |||
| \end{proof} | |||
| \end{enumerate} | |||
| \end{aufgabe} | |||
| \begin{aufgabe} | |||
| \begin{enumerate} | |||
| \item | |||
| \item | |||
| \item Beh.: Für $f : V \to V$ eine lineare Abbildung gilt | |||
| \begin{aufgabe} Sei $V$ ein Vektorraum, $U$ ein Untervektorraum und $W$ ein Komplement von $U$ in $V$. | |||
| \begin{enumerate}[(a)] | |||
| \item Beh.: Es existiert eine eindeutige lineare Abbildung $\pi\colon V \to V$, welche eingeschränkt | |||
| auf $U$ die Identität und eingeschränkt auf $W$ konstant null ist. | |||
| \begin{proof} | |||
| Sei $(v_i)_{i\in I}$ Basis von $U$ und $(v_j)_{j \in J}$ mit $J \cap U = \emptyset$ | |||
| Basis von $W$. | |||
| Damit ist $(v_i)_{i \in I \cup J}$ Basis von $V$. Definiere $\pi\colon V \to V$ linear mit | |||
| \[ | |||
| \pi(v_i) = \begin{cases} | |||
| v_i & \text{falls } i \in I \\ | |||
| 0 & \text{falls } i \in J | |||
| \end{cases} | |||
| .\] | |||
| Schränke nun $\pi$ auf $U$ ein: Dann ex. für alle $u \in U$ ein | |||
| $(\alpha_i)_{i \in I} \in K^{(I)}$, s.d. | |||
| $v = \sum_{i \in I} v_i$. Damit: | |||
| \[ | |||
| \pi(v) = \sum_{i \in I} \alpha_i \pi(v_i) = \sum_{i \in I} \alpha_i v_i = v | |||
| .\] | |||
| Schränke nun $\pi$ auf $W$ ein: Dann ex. für alle $w \in W$ | |||
| ein $(\alpha_j)_{j \in J} \in K^{(J)}$, s.d. | |||
| $v = \sum_{i \in J} \alpha_j v_j$. Damit | |||
| \[ | |||
| \pi(v) = \sum_{j \in J} \alpha_j \pi(v_j) = 0 | |||
| .\] | |||
| $\pi$ ist eindeutig, da eindeutig durch die Basisvektoren definiert. | |||
| \end{proof} | |||
| \item Beh.: Für dieses $\pi$ gilt: $\pi \circ \pi = \pi$. | |||
| \begin{proof} | |||
| Seien die Basen wie in (a). | |||
| Sei $v \in V$ beliebig. Dann ex. ein $(\alpha_i)_{i\in I} \in K^{(I)}$ und ein | |||
| $(\beta_j)_{j\in J} \in K^{(J)}$, s.d. | |||
| \[ | |||
| v = \sum_{i \in I} \alpha_i v_i + \sum_{j \in J} \beta_j v_j | |||
| .\] Damit gilt | |||
| \[ | |||
| \pi(v) = \sum_{i \in I} \alpha_i \pi(v_i) + \sum_{j \in J} \beta_j \pi(v_j) | |||
| = \sum_{i \in I} \alpha_i v_i | |||
| .\] $\implies$ | |||
| \[ | |||
| \pi(\pi(v)) = \pi\left( \sum_{i \in I} \alpha_i v_i\right) = | |||
| \sum_{i \in I} \alpha_i v_i = \pi(v) | |||
| .\] $\implies \pi = \pi \circ \pi$ | |||
| \end{proof} | |||
| \item Beh.: Für $\pi : V \to V$ eine lineare Abbildung gilt | |||
| \[ | |||
| V \stackrel{\sim }{=} \text{Bild}(f) \oplus \text{ker } f | |||
| V \stackrel{\sim }{=} \text{Bild}(\pi) \oplus \text{ker } \pi | |||
| .\] | |||
| \begin{proof} | |||
| Sei $(u_i)_{i\in I}$ Basis von $V \setminus \text{ker } f$. | |||
| Sei $U$ Komplement zu $\text{ker }\pi$ und $(u_i)_{i\in I}$ Basis von $U$. | |||
| Definiere $f: V / \text{ker } f \oplus \text{ker } f \to V$ mit | |||
| Wegen Homomorphiesatz gilt: $\text{Bild}(\pi) \stackrel{\sim }{=} V / \text{ker }(\pi)$. | |||
| Nach Blatt 6 gilt: $(u_i + \text{ker } \pi)_{i \in I}$ ist Basis von $V / \text{ker }(\pi)$. | |||
| Damit: | |||
| \[ | |||
| ([u_i + \text{ker }f], k) \mapsto u_i + k | |||
| .\] Wohldefiniert, da $([u_i + \text{ker }f])_{i \in I}$ | |||
| nach Blatt 6 Basis von $V / \text{ker }f$. | |||
| \text{Bild}(\pi) \stackrel{\sim }{=} V / \text{ker }(\pi) \stackrel{\sim }{=} U | |||
| .\] Daraus folgt direkt: | |||
| \[ | |||
| \text{Bild}(\pi) \oplus \text{ker }\pi \stackrel{\sim }{=} U \oplus \text{ker }\pi | |||
| \stackrel{\sim }{=} V | |||
| .\] | |||
| \end{proof} | |||
| \end{enumerate} | |||
| \end{aufgabe} | |||
| \begin{aufgabe} | |||
| \[ | |||
| A_1 := \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad | |||
| A_2 := \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}, \quad | |||
| A_3 := \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} | |||
| .\] | |||
| \begin{proof} | |||
| \begin{enumerate} | |||
| \item $A_1$ ist die Einheitsmatrix $\implies A_1 \cdot A_1 = A_1$ und | |||
| $A_1 \cdot (1,1)^{t} = (1,1)^{t}$. | |||
| \item \[ | |||
| A_2 \cdot A_2 = \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} | |||
| \cdot \begin{pmatrix} 0 & 1 \\ 0 & 1\end{pmatrix} | |||
| = \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} = A_2 | |||
| .\] \[ | |||
| A_2 \cdot (1,1)^{t} = \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} | |||
| \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix} | |||
| = \begin{pmatrix} 1 \\ 1 \end{pmatrix} = (1,1)^{t} | |||
| .\] | |||
| \item \[ | |||
| A_3 \cdot A_3 = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} | |||
| \cdot \begin{pmatrix} 1 & 0 \\ 1 & 0\end{pmatrix} | |||
| = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} = A_3 | |||
| .\] \[ | |||
| A_3 \cdot (1,1)^{t} = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} | |||
| \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix} | |||
| = \begin{pmatrix} 1 \\ 1 \end{pmatrix} = (1,1)^{t} | |||
| .\] | |||
| \end{enumerate} | |||
| \end{proof} | |||
| \end{aufgabe} | |||
| \end{document} | |||