Difference between revisions of "Section 1.10 Homework"

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(Created page with "'''3.''' Let <math>L: V \to W</math> be a linear map and <math>N \subset W</math> a subspace. Show that: <math>L^{-1}(N) = \{ x\in V: L(x) \in N\}</math> is a subspace of <ma...")
 
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'''12.''' Let <math>L: V \to W</math> be a linear map and consider the graph  
 
'''12.''' Let <math>L: V \to W</math> be a linear map and consider the graph  
 
<math>G_L = \{ (x,L(x)): x \in V\} \subset V \times W</math>
 
<math>G_L = \{ (x,L(x)): x \in V\} \subset V \times W</math>
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(a) Show that <math>G_L</math> is a subspace.<br />
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(a) Show that <math>G_L</math> is a subspace.<br />
 
<br />
 
<br />
 
''Proof:'' Suppose <math>(x_1,L(x_1),(x_2,L(x_2) \in G_L</math>. Then <math>(x_1,L(x_1)+(x_2,L(x_2) = (x_1+x_2,L(x_1)+L(x_2)) = (x_1+x_2, L(x_1+x_2) \in G_L</math>. Here I used the fact that <math>L</math> is linear which means <math>L(x_1)+L(x_2) = L(x_1+x_2)</math>. Thus, <math>G_L</math> is closed under vector addition. Now suppose <math>(x,L(x)) \in G_L</math> and <math>\alpha \in \mathbb{F}</math>. Then <math>\alpha (x,L(x)) = (\alpha x, \alpha L(x)) = (\alpha x, L(\alpha x) \in G_L</math>. Again I used the linearity property to conclude <math>\alpha L(x) = L(\alpha x)</math>. Hence <math>G_L</math> is closed under scalar multiplication. Therefore <math>G_L</math> is a subspace of <math>V \times W</math>.<br />
 
''Proof:'' Suppose <math>(x_1,L(x_1),(x_2,L(x_2) \in G_L</math>. Then <math>(x_1,L(x_1)+(x_2,L(x_2) = (x_1+x_2,L(x_1)+L(x_2)) = (x_1+x_2, L(x_1+x_2) \in G_L</math>. Here I used the fact that <math>L</math> is linear which means <math>L(x_1)+L(x_2) = L(x_1+x_2)</math>. Thus, <math>G_L</math> is closed under vector addition. Now suppose <math>(x,L(x)) \in G_L</math> and <math>\alpha \in \mathbb{F}</math>. Then <math>\alpha (x,L(x)) = (\alpha x, \alpha L(x)) = (\alpha x, L(\alpha x) \in G_L</math>. Again I used the linearity property to conclude <math>\alpha L(x) = L(\alpha x)</math>. Hence <math>G_L</math> is closed under scalar multiplication. Therefore <math>G_L</math> is a subspace of <math>V \times W</math>.<br />
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(b) Show that the map <math>V \to G_L</math> that sends <math>x</math> to <math>(x,L(x)</math> is an isomorphism.<br />
 
(b) Show that the map <math>V \to G_L</math> that sends <math>x</math> to <math>(x,L(x)</math> is an isomorphism.<br />
 
<br />
 
<br />
−
''Proof:'' Call this map <math>\hat{L}: V \to G_L</math>. That is <math>\hat{L}(x) = (x,L(x))</math>. First I will show this map is linear: <math>\begin{aligned}
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''Proof:'' Call this map <math>\hat{L}: V \to G_L</math>. That is <math>\hat{L}(x) = (x,L(x))</math>. First I will show this map is linear:  
 +
<math>
 
\hat{L}(x_1+x_2) & =(x_1+x_2, L(x_1+x_2)) = (x_1+x_2,L(x_1)+L(x_2)) = (x_1,L(x_1))+(x_2,L(x_2)) \\
 
\hat{L}(x_1+x_2) & =(x_1+x_2, L(x_1+x_2)) = (x_1+x_2,L(x_1)+L(x_2)) = (x_1,L(x_1))+(x_2,L(x_2)) \\
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& = \hat{L}(x_1)+\hat{L}(x_2)\end{aligned}</math> and <math>\hat{L}(\alpha x) =(\alpha x, L(\alpha x)) = (\alpha x,\alpha L(x)) = \alpha (x,L(x)) = \alpha \hat{L}(x)</math>. Thus <math>\hat{L}</math> is linear. Now to show <math>\hat{L}</math> is bijective. If <math>(x,L(x)) \in G_L</math>, then <math>\hat{L}(x) = (x,L(x))</math> so <math>\hat{L}</math> is trivially onto. In fact, we essentially chose to the codomain of our function <math>\hat{L}</math> to just be the image/range of the map to ensure it was onto. Now to show <math>\hat{L}</math> is one-to-one. Suppose <math>\hat{L}(x_1) = \hat{L}(x_2)</math>. Then <math>(x_1,L(x_1)) = (x_2,L(x_2)</math>. But two ordered pairs are equal if and only if both components are equal. That is, <math>x_1 = x_2</math>. Thus <math>\hat{L}</math> is one-to-one. Therefore <math>\hat{L}</math> is an isomorphism.
+
& = \hat{L}(x_1)+\hat{L}(x_2)</math>
 +
and <math>\hat{L}(\alpha x) =(\alpha x, L(\alpha x)) = (\alpha x,\alpha L(x)) = \alpha (x,L(x)) = \alpha \hat{L}(x)</math>. Thus <math>\hat{L}</math> is linear. Now to show <math>\hat{L}</math> is bijective. If <math>(x,L(x)) \in G_L</math>, then <math>\hat{L}(x) = (x,L(x))</math> so <math>\hat{L}</math> is trivially onto. In fact, we essentially chose to the codomain of our function <math>\hat{L}</math> to just be the image/range of the map to ensure it was onto. Now to show <math>\hat{L}</math> is one-to-one. Suppose <math>\hat{L}(x_1) = \hat{L}(x_2)</math>. Then <math>(x_1,L(x_1)) = (x_2,L(x_2)</math>. But two ordered pairs are equal if and only if both components are equal. That is, <math>x_1 = x_2</math>. Thus <math>\hat{L}</math> is one-to-one. Therefore <math>\hat{L}</math> is an isomorphism.

Revision as of 15:19, 12 November 2015

3. Let be a linear map and a subspace. Show that: is a subspace of .

Proof: Suppose Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1},x_{2}\in L^{-1}(N)} . Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L(x_{1}),L(x_{2})\in N} . But Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle N} is a subspace and so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L(x_{1})+L(x_{2})\in N} . But is linear so that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L(x_{1}+x_{2})=L(x_{1})+L(x_{2})\in N} so that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1}+x_{2}\in L^{-1}(N)} . Thus, is closed under vector addition. Now suppose Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x\in L^{-1}(N)} and . Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L(x)\in N} and since is a subspace, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha L(x)\in N} . But again is linear so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L(\alpha x)=\alpha L(x)\in N} . This means Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha x\in L^{-1}(N)} . Hence is closed under scalar multiplication. Therefore is a subspace of .

10. Show that if Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle M\subset V} and are subspaces, then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle M\times N\subset V\times W} is also a subspace.

Proof: Suppose Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (x_{1},y_{1}),(x_{2},y_{2})\in M\times N} . Then and . But Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle M} is a subspace and so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1}+x_{2}\in M} . Also is a subspace so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_1 + y_2 \in N} . This means Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x_1+x_2,y_1+y_2) \in M \times N} . On the other hand Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x_1,y_1)+(x_2,y_2) = (x_1+x_2,y_1+y_2)} . Thus, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M \times N} is closed under vector addition. Now suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x,y) \in M \times N} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha \in \mathbb{F}} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in M} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y \in N} . But Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N} are subspaces so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha x \in M} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha y \in N} . That means Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\alpha x, \alpha y) \in M \times N} . This means Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha (x,y) = (\alpha x, \alpha y) \in M \times N} . Hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M \times N} is closed under scalar multiplication. Therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M \times N} is a subspace of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V \times W} .

12. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L: V \to W} be a linear map and consider the graph Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G_L = \{ (x,L(x)): x \in V\} \subset V \times W} (a) Show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G_L} is a subspace.

Proof: Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x_1,L(x_1),(x_2,L(x_2) \in G_L} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x_1,L(x_1)+(x_2,L(x_2) = (x_1+x_2,L(x_1)+L(x_2)) = (x_1+x_2, L(x_1+x_2) \in G_L} . Here I used the fact that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} is linear which means Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L(x_1)+L(x_2) = L(x_1+x_2)} . Thus, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G_L} is closed under vector addition. Now suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x,L(x)) \in G_L} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha \in \mathbb{F}} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha (x,L(x)) = (\alpha x, \alpha L(x)) = (\alpha x, L(\alpha x) \in G_L} . Again I used the linearity property to conclude Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha L(x) = L(\alpha x)} . Hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G_L} is closed under scalar multiplication. Therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G_L} is a subspace of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V \times W} .

(b) Show that the map Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V \to G_L} that sends Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x,L(x)} is an isomorphism.

Proof: Call this map Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}: V \to G_L} . That is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}(x) = (x,L(x))} . First I will show this map is linear: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}(x_1+x_2) & =(x_1+x_2, L(x_1+x_2)) = (x_1+x_2,L(x_1)+L(x_2)) = (x_1,L(x_1))+(x_2,L(x_2)) \\ & = \hat{L}(x_1)+\hat{L}(x_2)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}(\alpha x) =(\alpha x, L(\alpha x)) = (\alpha x,\alpha L(x)) = \alpha (x,L(x)) = \alpha \hat{L}(x)} . Thus Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}} is linear. Now to show Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}} is bijective. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x,L(x)) \in G_L} , then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}(x) = (x,L(x))} so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}} is trivially onto. In fact, we essentially chose to the codomain of our function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}} to just be the image/range of the map to ensure it was onto. Now to show Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}} is one-to-one. Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}(x_1) = \hat{L}(x_2)} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x_1,L(x_1)) = (x_2,L(x_2)} . But two ordered pairs are equal if and only if both components are equal. That is, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_1 = x_2} . Thus Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}} is one-to-one. Therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{L}} is an isomorphism.

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