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>
(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)) \\
& = \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 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 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 \subset W} a subspace. 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 L^{-1}(N) = \{ x\in V: L(x) \in 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} .

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,x_2 \in L^{-1}(N)} . 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 L(x_1),L(x_2) \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 N} is a subspace and 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 L(x_1)+L(x_2) \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 L} is linear so 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(x_1+x_2) = L(x_1)+L(x_2) \in N} so 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 x_1+x_2 \in L^{-1}(N)} . 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 L^{-1}(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 \in L^{-1}(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 L(x) \in N} and since 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} is a subspace, 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) \in N} . But again 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 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 L(\alpha x) = \alpha L(x) \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 \alpha x \in L^{-1}(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 L^{-1}(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 L^{-1}(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} .

10. Show that 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 M \subset V} 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 \subset W} are subspaces, 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 M \times N \subset V \times W} is also 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,y_1),(x_2,y_2) \in M \times N} . 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,x_2 \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_1,y_2 \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} is a subspace and 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 x_1+x_2 \in M} . Also 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} 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 (a) Show that is a subspace.

Proof: Suppose . Then . Here I used the fact that is linear which means . Thus, is closed under vector addition. Now suppose and . Then . Again I used the linearity property to conclude . Hence is closed under scalar multiplication. Therefore is a subspace of .

(b) Show that the map that sends to is an isomorphism.

Proof: Call this map . That is . First I will show this map is linear: Failed to parse (syntax error): {\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 . Thus is linear. Now to show is bijective. If , then so is trivially onto. In fact, we essentially chose to the codomain of our function to just be the image/range of the map to ensure it was onto. Now to show is one-to-one. Suppose . Then . But two ordered pairs are equal if and only if both components are equal. That is, . Thus is one-to-one. Therefore is an isomorphism.