Difference between revisions of "Section 1.10 Homework"
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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: | ''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> | + | <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)+\hat{L}(x_2)</math> |
| − | \hat{L}(x_1+x_2) | ||
| − | |||
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. | 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 15:20, 12 November 2015
3. Let be a linear map and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle N\subset W}
a subspace. Show that:
is a subspace of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle V}
.
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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle N}
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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle N\subset W}
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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x_{1},x_{2}\in M}
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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle N}
is a subspace so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y_{1}+y_{2}\in N}
. This means Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (x_{1}+x_{2},y_{1}+y_{2})\in M\times N}
. On the other hand 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})=(x_{1}+x_{2},y_{1}+y_{2})}
. 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,y)\in M\times N}
and . Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x\in M}
and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y\in N}
. But Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle M}
and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle N}
are subspaces so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha x\in M}
and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha y\in N}
. That means . This means Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha (x,y)=(\alpha x,\alpha y)\in M\times N}
. Hence is closed under scalar multiplication. Therefore is a subspace of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle V\times W}
.
12. Let be a linear map and consider the graph
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle G_{L}=\{(x,L(x)):x\in V\}\subset V\times W}
(a) Show that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle G_{L}}
is a subspace.
Proof: Suppose . Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 is linear which means Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L(x_{1})+L(x_{2})=L(x_{1}+x_{2})}
. Thus, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle G_{L}}
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,L(x))\in G_{L}}
and . Then Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \alpha L(x)=L(\alpha x)}
. Hence Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle G_{L}}
is closed under scalar multiplication. Therefore Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle G_{L}}
is a subspace of .
(b) Show that the map Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle V\to G_{L}}
that sends to Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\hat {L}}(x)=(x,L(x))}
. First I will show this map is linear:
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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.
