Section 1.10 Homework - Revision history

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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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'''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 <math>V</math>. 
 
''Proof:'' Suppose <math>x_1,x_2 \in L^{-1}(N)</math>. Then <math>L(x_1),L(x_2) \in N</math>. But <math>N</math> is a subspace and so <math>L(x_1)+L(x_2) \in N</math>. But <math>L</math> is linear so that <math>L(x_1+x_2) = L(x_1)+L(x_2) \in N</math> so that <math>x_1+x_2 \in L^{-1}(N)</math>. Thus, <math>L^{-1}(N)</math> is closed under vector addition. Now suppose <math>x \in L^{-1}(N)</math> and <math>\alpha \in \mathbb{F}</math>. Then <math>L(x) \in N</math> and since <math>N</math> is a subspace, <math>\alpha L(x) \in N</math>. But again <math>L</math> is linear so <math>L(\alpha x) = \alpha L(x) \in N</math>. This means <math>\alpha x \in L^{-1}(N)</math>. Hence <math>L^{-1}(N)</math> is closed under scalar multiplication. Therefore <math>L^{-1}(N)</math> is a subspace of <math>V</math>. 
 
'''10.''' Show that if <math>M \subset V</math> and <math>N \subset W</math> are subspaces, then <math>M \times N \subset V \times W</math> is also a subspace. 
 
''Proof:'' Suppose <math>(x_1,y_1),(x_2,y_2) \in M \times N</math>. Then <math>x_1,x_2 \in M</math> and <math>y_1,y_2 \in N</math>. But <math>M</math> is a subspace and so <math>x_1+x_2 \in M</math>. Also <math>N</math> is a subspace so <math>y_1 + y_2 \in N</math>. This means <math>(x_1+x_2,y_1+y_2) \in M \times N</math>. On the other hand <math>(x_1,y_1)+(x_2,y_2) = (x_1+x_2,y_1+y_2)</math>. Thus, <math>M \times N</math> is closed under vector addition. Now suppose <math>(x,y) \in M \times N</math> and <math>\alpha \in \mathbb{F}</math>. Then <math>x \in M</math> and <math>y \in N</math>. But <math>M</math> and <math>N</math> are subspaces so <math>\alpha x \in M</math> and <math>\alpha y \in N</math>. That means <math>(\alpha x, \alpha y) \in M \times N</math>. This means <math>\alpha (x,y) = (\alpha x, \alpha y) \in M \times N</math>. Hence <math>M \times N</math> is closed under scalar multiplication. Therefore <math>M \times N</math> is a subspace of <math>V \times W</math>. 
 

'''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>
(a) Show that <math>G_L</math> is a subspace. 
 
''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>. 
 
(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. 
 
''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}
\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.

@@ Line 3: / Line 3: @@
 is a subspace of <math>V</math>.<br />
 <br />
-''Proof:'' Suppose <math>x_1,x_2 \in L^{-1}(N)</math>. Then <math>L(x_1),L(x_2) \in N</math>. But <math>N</math> is a subspace and so <math>L(x_1)+L(x_2) \in N</math>. But <math>L</math> is linear so that <math>L(x_1+x_2) = L(x_1)+L(x_2) \in N</math> so that <math>x_1+x_2 \in L^{-1}(N)</math>. Thus, <math>L^{-1}(N)</math> is closed under vector addition. Now suppose <math>x \in L^{-1}(N)</math> and <math>\alpha \in \mathbb{F}</math>. Then <math>L(x) \in N</math> and since <math>N</math> is a subspace, <math>\alpha L(x) \in N</math>. But again <math>L</math> is linear so <math>L(\alpha x) = \alpha L(x) \in N</math>. This means <math>\alpha x \in L^{-1}(N)</math>. Hence <math>L^{-1}(N)</math> is closed under scalar multiplication. Therefore <math>L^{-1}(N)</math> is a subspace of <math>V</math>.<br />
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 <br />
 '''10.''' Show that if <math>M \subset V</math> and <math>N \subset W</math> are subspaces, then <math>M \times N \subset V \times W</math> is also a subspace.<br />
 <br />
-''Proof:'' Suppose <math>(x_1,y_1),(x_2,y_2) \in M \times N</math>. Then <math>x_1,x_2 \in M</math> and <math>y_1,y_2 \in N</math>. But <math>M</math> is a subspace and so <math>x_1+x_2 \in M</math>. Also <math>N</math> is a subspace so <math>y_1 + y_2 \in N</math>. This means <math>(x_1+x_2,y_1+y_2) \in M \times N</math>. On the other hand <math>(x_1,y_1)+(x_2,y_2) = (x_1+x_2,y_1+y_2)</math>. Thus, <math>M \times N</math> is closed under vector addition. Now suppose <math>(x,y) \in M \times N</math> and <math>\alpha \in \mathbb{F}</math>. Then <math>x \in M</math> and <math>y \in N</math>. But <math>M</math> and <math>N</math> are subspaces so <math>\alpha x \in M</math> and <math>\alpha y \in N</math>. That means <math>(\alpha x, \alpha y) \in M \times N</math>. This means <math>\alpha (x,y) = (\alpha x, \alpha y) \in M \times N</math>. Hence <math>M \times N</math> is closed under scalar multiplication. Therefore <math>M \times N</math> is a subspace of <math>V \times W</math>.<br />
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 <br />
@@ Line 14: / Line 22: @@
 (a) Show that <math>G_L</math> is a subspace.<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 />
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 <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 />
-''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:
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 <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>
 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.

@@ Line 12: / Line 12: @@
 '''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>
- (a) Show that <math>G_L</math> is a subspace.<br />
+(a) Show that <math>G_L</math> is a subspace.<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 />
@@ Line 18: / Line 18: @@
 (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 />
-''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}
+''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:
 \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>