Difference between revisions of "Section 1.12 Homework"

Latest revision as of 00:06, 16 November 2015

8. Let $L:V\to W$ be a linear map.

(b) Show that if $x_{1},x_{2},...,x_{k}$ are linearly dependent, then $L(x_{1}),L(x_{2}),...,L(x_{k})$ are linearly dependent.

Proof:

Suppose that

x_{1},x_{2},...,x_{k}

are linearly dependent. Then there are scalars

c_{1},c_{2},...,c_{k}

, not all of which are zero that satisfy

c_{1}x_{1}+c_{2}x_{2}+\cdots +c_{k}x_{k}=0

. Now recall that for any linear transformation

L(0)=0

. So then

L(c_{1}x_{1}+\cdots +c_{k}x_{k})=L(0)=0

. But by linearity of

L

we have

L(c_{1}x_{1}+\cdots +c_{k}x_{k})=L(c_{1}x_{1})+L(c_{2}x_{2})+\cdots +L(c_{k}x_{k})=c_{1}L(x_{1})+c_{2}L(x_{2})+\cdots +c_{k}L(x_{k})

. Combining these facts gives that

c_{1}L(x_{1})+\cdots +c_{k}L(x_{k})=0

. In other words, we have a linear combination of

L(x_{1}),L(x_{2}),...,L(x_{k})

that gives zero and we know that not all of the

c_{1},...,c_{k}

are zero. Therefore

L(x_{1}),L(x_{2}),...,L(x_{k})

are linearly dependent.

(c) Show that if $L(x_{1}),L(x_{2}),...,L(x_{k})$ are linearly independent then $x_{1},x_{2},...,x_{k}$ are linearly independent.

Proof:

Note: This is the contrapositive statement of part (b). Hence since we proved (b), then (c) is also true as contrapositives are logically equivalent. However, we can prove this separately as follows.

Proof: Suppose that $L(x_{1}),L(x_{2}),...,L(x_{k})$ are linearly independent. To show that $x_{1},x_{2},...,x_{k}$ are linearly independent we consider any combination $c_{1}x_{1}+c_{2}x_{2}+\cdots +c_{k}x_{k}$ that gives 0. We want to show that this can only happen if all of $c_{1},c_{2},...,c_{k}=0$ . Since $c_{1}x_{1}+c_{2}x_{2}+\cdots +c_{k}x_{k}=0$ , then $L(c_{1}x_{1}+\cdots +c_{k}x_{k})=L(0)=0$ . As in the proof of part (b) we then have $c_{1}L(x_{1})+c_{2}L(x_{2})+\cdots +c_{k}L(x_{k})=0$ . That is, we have found a linear combination of $L(x_{1}),L(x_{2}),...,L(x_{k})$ that gives zero. But since $L(x_{1}),L(x_{2}),...,L(x_{k})$ are linearly independent, then we must have $c_{1}=c_{2}=\cdots =c_{k}=0$ . Therefore $x_{1},x_{2},...,x_{k}$ are linearly independent.

@@ Line 3: / Line 3: @@
 (b) Show that if <math>x_1,x_2,...,x_k</math> are linearly dependent, then <math>L(x_1),L(x_2),...,L(x_k)</math> are linearly dependent.<br />
 <br />
-''Proof:'' Suppose that <math>x_1,x_2,...,x_k</math> are linearly dependent. Then there are scalars <math>c_1,c_2,...,c_k</math>, not all of which are zero that satisfy <math>c_1 x_1 + c_2 x_2 + \cdots +c_k x_k = 0</math>. Now recall that for any linear transformation <math>L(0) = 0</math>. So then <math>L(c_1 x_1 + \cdots +c_k x_k) = L(0) = 0</math>. But by linearity of <math>L</math> we have <math>L(c_1x_1 + \cdots +c_k x_k) = L(c_1 x_1) + L(c_2 x_2) + \cdots +L(c_k x_k) = c_1 L(x_1) + c_2 L(x_2) + \cdots +c_k L(x_k)</math>. Combining these facts gives that <math>c_1 L(x_1) + \cdots +c_k L(x_k) = 0</math>. In other words, we have a linear combination of <math>L(x_1),L(x_2),...,L(x_k)</math> that gives zero and we know that not all of the <math>c_1,...,c_k</math> are zero. Therefore <math>L(x_1),L(x_2),...,L(x_k)</math> are linearly dependent.<br />
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+!Proof:
+|-
+|Suppose that <math>x_1,x_2,...,x_k</math> are linearly dependent. Then there are scalars <math>c_1,c_2,...,c_k</math>, not all of which are zero that satisfy <math>c_1 x_1 + c_2 x_2 + \cdots +c_k x_k = 0</math>. Now recall that for any linear transformation <math>L(0) = 0</math>. So then <math>L(c_1 x_1 + \cdots +c_k x_k) = L(0) = 0</math>. But by linearity of <math>L</math> we have <math>L(c_1x_1 + \cdots +c_k x_k) = L(c_1 x_1) + L(c_2 x_2) + \cdots +L(c_k x_k) = c_1 L(x_1) + c_2 L(x_2) + \cdots +c_k L(x_k)</math>. Combining these facts gives that <math>c_1 L(x_1) + \cdots +c_k L(x_k) = 0</math>. In other words, we have a linear combination of <math>L(x_1),L(x_2),...,L(x_k)</math> that gives zero and we know that not all of the <math>c_1,...,c_k</math> are zero. Therefore <math>L(x_1),L(x_2),...,L(x_k)</math> are linearly dependent.<br />
+|}
 <br />
 (c) Show that if <math>L(x_1),L(x_2),...,L(x_k)</math> are linearly independent then <math>x_1,x_2,...,x_k</math> are linearly independent.<br />
 <br />
-Note: This is the contrapositive statement of part (b). Hence since we proved (b), then (c) is also true as contrapositives are logically equivalent. However, we can prove this separately as follows.<br />
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+!Proof:
+|-
+|Note: This is the contrapositive statement of part (b). Hence since we proved (b), then (c) is also true as contrapositives are logically equivalent. However, we can prove this separately as follows.<br />
 <br />
 ''Proof:'' Suppose that <math>L(x_1),L(x_2),...,L(x_k)</math> are linearly independent. To show that <math>x_1,x_2,...,x_k</math> are linearly independent we consider any combination <math>c_1 x_1 + c_2 x_2 + \cdots +c_k x_k</math> that gives 0. We want to show that this can only happen if all of <math>c_1,c_2,...,c_k =0</math>. Since <math>c_1 x_1 + c_2 x_2 + \cdots +c_k x_k = 0</math>, then <math>L(c_1 x_1 + \cdots +c_k x_k) = L(0) = 0</math>. As in the proof of part (b) we then have <math>c_1 L(x_1) + c_2 L(x_2) + \cdots +c_k L(x_k) = 0</math>. That is, we have found a linear combination of <math>L(x_1),L(x_2),...,L(x_k)</math> that gives zero. But since <math>L(x_1),L(x_2),...,L(x_k)</math> are linearly independent, then we must have <math>c_1 = c_2 = \cdots = c_k = 0</math>. Therefore <math>x_1,x_2,...,x_k</math> are linearly independent.
+|}

Difference between revisions of "Section 1.12 Homework"

Latest revision as of 00:06, 16 November 2015

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