Grad at 07:06, 16 November 2015

2015-11-16T07:06:47Z

Grad: Created page with "'''8.''' Let $L: V \to W$ be a linear map.

(b) Show that if $x_1,x_2,...,x_k$ are linearly dependent, then $L(x_1),L(x_2),...,L(x_k)$
2015-11-12T23:38:56Z

Created page with "'''8.''' Let <math>L: V \to W</math> be a linear map. (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)</m..."

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'''8.''' Let <math>L: V \to W</math> be a linear map. 
 
(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. 
 
''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. 
 
(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. 
 
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 <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.

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 (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;"
 <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;"
 <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.

Section 1.12 Homework - Revision history

Grad at 07:06, 16 November 2015