# Section 1.12

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.