# Section 1.12

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

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

Proof:
Suppose that ${\displaystyle x_{1},x_{2},...,x_{k}}$ are linearly dependent. Then there are scalars ${\displaystyle c_{1},c_{2},...,c_{k}}$, not all of which are zero that satisfy ${\displaystyle c_{1}x_{1}+c_{2}x_{2}+\cdots +c_{k}x_{k}=0}$. Now recall that for any linear transformation ${\displaystyle L(0)=0}$. So then ${\displaystyle L(c_{1}x_{1}+\cdots +c_{k}x_{k})=L(0)=0}$. But by linearity of ${\displaystyle L}$ we have ${\displaystyle L(c_{1}x_{1}+\cdots +c_{k}x_{k})=L(c_{1}x_{1})+L(c_{2}x_{2})+\cdots +L(c_{k}x_{k})=c_{1}L(x_{1})+c_{2}L(x_{2})+\cdots +c_{k}L(x_{k})}$. Combining these facts gives that ${\displaystyle c_{1}L(x_{1})+\cdots +c_{k}L(x_{k})=0}$. In other words, we have a linear combination of ${\displaystyle L(x_{1}),L(x_{2}),...,L(x_{k})}$ that gives zero and we know that not all of the ${\displaystyle c_{1},...,c_{k}}$ are zero. Therefore ${\displaystyle L(x_{1}),L(x_{2}),...,L(x_{k})}$ are linearly dependent.

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