Georgi E Shilov Linear Algebra P44 $P44$">
Hi, I am reading Georgi E. Shilov's Linear Algebra. The attached image is page 44. 
Can you explain why if $\alpha_{l+1} = 0$, the vectors $f_1, f_2, ... f_l$ would be linearly dependent? (Location: see the second line below the displayed formula.)
Thank you.
 A: When the author assumes that there is a relation of the form$$\alpha_1f_1+\alpha_2f_2+\cdots+\alpha_lf_l+\alpha_{l+1}f_{l+1}=0,\tag{1}$$he assumes that not all $\alpha_k$'s are equal to $0$. First he deals with the case $\alpha_{l+1}\neq0$. Then, if $\alpha_{l+1}=0$, some $\alpha_k$ is different from $0$ and therefore, by the definition of linear independence and by $(1)$, the vectors $f_1,f_2,\ldots,f_l$ are linearly dependent.
A: This is proof by contradiction. It goes as follows. Suppose $f_1, f_2, ..., f_{l+1}$ are linearly dependent, which means there exist coefficients $\alpha_1, \alpha_2, ..., \alpha_{l+1}$, not all equal to zero, satisfying $$\alpha_1 f_1 + \alpha_2 f_2 + \cdots + \alpha_l f_l + \alpha_{l+1} f_{l+1} = 0 \tag{1}.$$ Then, if $\alpha_{l+1} \neq 0$, contradiction results. Otherwise, if $\alpha_{l+1} = 0$, $(1)$ becomes $$\alpha_1 f_1 + \alpha_2 f_2 + \cdots + \alpha_l f_l = 0 \tag{2},$$ where at least one coefficient is non-zero. But this means $f_1, f_2, ..., f_l$ are linearly dependent, which contradicts the fact, and thereby proves the assertion.
