4

Suppose I have the vectors $\underline{a}_1, \underline{a}_2,\ldots,\underline{a}_k$ and $\underline{b} \neq 0$ in $\mathbf{R}^n$. Also, $\underline{a}_1 \neq \underline{a}_2 \neq \ldots\neq \underline{a}_k$.

Say that the equation $x_1 \underline{a}_1 + \ldots + x_k \underline{a}_k = \underline{b}$ has an infinite number of solutions.

Is it true that the set $\{\underline{a}_1, \underline{a}_2,\ldots,\underline{a}_k\}$ is linearly dependent?

I feel like the set is indeed linearly dependent, but I fail to prove it. Here's my reasoning: if the given equation has infinitely many solutions, then the coefficient matrix of the system (in its reduced row echelon form) has some free variables in it. If the vectors were linearly independent, the matrix would've had no free variables at all. Thus the vectors are linearly dependent.

Hagen von Eitzen
  • 374,180
user187240
  • 45

1 Answers1

4

It would be enough to have two distinct solutions $x_1 \underline{a}_1 + \ldots + x_k \underline{a}_k = \underline{b}$ and $y_1 \underline{a}_1 + \ldots + y_k \underline{a}_k = \underline{b}$. If we subtract them, we get $(x_1-y_1) \underline{a}_1 + \ldots + (y_k-y_k) \underline{a}_k = \underline{0}$ and not all coefficients $x_i-y_i$ are zero. Hence the vectors are linearly dependant.

Hagen von Eitzen
  • 374,180
  • Your solution makes me feel stupid. Thank you and sorry. I will accept the answer as soon as the system allows me (in 9 minutes or so). – user187240 Oct 25 '14 at 15:17