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Suppose $a_1,...,a_k$ and $b$ are vectors in $\mathbb{R}^n$, such that $b \ne 0$, and $a_1,...,a_k$ are different from each other.

Also, suppose the equation $x_1a_1+...+x_ka_k = b$ has an infinite amount of solutions.

Now, assume $k \ge n+1$.

Can we say that the set $\{a_1,...,a_k\}$ spans $\mathbb{R}^n$?

I think the answer is positive. The reason is as follows:

$k \ge n+1$ implies a matrix with more columns than rows. Given that $x_1a_1+...+x_ka_k = b$ has an infinite amount of solutions, we can build a matrix that comprises of this exact linear combination and be sure we'd get an infinite amount of solutions. In that matrix, $b$ represents any vector in $\mathbb{R}^n$, and since we'd get an infinite amount of solutions, it means that any vector in $\mathbb{R}^n$ is a linear combination of $\{a_1,...,a_k\}$, hence $\{a_1,...,a_k\}$ spans $\mathbb{R}^n$.

However, I'm not sure that I'm right, and in any case it doesn't seem like a well-structured proof to me. So, I'd like to know if I'm right, and if so, how can I build a structured proof out of this?

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  • $\begingroup$ Here's the big mistake in your reasoning. $b$ is some fixed vector. In fact, if there are solutions for every $b$, then you can conclude that $a_1,\dots,a_k$ span $\Bbb R^k$, but then how can you have infinitely many solutions unless $k<n$? $\endgroup$ – Ted Shifrin Nov 8 '18 at 0:52
  • $\begingroup$ @TedShifrin, did you write $\mathbb{R}^k$ on purpose, or should it be $\mathbb{R}^n$? If it's indeed on purpose, what's $n$? $\endgroup$ – HeyJude Nov 8 '18 at 1:00
  • $\begingroup$ My apology. I meant $\Bbb R^n$. $\endgroup$ – Ted Shifrin Nov 8 '18 at 1:43
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This is false. Here is a simple counterexample: let $n=3$, $k=4$, $a_1=(1,0,0)$, $a_2=(0,1,0)$, $a_3=(1,1,0)$, $a_4=(-1,-1,0)$ and $b=(1,0,0)$. Clearly $\{a_1,a_2,a_3,a_4\}$ does not span $\mathbb{R}^3$ and $\sum_{i=1}^4 x_ia_i=b$ has infinitely many solutions (in fact, $x=(1,0,\lambda,\lambda)$ is a solution for every $\lambda$).

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  • $\begingroup$ Can you please explain why these vectors don't span $\mathbb{R}^3$? I thought that the fact we have free variables (and hence infinite solutions), it means that any vector in the field can be represented as a linear combination of $a_1...a_k$. Why isn't that true? $\endgroup$ – HeyJude Nov 8 '18 at 1:13
  • $\begingroup$ @HeyJude What linear combination of the four vectors equals $(0,0,1)$? $\endgroup$ – amd Nov 8 '18 at 1:15
  • $\begingroup$ @amd, none. But what I'd really like to know is where do I get it wrong in my understanding of a span\proving a span? i.e., when I commented: I thought that the fact we have free variables (and hence infinite solutions), it means that any vector in the field can be represented as a linear combination of a1...ak. Why isn't that true? $\endgroup$ – HeyJude Nov 8 '18 at 1:19
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    $\begingroup$ @HeyJude Having free variables only means that there are redundant vectors among the set. It says nothing useful about what their span might be, although the number of free variables does. In this example, their span is only two-dimensional. A more extreme example: the single equation $x_1=0$ has an infinite number of solutions in $R^n$ $(n>1)$, but the vector $(1,0,\dots,0)$ obviously doesn’t span the enclosing space. $\endgroup$ – amd Nov 8 '18 at 1:23
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    $\begingroup$ @HeyJude Every vector with a first coordinate equal to zero is a solution. The equation doesn’t constrain any of the other coordinates. $\endgroup$ – amd Nov 8 '18 at 1:34

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