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Let \begin{align} W_1&=\{(a_1,a_2,...,a_n)\in \mathbb C^n \mid a_1=\mu a_n, \text{for some fixed $\mu \in \mathbb C$} \}\\ W_2&=\{(a_1,a_2,...,a_n)\in \mathbb C^n \mid \frac{a_1}{a_2}=\mu , \text{for some fixed $\mu \in \mathbb C$}\}. \end{align} What is the difference between two subsets?

I know that $W_1$ is a subspace with dimension $n-1$. Since $W_1=\{(\mu a_n,a_2,...,a_n):$for some fixed $\mu \in \mathbb C\}=span\{(\mu,0,0,..,1),e_2,e_3,...,e_{n-1}\}$

What is the problem with $W_2$? Are $W_1$ and $W_2$ same? How they are different?In the answer key, It is given that $W_2$ is not a vector space. Answer given as $(0,0,...,0)\notin W_2$. Why? For $(0,0,...,0)$ vector, $a_1=0,a_n=0 \implies a_1=\mu a_n=0=\mu 0$. Satisfying the condition. right?

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  • $\begingroup$ I've reformatted to make the definitions of $W_1$ and $W_2$ line up. Unfortunately, they're identical, so it's tough to see what you're asking. Is there a typo perhaps? $\endgroup$ – John Hughes May 15 '19 at 2:42
  • $\begingroup$ I have edited my typo. Sorry for the typo $\endgroup$ – Math geek May 15 '19 at 2:50
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If $a_1/a_2 = \mu$, then $a_2$ cannot be zero. Hence the zero-vector is not in $W_2$.

This is more or less a trick question, alas.

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  • $\begingroup$ But Can't we write $a_1=\mu a_2$? $\endgroup$ – Math geek May 15 '19 at 2:59
  • $\begingroup$ Okay. $a_1/a_2$ means $a_1.a_2^{-1}=\mu$. $0$ has no inverse. so $a_2 a_2^{-1}$ need not exist. Am I giving correct explanation? $\endgroup$ – Math geek May 15 '19 at 3:03
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    $\begingroup$ Well...I'd simply say "Suppose that $(0, 0, \ldots) \in W_2$. Then $0/0 = \mu$, which is impossible." $\endgroup$ – John Hughes May 15 '19 at 12:27
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$$ W_1=\{(a_1,a_2,...,a_n)\in \mathbb C^n \mid a_1=\mu a_n, \text{for some fixed $\mu \in \mathbb C$} \}\\ $$

is not the same as $$W_2=\{(a_1,a_2,...,a_n)\in \mathbb C^n \mid \frac{a_1}{a_2}=\mu , \text{for some fixed $\mu \in \mathbb C$}\}. $$

Note that for $(0,0,...,0)$ while $a_1 =\mu a_2$ for every $\mu$, we do not have $$\frac {a_1}{a_2} = \mu$$ because the expression $\frac {0}{0}$ is undefined.

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