# Are $W_1$ and $W_2$ same? How they are different?

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?

• 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? – John Hughes May 15 '19 at 2:42
• I have edited my typo. Sorry for the typo – Math geek May 15 '19 at 2:50

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.

• But Can't we write $a_1=\mu a_2$? – Math geek May 15 '19 at 2:59
• 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? – Math geek May 15 '19 at 3:03
• Well...I'd simply say "Suppose that $(0, 0, \ldots) \in W_2$. Then $0/0 = \mu$, which is impossible." – John Hughes May 15 '19 at 12:27

$$W_1=\{(a_1,a_2,...,a_n)\in \mathbb C^n \mid a_1=\mu a_n, \text{for some fixed \mu \in \mathbb C} \}\\$$

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.