2
$\begingroup$

I'm struggling with this exercise. Please help me out. In the following 1. will refer to the existence of the zero vector, 2. to the closure under addition with $a:=(a_1,a_2,a_3)^T$ and $b:=(b_1,b_2,b_3)^T$, 3. to the closure under multiplication with $0\neq c\in\mathbb{R}$ as the scalar.


$V_1:=\{x_1,x_2,x_3\in\mathbb{R}^3|2x_1+5x_2=x_3 \}$.

$V_2:=\{x_1,x_2,x_3\in\mathbb{R}^3|x_1^2+x_2^2=2x_1x_2\}$.

$V_3:=\{x_1,x_2,x_3\in\mathbb{R}^3|x_1=0\lor x_2=0\}$.

$V_4:=\{x_1,x_2,x_3\in\mathbb{R}^3|3x_1+x_2=0\land x_2+2x_3=0\}$.

$V_5:=V_1\cup V_2$.


$V_1$.

  1. $2\cdot 0 + 5\cdot 0 = 0$.

  2. $a+b=2a+5b=2a_1+2b_1+5a_2+5b_2=(2a_1+5a_2)+(2b_1+5b_2)=a_3+b_3$.

  3. $c2x_1+5cx_2=cx_3 \Leftrightarrow c(2x_1+5x_2)=cx_3\Leftrightarrow 2x_1+5x_2=x_3$.

$V_2$.

  1. $0^2 + 0^2=0=2\cdot0\cdot0$.

  2. $a+b=a_1^2+b_1^2+a_2^2+b_2^2=(a_1^2+a_2^2)+(b_1^2+b_2^2)=2a_1a_2 + 2b_1b_2 = 2(a_1b_1 + a_2b_2)$.

  3. $cx_1^2+cx_2^2=c2x_1x_2\Leftrightarrow c(x_1^2+x_2^2)=c(2x_1x_2) \Leftrightarrow x_1^2+x_2^2=2x_1x_2$.

$V_3$.

This set is not a subspace since it is not closed under addition, e.g. for $a_1=0$ and $b_1\neq0$, $a+b$ is only closed iff $a_1=b_1 \land a_2=b_2$.

$V_4$.

Here we have a linear equation that is solved by $x_1=2/3x_3, x_2=-2x_3, x_3=x_3$.

  1. This is true for $x_3=0$.

I'm not sure what to do with 2. and 3.

$V_5$.

Is this equivalent to $V_5:=\{x_1,x_2,x_3\in\mathbb{R}^3|2x_1+5x_2=x_3 \land x_1^2+x_2^2=2x_1x_2\}$? If yes I'm similarly not sure what to do as with $V_4$.

$\endgroup$
1
  • $\begingroup$ I am not following your algebra in case $V_2$, $(a+b) = (a_1+b_1)^2 + (a_2+b_2)^2 = 2(a_1+b_1)(a_2+b_2) \cdots$ but this looks like it is going to be a mess.... however $x_1^2 + x_2^2 = 2x_1x_2 \implies (x_1-x_2)^2 = 0 \implies x_1-x_2 = 0$ is easy to work with. $\endgroup$
    – Doug M
    Commented Jun 18, 2018 at 21:30

1 Answer 1

1
$\begingroup$

Hand waving hint: For $V_4$, it looks like a subspace. Just check that two vectors for which the equations given are satisfied give you another such when added. I can see it just by looking at it. Because if, for instance, $3x_1+x_2=0 \land 3x_1'+x_2'=0$, then $3(x_1+x_1')+(x_2+x_2')=3x_1+x_2+3x_1'+x_2'=0+0=0$... Similarly for the other equation. (Uses $0+0=0$) The scalar property ($3$) follows too, since $\alpha\cdot0=0$ ($\alpha$ factors out...)

Note: we get the span of $(\frac23,-2,1)$...

Hint on $V_5$: Replace $\land$ by $\lor$... I have a feeling this isn't a subspace...

Also, $V_2$ isn't a subspace, i don't think ... Because $$(a_1+b_1)^2+(a_2+b_2)^2=a_1^2+b_1^2+2a_1b_1+a_2^2+b_2^2+2a_2b_2=2a_1b_1+2a_1b_1+2a_2b_2+2a_2b_2=4a_1b_1+4a_2b_2 \not = 2(a_1+b_1)(a_2+b_2)$$ in general...

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .