Determine whether Subsets are Subspaces

[Saladsamurai]

Here we go...wheeeee

Homework Statement

For each of the following subsets of F³, determine whether it is a subspace of F³

(a)$${(x_1, x_2, x_3) \in \mathbf{F}^3: x_1+2x_2+3x_3=0}$$

(b)$${(x_1, x_2, x_3) \in \mathbf{F}^3: x_1+2x_2+3x_3=4}$$

(c)$${(x_1, x_2, x_3) \in \mathbf{F}^3: x_1x_2x_3=0}$$

(d)$${(x_1, x_2, x_3) \in \mathbf{F}^3: x_1=5x_3}$$

My attempt

(a)
additive ID:
the additive ID exists in the subset since if (x1,x2,x3)=0 then the criteria that x_1+2x_2+3x_3=0 holds true.

closure under addition:
given
$${(x_1, x_2, x_3) \in \mathbf{F}^3: x_1+2x_2+3x_3=0}\text{ and }$$
$${(y_1, y_2, y_3) \in \mathbf{F}^3: y_1+2y_2+3y_3=0}$$

then,

$$x_1=-(2x_2+3x_3)\text{ and }y_1=-(2y_2+3y_3)$$

then $$(x_1, x_2, x_3)+(y_1, y_2, y_3)=(x_1+y_1,x_2+y_2,x_3+y_3)$$

$$=(-(2x_2+3x_3)-(2y_2+3y_3),x_2+y_2,x_3+y_3)$$

Applying the condition that $$x_1+2x_2+3x_3=0$$

We have:$$-(2x_2+3x_3)-(2y_2+3y_3)+2(x_2+y_2)+3(x_3+y_3)=0$$

Closure under Scalar multiplication: given
$${a\in\mathbf{F}\text{ and }(x_1, x_2, x_3) \in \mathbf{F}^3: x_1+2x_2+3x_3=0}$$

$$a(x_1,x_2,x_3)=(ax_1,ax_2,ax_3)$$

Applying the condition we have:

$$ax_1+2ax_2+3ax_3=0=a(x_1+2x_2+3x_3)$$

Thus the subset is a subspace. Look okay?

(b)
immediately we can see that if (x1,x2,x3)=0 than the condition that $x_1+2x_2+3x_3=4$ does not hold. Not a subspace.(c)
I am having a little trouble seeing this one. if (x1,x2,x3)=0 than the condition that $x_12x_23x_3=0$ is true confirming that the additive ID is included.

Under addition we have

$$(x_1+y_1,x_2+y_2,x_3+y_3)$$ applying the condition that the product=0 we have

$(x_1+y_1)(x_2+y_2)(x_3+y_3)=0$

I am not sure what to say about this??
(d) ... working on it

 

[tiny-tim]

Hi Saladsamurai!

Here we go...wheeeee

(no leaning over the side! )

Yup, your (a) and (b) are fine.

Hint for (c): what if x₁ = y₂ = 0?

and try rewriting (d) as x₁ - 5x₃ = 0

 

[Dick]

For (c) it may be clearer to look at a specific example. a=(1,1,0) and b=(0,1,1) are in your subset, right? What about a+b?

 

[Saladsamurai]

For (c) it may be clearer to look at a specific example. a=(1,1,0) and b=(0,1,1) are in your subset, right? What about a+b?

Then a+b=(1+0, 1+1, 0+1) and the product of these elements $\ne 0$ therefore all a+b $\not\in$ U (where U is the subset in question)

So c is not a subspace.

 

[Saladsamurai]

For d : If U is the subset in question: The additive identity is on U since if x1=x2=x3=0 then x1=5*x3=0

Closure under addition:
Let a=(x1,x2,x3) where x1=5*x3 and b=(y1,y2,y3) where y1=5*y3
=> a+b = ((5*x3+5*y3), x2+y2, x3+y3)

=(5(x3+y3), x2+y2, x3+y3) => closed.

Closure under scalar multiplication: if r is a scalar and U= {(x1,x2,x3): x1=5*x3} then,

r(x1,x2,x3)=(5r*x3,r*x2,r*x3) => closed under mult.

So d is a subspace

 

[tiny-tim]

Yup … your c and d are fine now!

 

[Saladsamurai]

This isn't so bad

:notices he has 10 more chapters to go: