Homework Help: Determine whether Subsets are Subspaces

1. Jul 8, 2009

Here we go....wheeeee

1. The problem statement, all variables and given/known data
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)
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.

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.

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

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

(d) .... working on it

2. Jul 8, 2009

tiny-tim

(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

3. Jul 8, 2009

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?

4. Jul 8, 2009

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.

Last edited: Jul 8, 2009
5. Jul 8, 2009

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

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

6. Jul 8, 2009

tiny-tim

Yup … your c and d are fine now!

7. Jul 8, 2009