# Determine whether Linear Subspace

#### Opalg

##### MHB Oldtimer
Staff member
Re: Subspace

Hi guys, back again....

Any help on this question would be appreciated.

Thanks in advance View attachment 803
The test for a subspace is that if it contains $x$ and $y$ then it also contains $x+y$ and $cx$ (where $c$ is any constant). Apply that test to each of those two subsets.

#### Kaspelek

##### New member
Re: Subspace

The test for a subspace is that if it contains $x$ and $y$ then it also contains $x+y$ and $cx$ (where $c$ is any constant). Apply that test to each of those two subsets.
Hey, these are the laws closed under vector addition and multiplication?

However perhaps you can provide an example in this situation, not really sure how to go about it #### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Subspace

First find all polynomials whose second derivative is 0 , do you know how to do that ?

#### Kaspelek

##### New member
Re: Subspace

First find all polynomials whose second derivative is 0 , do you know how to do that ?
Doesn't it have to be a second degree polynomial?

I know that for the one where derivative = 1 1/2(x^2) '' = 1

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Subspace

Doesn't it have to be a second degree polynomial?

I know that for the one where derivative = 1 1/2(x^2) '' = 1
Start by the general form $$\displaystyle P(x)=ax^2+bx+c$$ then try to find the constants , what do you get ?

#### Kaspelek

##### New member
Re: Subspace

Start by the general form $$\displaystyle P(x)=ax^2+bx+c$$ then try to find the constants , what do you get ?
So if you double differentiate, 2a=0 for the second derivative to yield 0?

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So if you double differentiate, 2a=0 for the second derivative to yield 0?

There is no second degree polynomial that yields a second derivative =0?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Subspace

you can use integration

$$\displaystyle p''(x)=1$$

$$\displaystyle p'(x)= x +b$$

$$\displaystyle p(x)= \frac{x^2}{2} +bx+c$$

[HR][/HR]

$$\displaystyle p''(x)=0$$

$$\displaystyle p'(x)= a$$

$$\displaystyle p(x)= ax+c$$

Now test each one to see whether it forms a subspace .

#### Kaspelek

##### New member
Re: Subspace

you can use integration

$$\displaystyle p''(x)=1$$

$$\displaystyle p'(x)= x +b$$

$$\displaystyle p(x)= \frac{x^2}{2} +bx+c$$

[HR][/HR]

$$\displaystyle p''(x)=0$$

$$\displaystyle p'(x)= a$$

$$\displaystyle p(x)= ax+c$$

Now test each one to see whether it forms a subspace .

I'm not trying to leech here, i'm just legitimately confused.

So i'm thinking of testing if its closed under multiplication first.

Am i correct in saying (assume a=alpha)

a* $$\displaystyle p(x)= \frac{x^2}{2} +bx+c$$ does not equal a*p''(x) ?

How do i correctly show this proof?

#### Kaspelek

##### New member
Re: Subspace

I'm not trying to leech here, i'm just legitimately confused.

So i'm thinking of testing if its closed under multiplication first.

Am i correct in saying (assume a=alpha)

a* $$\displaystyle p(x)= \frac{x^2}{2} +bx+c$$ does not equal a*p''(x) ?

How do i correctly show this proof?

Does this correctly answer the question?

Test if subsets are closed under multiplication.

a*($$\displaystyle \frac{x^2}{2} +bx+c$$)

d^2/dx^2(a*($$\displaystyle \frac{x^2}{2} +bx+c$$) = a

Therefore since a*p(x) does not equal p''(x) not closed. Hence not a subspace.

a*($$\displaystyle ax+c$$)

d^2/dx^2(a*($$\displaystyle ax+c$$)) = 0

Therefore since a*P(x)=P''(x) this subset is a subspace.

Thoughts?

#### Opalg

##### MHB Oldtimer
Staff member
I think I may have misled you in my first comment by using bad notation. The situation is that if you have $P(x)$ and $Q(x)$, and $c$ is a constant, then you need to test whether $P(x)+Q(x)$ and $cP(x)$ satisfy the appropriate condition.

To avoid doing the exact same problem for you, suppose for example that you were given the set $U$ of all polynomials whose third derivative is 3: $U = \{P(x)\in\mathcal{P}_3|P'''(x)=3\}$. If $P(x)$ and $Q(x)$ satisfy that condition then $P'''(x) = Q'''(x)=3$, so $P'''(x)+Q'''(x) = 3+3=6$. So the third derivative of $P(x)+Q(x)$ is $6$. But $6\ne3$, so $P(x)+Q(x)$ does not satisfy the condition for being in $U$. Therefore $U$ is not a subspace. Incidentally, the third derivative of $cP(x)$ is $3c$, and since $3c\ne3$ (unless the constant $c$ happens to be equal to $1$), $U$ does not satisfy either of the conditions for being a subspace.

Edit. Your previous comment #10 is a mostly correct solution. It shows that the first set is not a subspace. But to show that the second set is a subspace you need to check that it satisfies both of the conditions for a subspace (vector addition as well as scalar multiplication). But you might also like to see the method I have used in this example above.

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
You can use the method Oplag illustrated or you can do the following

we have proved that all the polynomials of degree 2 that satisfy the conditions must be of the form

$$\displaystyle P(x) = \frac{x^2}{2}+bx+c$$ were $$\displaystyle b$$ and $$\displaystyle c$$ are arbitrary numbers .

So the set contains only second degree polynomials with a leading coefficient $$\displaystyle =\frac{1}{2}$$ .

Now , if we choose two polynomials in the set ,say, $$\displaystyle R(x) = \frac{x^2}{2}+x+1$$ and $$\displaystyle Q(x) = \frac{x^2}{2}+x+2$$

We know by definition that a subspace must be closed under addition . Now, if we add the two polynomials , do we get a polynomial that is still in the set ?

#### Kaspelek

##### New member
You can use the method Oplag illustrated or you can do the following

we have proved that all the polynomials of degree 2 that satisfy the conditions must be of the form

$$\displaystyle P(x) = \frac{x^2}{2}+bx+c$$ were $$\displaystyle b$$ and $$\displaystyle c$$ are arbitrary numbers .

So the set contains only second degree polynomials with a leading coefficient $$\displaystyle =\frac{1}{2}$$ .

Now , if we choose two polynomials in the set ,say, $$\displaystyle R(x) = \frac{x^2}{2}+x+1$$ and $$\displaystyle Q(x) = \frac{x^2}{2}+x+2$$

We know by definition that a subspace must be closed under addition . Now, if we add the two polynomials , do we get a polynomial that is still in the set ?

Both methods make sense to me, cheers for the help guys!