Determine whether Linear Subspace

[Kaspelek]

Hi guys, back again...Any help on this question would be appreciated.

Thanks in advance :)View attachment 803

 

[Opalg]

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]

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 :)

 

[alyafey22]

Re: Subspace

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

 

[Kaspelek]

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

 

[alyafey22]

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]

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?

 

[alyafey22]

Re: Subspace

you can use integration

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

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

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

---

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

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

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

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

 

[Kaspelek]

Re: Subspace

you can use integration

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

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

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

---

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

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

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

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]

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*(\(\displaystyle \frac{x^2}{2} +bx+c\)) = aTherefore 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]

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.

 

[alyafey22]

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]

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!