Kernel, Basis, Rank: Hints & Answers

[victoranderson]

Please see attached question
In my opinion this question is conceptional and abstract..

For part a and b,
I think dim(Ker(D)) = 1 and Rank(D) = n
but I do not know how to explain them

For part c
What I can think of is if we differentiate f(x) by n+1 times
then we will get 0

Can somebody give me some hints, please?

 

[Dick]

Please see attached question
In my opinion this question is conceptional and abstract..

For part a and b,
I think dim(Ker(D)) = 1 and Rank(D) = n
but I do not know how to explain them

For part c
What I can think of is if we differentiate f(x) by n+1 times
then we will get 0

Can somebody give me some hints, please?

dim(ker(D))=1, yes. Rank(D)=n, also right. I think you should start by writing down a basis for $P_n$. Use that to try and explain.

 

[PeroK]

Why is Rank(D) not = n? Pn has dimension n+1.

 

[Dick]

Why is Rank(D) not = n? Pn has dimension n+1.

It does. I miscounted. Sorry. I edited the original answer.

 

[Mandelbroth]

Please see attached question
In my opinion this question is conceptional and abstract.

I don't think that's a bad thing, is it? :tongue:

For part a and b,
I think dim(Ker(D)) = 1 and Rank(D) = n
but I do not know how to explain them

For part (a), what are the only polynomials with derivative zero?

For part (b), we have a couple options. The easiest way is by our rank-nullity theorem.

For part c
What I can think of is if we differentiate f(x) by n+1 times
then we will get 0

Again, look at the kernel.

 

[victoranderson]

Just figure out what a basis is..

Basis of P = ${x^n,x^{n-1},...,x,1}$

Kernel is everything that gets mapped to 0
The only polynomial with derivative 0 is $a_0$, with basis 1
so basis for kernel D is 1

Is this a correct explanation?

 

[victoranderson]

(b) null (D) = 1
since dim($P_n$)=n+1
by rank-nullity theorem, rank (D) = n
Pretty sure this explanation is correct

 

[victoranderson]

For part (c)

I think the matrix D is look like (see attached)
but how does we relate to D^(n+1)?

 

[Dick]

Just figure out what a basis is..

Basis of P = ${x^n,x^{n-1},...,x,1}$

Kernel is everything that gets mapped to 0
The only polynomial with derivative 0 is $a_0$, with basis 1
so basis for kernel D is 1

Is this a correct explanation?

Best to say the basis for kernel(D) is {1} because it's a set. But yes, correct.

(b) null (D) = 1
since dim($P_n$)=n+1
by rank-nullity theorem, rank (D) = n
Pretty sure this explanation is correct

Also correct.

For part (c)

I think the matrix D is look like (see attached)
but how does we relate to D^(n+1)?

What the matrix looks like depends on which components of the vector correspond to which basis elements. Yours is where the topmost element of the vector corresponds to the coefficient of x^n. But, yes, it's fine. D^(n+1) is just that matrix multiplied by itself n+1 times. What's the result?

 

[victoranderson]

Best to say the basis for kernel(D) is {1} because it's a set. But yes, correct.

Also correct.

What the matrix looks like depends on which components of the vector correspond to which basis elements. Yours is where the topmost element of the vector corresponds to the coefficient of x^n. But, yes, it's fine. D^(n+1) is just that matrix multiplied by itself n+1 times. What's the result?

Obviously, D^(n+1) = 0
but how to show it? Is there any suitable method except showing by induction?

 

[Dick]

Obviously, D^(n+1) = 0
but how to show it? Is there any suitable method except showing by induction?

You could argue it's true just by using what you know about derivatives. Also your matrix has zeros along the diagonal and is only nonzero along the first subdiagonal. If you look at D^2 that's zero everywhere except along the second subdiagonal. By the time you get to D^(n+1) you'll run out a subdiagonals. I don't think it needs a formal induction argument.