# Kernel, Basis, Rank: Hints & Answers

• victoranderson
In summary, a kernel in linear algebra is the set of all inputs that produce an output of zero when applied to a linear transformation. The basis of a vector space is defined as a set of linearly independent vectors that span the entire space. The rank of a matrix refers to the maximum number of linearly independent rows or columns in the matrix. Some properties of the kernel of a matrix include containing the zero vector, being a subspace of the domain, and being closed under scalar multiplication and vector addition. To find the basis of a kernel, you can solve the homogeneous system of equations represented by the matrix or use row reduction techniques.
victoranderson
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?

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victoranderson said:
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.

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

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

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

victoranderson said:
In my opinion this question is conceptional and abstract.
I don't think that's a bad thing, is it?

victoranderson said:
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.

victoranderson said:
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.

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?

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

For part (c)

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

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victoranderson said:
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.

victoranderson said:
(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.

victoranderson said:
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?

Dick said:
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?

victoranderson said:
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.

## 1. What is a kernel in linear algebra?

A kernel in linear algebra refers to the set of all inputs that produce an output of zero when applied to a linear transformation. It is also known as the null space and is an important concept in understanding linear transformations.

## 2. How is the basis of a vector space defined?

The basis of a vector space is defined as a set of linearly independent vectors that span the entire vector space. This means that any vector in the space can be written as a unique linear combination of the basis vectors. The number of vectors in the basis is equal to the dimension of the vector space.

## 3. What is the rank of a matrix?

The rank of a matrix refers to the maximum number of linearly independent rows or columns in the matrix. It is also equivalent to the dimension of the column space or row space of the matrix. The rank of a matrix is an important concept in determining the invertibility and solutions of linear systems.

## 4. What are some properties of the kernel of a matrix?

Some properties of the kernel of a matrix include: it always contains the zero vector, it is a subspace of the domain of the matrix, and it is closed under scalar multiplication and vector addition. Additionally, the dimension of the kernel is related to the rank of the matrix through the rank-nullity theorem.

## 5. How do I find the basis of a kernel?

To find the basis of a kernel, you can solve the homogeneous system of equations represented by the matrix and set the free variables to any arbitrary values. The resulting vectors will form the basis of the kernel. You can also use row reduction techniques to find a basis for the kernel, as the non-pivot columns in the reduced row-echelon form of the matrix will correspond to basis vectors for the kernel.

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