Linear independence of polynomials of different degree

[Mr Davis 97]

Homework Statement

Let S be a set of nonzero polynomials. Prove that if no two have the same degree, then S is linearly independent.

Homework Equations

The Attempt at a Solution

We will proceed by contraposition.

Assume that S is a linearly dependent set. Thus there exists a linear dependence relation $a_1p_1 + \cdots + a_np_n = 0$ such that $a_1,...,a_n$ are not all zero. Now, the RHS of the relation has degree -1. Hence, the LHS must have degree -1 also. In this case, assume that no two polynomials have the same degree. Then the LHS would have the degree of the polynomial with largest degree, which is other than -1 since there are no zero polynomials in the set S. This is a contradiction, because the LHS must have degree -1. Hence, there must exist at least two polynomials with the same degree.

In this correct? Am I begging the question if I claim that the LHS must have degree -1 also?

 

[andrewkirk]

I would start by proving that the sum of two polynomials of different degree has the higher of the two degrees. By induction it will then follow that, if there is at least one polynomial, the LHS must have degree at least 0.

 

[Stephen Tashi]

Now, the RHS of the relation has degree -1.

How do your text materials define the degree of a polynomial that is a constant function, such as P(x) = 0 ?

 

[Mr Davis 97]

How do your text materials define the degree of a polynomial that is a constant function, such as P(x) = 0 ?

The degree of a constant function is 0, while the degree of the zero polynomial is defined to be -1.

 

[Mark44]

The degree of a constant function is 0, while the degree of the zero polynomial is defined to be -1.

I've never seen this (zero polynomial defined to be degree -1). If f(x) = k is of degree zero, why should the degree change to -1 in the case where k = 0?

 

[andrewkirk]

I've never seen this (zero polynomial defined to be degree -1). If f(x) = k is of degree zero, why should the degree change to -1 in the case where k = 0?

I also have never seen that notation. It seems quite a good idea, because it recognises the important distinction between the space generated by the zero polynomial, which is {0} and has dimension 0, and the space generated by a constant, non-zero polynomial, which is one-dimensional. The rule would be something like that the degree is one less than the dimension of the vector space spanned by all polynomials of that degree.

 

[Stephen Tashi]

I've never seen this (zero polynomial defined to be degree -1).

It's a controversial subject: http://math.stackexchange.com/quest...egree-of-the-zero-polynomial-and-why-is-it-so

If f(x) = k is of degree zero, why should the degree change to -1 in the case where k = 0?

That's my first reaction, too. However, the weight of authority is against us.

 

[Stephen Tashi]

In this correct? Am I begging the question if I claim that the LHS must have degree -1 also?

Whether the proof is formally correct depends on what theorems have previously been proven and the audience for the proof. The crucial step is the assertion that if none of a finite number of polynomials has the same degree then the polynomial that is their sum has the degree of polynomial in the summands that has the highest degree. If you can cite theorem that proves this then you should cite it. If you are presenting a proof to an audience that accepts the assertion without formal proof then your proof passes. If you are expected to back up that assertion with an detailed demonstration, then you have more work to do.