Linear algebra: Vector subspaces

[cesaruelas]

Homework Statement

Is the subset of
P= {a₀ + a₁x + a₂x² + ... + a_nxⁿ}
formed only by the polynomials that satisfy the condition:
a₁a₃≤0
a vector subspace?

Homework Equations

I already proved the subset is not closed under addition so I know it's not a vector subspace, however, the answer my teacher marked as correct reads: "No, it's closed under the product but not under addition."

The Attempt at a Solution

Let P_b and P_c be elements of a subset of P= {a₀ + a₁x + a₂x² + ... + a_nnⁿ} with their coefficients given by any series in function of n (this is an assumption I'm not sure I can make but I find no other way to tackle the product step without making it).

Addition:
P_b + P_c = {b₀+c₀ + (b₁+c₁)x + ... + (b₃+c₃)x³ + ... + (b_n+c_n)xⁿ}
where
(b₁+c₁)(b₃+c₃)≤0
if
(b₁+c₁)≤0 and (b₃+c₃)≥0
or
(b₁+c₁)≥0 and (b₃+c₃)≤0

Therefore, it is not closed under addition.

Product:
P_b * P_c = {b₀c₀ + (b₁c₁)x + ... + (b₃c₃)x³ + ... + (b_nc_n)xⁿ}
where
(b₁c₁)(b₃c₃)≤0
if
(b₁c₁)= 0
and/or
(b₃c₃)= 0
(because of their positions in the series).

Therefore, it is not closed under the product.

Was the assumption mistaken? If so, can you give me a clue how to prove that it is/isn't closed under the product?

 

[vela]

Homework Statement

Is the subset of
P= {a₀ + a₁x + a₂x² + ... + a_nxⁿ}
formed only by the polynomials that satisfy the condition:
a₁a₃≤0
a vector subspace?

Homework Equations

I already proved the subset is not closed under addition so I know it's not a vector subspace, however, the answer my teacher marked as correct reads: "No, it's closed under the product but not under addition."

The Attempt at a Solution

Let P_b and P_c be elements of a subset of P= {a₀ + a₁x + a₂x² + ... + a_nnⁿ} with their coefficients given by any series in function of n (this is an assumption I'm not sure I can make but I find no other way to tackle the product step without making it).

Addition:
P_b + P_c = {b₀+c₀ + (b₁+c₁)x + ... + (b₃+c₃)x³ + ... + (b_n+c_n)xⁿ}
where
(b₁+c₁)(b₃+c₃)≤0
if
(b₁+c₁)≤0 and (b₃+c₃)≥0
or
(b₁+c₁)≥0 and (b₃+c₃)≤0

Therefore, it is not closed under addition.

I don't get what your logic is here.

Product:
P_b * P_c = {b₀c₀ + (b₁c₁)x + ... + (b₃c₃)x³ + ... + (b_nc_n)xⁿ}
where
(b₁c₁)(b₃c₃)≤0
if
(b₁c₁)= 0
and/or
(b₃c₃)= 0
(because of their positions in the series).

Therefore, it is not closed under the product.

Was the assumption mistaken? If so, can you give me a clue how to prove that it is/isn't closed under the product?

You want to show that subset P has to be closed under scalar multiplication. That is, if p ∈ P and r ∈ R, then (rp) ∈ P.

 

[cesaruelas]

Thank you for replying. I know, I realized after two hours that I needed to use scalar product and not whatever I was doing, lol. How would you do the first part, though? I suck at this type of problems :/.

 

[vela]

Since you're trying to disprove a statement, all you need to do is find a single counterexample. So first assume
\begin{align*}
p_b &= b_0 + b_1x + b_2x^2 + \cdots + b_n x^n \in P \\
p_c &= c_0 + c_1x + c_2x^2 + \cdots + c_n x^n \in P \\
\end{align*} which means $b_1b_3 \le 0$ and $c_1c_3 \le 0$. Then form the sum
$$ p = p_b + p_c = (b_0+c_0) + (b_1+c_1)x + (b_2+c_2)x^2 + \cdots + (b_n+c_n)x^n.$$ You want to show there exists a case where $(b_1+c_1)(b_3+c_3) \gt 0$ which would imply that p is not an element of P. In other words, find specific numbers for b₁, b₃, c₁, and c₃ that satisfy those three conditions.