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Information:
Let $ \mathbb{C}[z] $ denote the set of all polynomials with complex coefficients. Let $ \mathbb{C}_5[z] $ denote the set of all polynomials in $ \mathbb{C}[z] $ with degree at most 5.

Question:
Given that $ \mathbb{C}[z] $ is a vector space, show that $ \mathbb{C}_5[z] $ is a vector space.

Approach:
Since $ \mathbb{C}_5[z] $ is in $ \mathbb{C}[z] $ we evaluate if $ \mathbb{C}_5[z] $ is a subspace of $ \mathbb{C}[z] $.

1) Additive Identity:
$$ 0 \cdot p^n = 0 \in \mathbb{C}_5[z] $$ where $ 0 \leq n \leq 5 $

2) Closure under Addition:
Given two polynomials $$ f(z), g(z) \in \mathbb{C}_5[z] $$ where $$ f(1) = g(1) = 0 $$ then $$ f(1) + g(1) = 0 + 0 = 0 $$ demonstrating closure under addition

3) Closure under Scalar Multiplication
Given $$ \alpha \cdot f(1) = \alpha \cdot 0 = 0 \in \mathbb{C}_5[z] $$ for any $$ \alpha \in \mathbb{F} $$ demonstrating closure under scalar multiplication

Since $ \mathbb{C}_5[z] $ is a subspace of $ \mathbb{C}[z] $, it follows that $ \mathbb{C}_5[z] $ is a vector space.

Is this a correct proof for the given question, or is there a logical leap that I am making somewhere?

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  • $\begingroup$ Why do you say $f(1)=0$? That is not the defining property here. Instead you should compare the degree of $f$, the degree of $g$, and the degree of $f+g$. $\endgroup$
    – vadim123
    Oct 19, 2017 at 5:13
  • $\begingroup$ @vadim123 Essentially I should show that a polynomial with max degree 5 added with another polynomial with max degree 5 would still be max degree 5 thus, being closed under addition? $\endgroup$
    – deko
    Oct 19, 2017 at 5:16
  • $\begingroup$ That is correct, to prove closure under addition. $\endgroup$
    – vadim123
    Oct 19, 2017 at 5:17

2 Answers 2

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No, but it is easy to show that sum of two polynomials of degree $\leq5$ is a polynomial of degree $\leq5$ and multiplying a polynomial of degree $\leq5$ by a number gives a polynomial of degree $\leq5$.

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First of all, in $1$, you are taking a specific polynomial, and showing the result through that, which makes your argument invalid.

Secondly, in $2$, you need to show that if you add to polynomials, that will result in an again a polynomial of degree less than 5, so what you are doing again shows nothing.

Thirdly, you are using something like $f(1)$, which is even not defined in the sense of polynomials. In other words, polynomials are not functions, they are just expressions in their pure form.

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