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

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 $\mathbb{C}[z]$.

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

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?

• 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 $f+g$. Oct 19, 2017 at 5:13
• @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?
– deko
Oct 19, 2017 at 5:16
• That is correct, to prove closure under addition. Oct 19, 2017 at 5:17

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 $\leq5$.
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.