Proving subspace of equation with first and second derivatives

Question

QUESTION

Let $P_n$ be the set of real polynomials of degree at most $n$ , and write $p′$ and $p′′$ for the first and second derivatives of $p$ . Show that:

$S = \{p \in P_4 : p''(-7)+7p'(-7) = 0\}$

is a subspace of $P_4$.

My issue

I know to prove it is a subspace of $P_4$ I need to prove it has zero vector, is closed under addition vector and closed under scalar vector. However, how do I do this with first and second derivatives? I've only done it with matrices so far

Answer 1

Setting $$S=\left\{p(x)\in P_{4}(\mathbb{R}): p''(-7)+ 7p'(-7)=0 \right\}\subseteq P_{4}(\mathbb{R}).$$ We need to prove that $S$ is a subspace vector of $P_{4}(\mathbb{R})$ so you need to prove that

1. $S\not=\emptyset$.
2. $\forall s_{1},s_{2}\in S: s_{1}\oplus s_{2}\in S$.
3. $\forall \alpha \in \mathbb{R}, \forall s\in S: \alpha \odot s\in S$.

Sketch: Let $p(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}\in P_{4}(\mathbb{R})$, so we have that $$p'(x)=a_{1}+2a_{2}x+3a_{3}x^{2}+4a_{4}x^{3} \implies p'(-7)=a_{1}+2\cdot(-7)a_{2}+3\cdot(-7)^{2}a_{3}+4\cdot(-7)^{3}a_{4},$$ and $$p''(x)=2a_{2}+6a_{3}x+12a_{4}x^{2} \implies p''(-7)=2a_{2}+6\cdot(-7)a_{3}+12\cdot(-7)^{2}a_{4}$$ Thus $$S=\left\{a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}\in P_{4}(\mathbb{R}): 2a_{2}+6\cdot(-7)a_{3}+12\cdot(-7)^{2}a_{4}+7\left( a_{1}+2\cdot(-7)a_{2}+3\cdot(-7)^{2}a_{3}+4\cdot(-7)^{3}a_{4} \right)=0\right\} $$

It's clear that $S\not=\emptyset$ because $\vec{0}_{P_{4}(\mathbb{R})}\in S$ since that $$\vec{0}_{P_{4}(\mathbb{R})}:=0+0x+0x^{2}+0x^{3}+0x^{4},$$ satisfies the condition in $S$.

Is closed under the addition, since that $$s_{1}(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\in S$$ and $$ s_{2}(x)=b_{0}+b_{1}x+b_{2}x^{2}+b_{3}x^{3}+b_{4}x^{4}\in S$$implies $s_{1}(x)+s_{2}(x)\in S$ because the sum satisfies the condition in $S$.

Is closed under the product of scalar vector, since that for all $\alpha \in \mathbb{R}$ and $$s(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}\in S$$ implies that $\alpha \odot s(x)\in S$ because satisfies the condition in $S$.

Of course you need to complete the details in the sketch for complete the proof. So, you should ty to complete of details and if you have any questions you can write in the comments to help you.

Answer 2

1. It should be clear that the zero polynomial $p=0$ satiesfies $p''(-7)+ 7p'(-7)=0.$

2. Let $p_1,p_2 \in S$ and $ \alpha \in \mathbb R.$ Let $p=p_1+p_2$ and $q= \alpha p.$ With the elementary rules of differentiation show that

$ \quad \quad p''(-7)+ 7p'(-7)=0$ and $q''(-7)+ 7q'(-7)=0.$