Which of these subsets are subspaces? 
I will always have a go at questions and post up my workings here when asking questions however in this case I haven't a clue what to do as we never covered this in class or tutorials. Can someone tell me what is being actually asked here and maybe show me how to do one or two so I can have a go at the rest? Does P stand for polynomial or something?
 A: I presume $P_2$ stands for the set of all polynomials of degree less than 2 over some already specified field, such as the real numbers for example. 
They are asking you to determine whether the subsets of $P_2$ you have listed above satisfy the criterion for being a vector subspace (essentially, a vector space within a vector space). To do this you need to know what properties a vector space must satisfy and check whether these subsets satisfy these properties. Here goes: 
A vector space is a set $V$, whose elements are called vectors, on which we have two operations, namely the binary operation of vector addition and scalar multiplication in the field $F$, satisfying the following axioms:
$(V,+)$ Axioms:
(1) $\vec{v}+\vec{w}=\vec{w}+\vec{v}$ for all $\vec{v},\vec{w} \in V$.
(2) $(\vec{u}+\vec{v})+\vec{w}=\vec{u}+(\vec{v}+\vec{w})$ for all $\vec{u},\vec{v},\vec{w} \in V$. 
(3) There exists a vector $\vec{0} \in V$ such that $\vec{v}+\vec{0}=\vec{v}$ for all $\vec{v} \in V$.
(4) For every $\vec{v} \in V$ there exists a vector $-\vec{v}$ such that $\vec{v}+(-\vec{v})=\vec{0}$.  
Scalar Multiplication Axioms:
(5) For all $\vec{v} \in V$, we have $1\cdot \vec{v}=\vec{v}$. 
(6) $(ab)\vec{v}=a(b\vec{v})$ for all $a,b \in F$ and $\vec{v} \in V$.
Distributive Axioms:
(7) $(a+b)\vec{v}=a\vec{v}+b\vec{v}$ for all $a,b \in F$ and $\vec{v} \in V$.
(8) $a(\vec{v}+\vec{w})=a\vec{v}+a\vec{w}$ for all $a\in F$ and $\vec{v},\vec{w} \in V$. 
Written using group theory terminology, a vector space is an abelian group $V$ acted upon by elements of a field. The nature of the actions are expressed in the scalar multiplication and distributive axioms.
Verifying that a subset is a subspace is not as tedious as checking all of the axioms above, because your subset automatically inherits all of the properties from the larger set, except for closure under the vector space operations. This is what you must check. 
A: Once you learn about linear transformations, the following characterization is often useful:

Let $\mathbf{V}$ be a vector space. A subset $S\subseteq \mathbf{V}$ is a subspace of $\mathbf{V}$ if and only if there is a linear transformation $T$ with domain $\mathbf{V}$ such that $\mathrm{ker}(T)=S$.

This is the vector space version of "$N\subset G$ is a normal subgroup of the group $G$ if and only if $N$ is the kernel of a group homomorphism with domain $G$", and "$I\subseteq R$ is an ideal of the ring $R$ if and only if $I$ is the kernel of a ring homomorphism with domain $R$"; if you don't know about groups and rings, don't worry about that.
In any case, it is easy to see that the kernel of a linear transformation must be a subspace. For your sets:


*

*$S=\{p(t)\in\mathbf{P}_2(t)\mid p'(4)=p(2)\}$. This is the kernel of the map $T\colon\mathbf{P}_2(t)\to\mathbb{R}$ by $T(p) = p'(4)-p(2)$. (Evaluation is linear, as is the derivative; the composition of linear maps is linear, and the difference of linear maps is linear, so $T$ is linear).

*$S=\left\{p(t)\,\left|\,\int_0^7p(t)\,dt=0\right\}\right.$ is the kernel of $T\colon\mathbf{P}_2(t)\to\mathbb{R}$ given by $T(p) = \int_0^tp(t)\,dt$, which is linear.

*$\{p(t)\in\mathbf{P}_2\mid p'(t)+7p(t)+2=0\}$ is not the kernel of any linear transformation, since it does not include $\mathbf{0}$.

*$\{p(t)\in\mathbf{P}_2\mid p'(t)\text{ is constant}\}$ is the same as $\{p(t)\in\mathbf{P}_2\mid p''(t)=0\}$. Since $p\mapsto p''$ is linear, this is a kernel hence a subspace.

*$\{p(t)\in\mathbf{P}_2\mid p(-t)=-p(t)\}$ is the kernel of $T(p) = p(t)+p(-t)$.

*$\{p(t)\in\mathbf{P}_2\mid p(1)=5\}$ is not the kernel of a linear transformation, since $\mathbf{0}$ is not in the set.
A: To check that a subset of a vector space is a subspace, we must check that it is a vector space in its own right. Most of the usual axioms hold true (inherited), but we must check that the set is closed under linear combinations. That is, if $a,b\in\mathbb{F}$ (coefficients in the field under consideration, usually the reals $\mathbb{R}$), and $u,v\in S$ are elements of the subset, then $au+bv\in S$ too. Options C and F can be ruled out instantly because they have don't have $0$ (that is, the zero polynomial); if they did, we would have absurdities like $2=0$ or $0=5$. All the others you can check are subspaces by linearity; for example if $u'(4)=u(2)$ and $v'(4)=v(2)$ (that is, $u,v$ are in (A)'s subset), then $w=au(t)+bv(t)$ has as its derivative $w'=au'+bv'$, which evaluated at $4$ is $au'(4)+bv'(4)$. Using our assumptions, we can substitute for the derivatives at $4$ and obtain $au(2)+bv(2)$, which is exactly $w(2)$. Thus we know $w'(4)=w(2)$, proving $w=au+bv$ is in the subset, thus proving the subset is a subspace.
A: To prove this you could use the Subspace Criterion (also known as the Subspace Test).
Let $\left(X,\mathbb{P}\right)$ be a linear space over a field $\mathbb{P}$.
$L\subseteq X ,  L\neq\emptyset$
$L$ is a subspace of $X$ $\Leftrightarrow \forall a,b \in X, \forall\alpha,\beta\in \mathbb{P} :  \alpha a + \beta b \in L$
So, basically you take two vectors (polynomials in this case) with a certain property and check if the linear combination of these vectors also has this property. If so, then subset of such vectors is going to be a linear subspace.
For example, let's solve number 2:
$L = \left\{p(t) | \int_{0}^{7} p(t) \text{d}t = 0 \right\}$
We take two polynomials $p_{1}(t), p_{2}(t) \in L$ and check if the integral from 0 to 7 of the linear combination equals 0.
$\int_{0}^{7} (\alpha p_{1}(t) + \beta p_{2}(t) ) \text{d}t = \alpha\int_{0}^{7} p_{1}(t) \text{d}t + \beta\int_{0}^{7} p_{2}(t) \text{d}t = \alpha \cdot 0 + \beta\cdot0 = 0$
(note: here we used additivity and homogeneity of degree 1 of a definite integral) 
So, $L$ is a subspace of $\mathbb{P}_{2}[t]$
All the other problems of this kind can be solved exactly the same way, be it vectors, polynomials, matrices, etc. You just use the criterion.
However, when dealing with polynomials there are a couple of things to know:


*

*If you are given a root, say $\lambda$ so $p(\lambda) = 0$ then you could use the polynomial remainder theorem which says that if $\lambda$ is a root then $p(t) = q(t)(t-\lambda )$.

*If you are given that polynomials are divisible by some other polynomial q(t), you could rewrite them as $p(t) = g(t)q(t)$.
