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Problems #6 and #32 on pg.196-197 will be graded.

#6,pg.196: Show that the set of all polynomials of the form

$p(t)=a+t^2$ where

$a \in \mathbb{R}$ is a subspace of

$\mathbb{P}_2$ , where

$\mathbb{P}_2$ denotes the vector space of all polynomials of degree

$2$ or lower.

Solution: We can write the set in question as

$S=\{p \in \mathbb{P}_2 \colon p(t)=a+t^2 \hspace{2pt} \mathrm{and} \hspace{2pt} a \in \mathbb{R} \}$ . We must answer the following three questions:

Is $S \subset \mathbb{P}_2$ ?
Is $\vec{0}_{\mathbb{P}_2} \in S$ (i.e. is the zero vector of $\mathbb{P}_2$ also an element of $S$ ?)
Is $S$ closed under addition and scalar multiplication?

To answer 1, notice that if

$q$ is an element of

$\mathbb{P}_2$ then

$q$ can be written as

$q(t)=a_0+a_1t+a_2t^2$ for some numbers

$a_0,a_1,$ and

$a_2$ . If we can express all elements of

$S$ in the same way, then we will see that

$S \subset \mathbb{P}_2$ . To that end pick any

$p \in S$ -- by definition of

$S$ we know that

$p(t)=a+t^2$ for some number

$a$ . But notice we can force

$p=q$ if we pick

$a_0=a, a_1=0,$ and

$a_2=1$ for the polynomial

$q$ . This shows that any polynomial in

$S$ is also in

$\mathbb{P}_2$ . (The "quick way" to do this is to simply say that

$\mathbb{P}_2$ contains all 2nd degree polynomials and point out that all members of

$S$ are 2nd degree polynomials.)

To answer 2, realize that the zero vector of

$\mathbb{P}_2$ is simply the polynomial

$p(t)=0$ . We must ask: is this polynomial in

$S$ ? The answer is no: as we saw earlier, if

$p \in S$ then

$p(t)=a+t^2$ for some

$a \in \mathbb{R}$ . We only have freedom to pick the value of

$a$ , because the variable "

$t$ " is part of the polynomial itself.

Therefore we must conclude that

$S$ is not a subspace of

$\mathbb{P}_2$ .

#32, pg.197: Let

$H$ and

$K$ be subspaces of a vector space

$V$ . The intersection of

$H$ and

$K$ , denoted

$H \cap K$ , is the set of all vectors

$v \in V$ that lie in both

$H$ and

$K$ (i.e. we can express this set as

$H \cap K = \{v \in V \colon v \in H \hspace{2pt} \mathrm{and} \hspace{2pt} v \in K\}$ ). Show that

$H \cap K$ is a subspace of

$V$ .

Solution: Let

$S=H \cap K$ . We would like to show that

$S$ is a subspace of

$V$ , i.e. it is a subset that is also a vector space. We again must answer the three questions:

Is $S \subset V$ ?
Is $\vec{0}_{V} \in S$ ?
Is $S$ closed under addition and scalar multiplication?

To answer 1, just observe the definition of

$H \cap K$ which begins by assuming it contains vectors in

$V$ . This is sufficient information to show that

$H \cap K \subset V$ and hence the answer to question 1 is yes.

To answer 2, notice that we assumed that

$H$ and

$K$ are subspaces of

$V$ . Since subspaces are just subsets that are also vector spaces and (by definition) ALL vector spaces contain a zero vector, we note that

$H$ and

$K$ both contain

$\vec{0}_V$ . (note: part of being a subspace is that you use the same operations as

$V$ and so the zero vector in

$H$ and

$K$ must be

$\vec{0}_V$ and couldn't be anything else). Thus the answer to question 2 is yes.

To answer 3, we must show that if

$a$ and

$b$ are vectors in

$H \cap K$ and

$\alpha,\beta$ are scalars, then

$\alpha a + \beta b \in H \cap K$ . To see this again recall that

$H$ and

$K$ are subspaces of

$V$ (and hence vector spaces) and so since

$a$ and

$b$ are in

$H$ , then

$\alpha a + \beta b$ is also in

$H$ (else

$H$ wouldn't be a vector space). Similarly, since

$a$ and

$b$ are in

$K$ , then

$\alpha a + \beta b$ is also in

$K$ . We have shown that

$\alpha a + \beta b$ is in both

$H$ and

$K$ and so by the definition of

$H \cap K$ , we see that

$\alpha a + \beta b \in H \cap K$ , and hence we must answer yes to question 3.

We have answered yes to all three questions and so we must conclude that if

$H$ and

$K$ are subspaces of

$V$ , then

$H \cap K$ is a subspace of

$V$ .