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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 $V$.
