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Problems #15 and #30 pg.206-207 are graded.
#15,pg.206-207: Find a matrix $A$ such that the given set $X$ equals $\mathrm{Col} \hspace{2pt} A$:
$$X=\left\{ \begin{bmatrix} 2s+t \\ r-s+2t \\ 3r+s \\ 2r-s-t \end{bmatrix} \colon r,s,t \in \mathbb{R} \right\}.$$
Solution: Recall that if $A$ is expressed by column vectors $\begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \ldots & \vec{a}_n \end{bmatrix}$, then $\mathrm{Col} \hspace{2pt} A=\mathrm{span}\{\vec{a}_1,\ldots,\vec{a}_n\}$. We will express $X$ as a span of vectors, take these vectors as $\vec{a}_1,\ldots,\vec{a}_n$, and then construct $A$ from them. Matrix algebra tells us that
$$\begin{bmatrix} 2s+t \\ r-s+2t \\ 3r+s \\ 2r-s-t \end{bmatrix} = \begin{bmatrix} 2s \\ -s \\ s \\ -s \end{bmatrix} + \begin{bmatrix} t \\ 2t \\ 0 \\ -t \end{bmatrix} + \begin{bmatrix} 0 \\ r \\ 3r \\ 2r \end{bmatrix}=s \begin{bmatrix} 2 \\ -1 \\ 1 \\ -1 \end{bmatrix} + t \begin{bmatrix} 1 \\ 2 \\ 0 \\ -1 \end{bmatrix} + r \begin{bmatrix} 0 \\ 1 \\ 3 \\ 2 \end{bmatrix}.$$
Let us write $\vec{a}_1 = \begin{bmatrix} 2 \\ -1 \\ 1 \\-1 \end{bmatrix}, \vec{a}_2 = \begin{bmatrix} 1 \\ 2 \\ 0 \\ -1 \end{bmatrix},$ and $\vec{a}_3 = \begin{bmatrix} 0 \\ 1 \\ 3 \\ 2 \end{bmatrix}$. Hence we may express $X$ as
$$X = \left\{ s \vec{a}_1 + t \vec{a}_2 + r \vec{a}_3 \colon r,s,t \in \mathbb{R} \right\} = \mathrm{span} \{ \vec{a}_1, \vec{a}_2, \vec{a}_3 \}.$$
Therefore by the definition of $\mathrm{Col}$ if we write
$$A = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \vec{a}_3 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 0 \\ -1 & 2 & 1 \\ 1&0&3 \\ -1 & -1 & 2 \end{bmatrix},$$
then we have
$$\mathrm{Col} \hspace{2pt} A = \mathrm{span}\{\vec{a}_1,\vec{a}_2,\vec{a}_3\} = X.$$
#30, pg.206-207: Let $T \colon V \rightarrow W$ be a linear transformation from a vector space $V$ into a vector space $W$. Prove that the range of $T$ is a subspace of $W$.
Solution: Write $S=\mathrm{Range}(T)=\{T(x) \colon x \in V\}$. To prove that $S$ is a subspace of $W$ we must verify that the answer to the following three questions is "yes":
- Is $S \subset W$?
- Does $S$ contain the zero vector of $W$, $\vec{0}_W$?
- Is $S$ closed under vector addition and scalar multiplication?
To answer 1, notice that $S$ contains points of the form $T(x)$ for $x \in V$. By the definition of a function, we know that $T(x) \in W$ because it must lie in the codomain of $T$. Hence all points in $S$ lie in $W$.
To answer 2, recall that that $T(\vec{0}_V)=\vec{0}_W$ which is a consequence of the following calculation (where we use the fact that $T$ is a linear transformation):
$$T(\vec{0}_V) = T(0\cdot \vec{0}_V) = 0T(\vec{0}_V) = \vec{0}_W.$$
To answer 3, let $p,q \in S$ and $\alpha,\beta$ be scalars. We must argue that $\alpha p + \beta q \in S$. Since $p,q \in S$ we know there exist vectors $x_p,x_q \in V$ such that $T(x_p)=p$ and $T(x_q)=q$ and since $V$ is a vector space, we also know that $\alpha x_p + \beta x_q \in V$. Therefore compute (using the fact that $T$ is a linear transformation),
$$T(\alpha x_p + \beta x_q)=\alpha T(x_p) + \beta T(x_q) = \alpha p + \beta q,$$and hence we see that $\alpha p + \beta q \in S$.
Since we have answered yes to all three questions, we may conclude that $\mathrm{Range}(T)$ is a subspace of $W$.
