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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

$W$ .