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Problems #9,24, pg.229 are graded.
#9, pg.229: Find the dimension of the subspace of all vectors in $\mathbb{R}^{3 \times 1}$ whose first and third entries are equal.
Solution: Call the space in question $S$. We see that
$$S = \left\{ \begin{bmatrix} b_1 \\ b_2 \\ b_1 \end{bmatrix} \colon b_1,b_2 \in \mathbb{R} \right\}.$$
Claim: The set $\mathscr{B}= \left\{ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right\}$ is a basis for $S$.
First notice that $S = \mathrm{span} \mathscr{B}$, which is clear because we can write
$$S = \left\{ b_1 \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + b_2 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \colon b_1,b_2 \in \mathbb{R} \right\}.$$
Is the set $\mathscr{B}$ independent? Consider the vector equation
$$x_1 \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + x_2 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0\\ 0 \end{bmatrix}.$$
This vector equation is equivalent to the system of equations
$$\left\{ \begin{array}{ll}
x_1 = 0 \\
x_2 = 0 \\
x_1 = 0
\end{array} \right.$$
and so we see the vector equation has only trivial solution. Hence $\mathscr{B}$ is, by definition, an independent set of vectors. We therefore conclude that $\mathscr{B}$ is a basis for $S$. Finally, we conclude from this that the dimension of $S$ is 2.
#24, pg.229: The first three Laguerre polynomials are $L_0(x)=1, L_1(x)=-t+1,$ and $L_2(x)=t^2-4t+2$. It is known that $\mathscr{B}=\{L_0,L_1,L_2\}$ forms a basis of $\mathbb{P}_2$. Find the coordinate vector of the vector $\vec{p}(t)=-2t^2+5t+5$ with respect to $\mathscr{B}$, taking $b_0=L_0, b_1=L_1, b_2=L_2$.
Solution: Our goal here is to compute the column vector $[\vec{p}]_{\mathscr{B}}$. This is done by finding the weights $c_0,\ldots,c_3$ such that
$$c_0 L_0 + c_1 L_1 + c_2 L_2 = \vec{p},$$
or in other words
$$c_0 + c_1(-t+1) + c_2(t^2-4t+2)=-2t^2+5t+5.$$
Simplify the left hand side to get
$$t^2 c_2 + t(-c_1 -4c_2) + (c_0 + c_1 +2c_2) = -2t^2 +5t + 5.$$
Equating coefficients yields
$$\left\{ \begin{array}{ll}
c_2 &= -2 \\
-c_1-4c_2 &= 5 \\
c_0+c_1+2c_2 &= 5
\end{array} \right.$$
yielding
$$\left\{ \begin{array}{ll}
c_2 &= -2 \\
c_1 &= 3 \\
c_0 &= 6.
\end{array} \right.$$
Hence
$$[\vec{p}]_{\mathbb{B}} = \begin{bmatrix} -2 \\ 3 \\ 6 \end{bmatrix}.$$
