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Problems #1 and #34 on pg.213 are graded.
#1,pg.213: Determine if the following set of vectors is or is not a basis for $\mathbb{R}^{3 \times 1}$. If it is not a basis, is it linearly independent or does it span $\mathbb{R}^{3 \times 1}$?
$$\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1\\ 1 \end{bmatrix} \right\}.$$
Solution: We will check to see if this set is an independent set of vectors. Consider the vector equation
$$x_1 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \vec{0}.$$
To solve this vector equation consider its augmented matrix and compute its reduced echelon form
$$\begin{bmatrix} 1&1&1&0 \\ 0&1&1&0 \\ 0&0&1&1 \end{bmatrix} \sim \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \end{bmatrix}.$$
From this we see that the only solution of the vector equation is the trivial solution $x_1=x_2=x_3=0$. Therefore the set is an independent set of vectors.
Now we must check to see if the span of the set equals $\mathbb{R}^{3 \times 1}$. We could do this directly by picking an arbitrary $\begin{bmatrix} a \\ b \\ c \end{bmatrix} \in \mathbb{R}^{3 \times 1}$ and then explicitly find weights $x_1,x_2,x_3$ such that
$$x_1 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 1 \\ 1 \\
0 \end{bmatrix} + x_3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} a\\ b \\ c \end{bmatrix}$$
by solving the vector equation. Another way to do this is to note that since the matrix $\begin{bmatrix} 1&1&1 \\ 0&1&1 \\ 0&0&1 \end{bmatrix}$ is invertible (to see that, just compute $\det \begin{bmatrix} 1&1&1 \\ 0&1&1 \\ 0&0&1 \end{bmatrix} =1 \neq 0$), it follows from the invertible matrix theorem that its column vectors span $\mathbb{R}^{3 \times 1}$.
Therefore we have shown that $\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1\\ 1 \end{bmatrix} \right\}$ is an independent set of vectors that spans $\mathbb{R}^{3 \times 1}$. Hence $\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1\\ 1 \end{bmatrix} \right\}$ is a basis for $\mathbb{R}^{3 \times 1}$.
#34,pg.213: Consider polynomials $\vec{p}_1(t)=1+t, \vec{p}_2(t)=1-t$, and $\vec{p}_3(t)=2$ (constant polynomial). By inspection, write a linear dependence relation among $\vec{p}_1,\vec{p}_2,\vec{p}_3$. Then find a basis for $S=\mathrm{span}\{\vec{p}_1,\vec{p}_2,\vec{p}_3\}$.
Solution: It is clear that
$$\vec{p}_1(t)+\vec{p}_2(t)=\vec{p}_3(t).$$
Therefore we know that $S = \mathrm{span}\{\vec{p}_1,\vec{p}_2,\vec{p}_3\} = \mathrm{span} \{\vec{p}_1,\vec{p}_2\}$. We must now ask: is the set $\{\vec{p}_1,\vec{p}_2\}$ linearly independent?
To answer this question consider the vector equation
$$c_1 \vec{p}_1 + c_2 \vec{p}_2 = 0.$$
Substitution of the definitions of $\vec{p}_1$ and $\vec{p}_2$ yields
$$c_1(1+t) + c_2(1-t) = 0.$$
Rearrangement of the left side yields
$$(c_1+c_2) + (c_1-c_2)t = 0.$$
Equating coefficients yields the the following system of equations:
$$\left\{ \begin{array}{ll}
c_1+c_2 &= 0 \\
c_1 - c_2 &= 0.
\end{array} \right.$$
And so we compute the reduced echelon form of the augmented matrix to see
$$\begin{bmatrix} 1 & 1 & 0 \\ 1 & -1 & 0\end{bmatrix} \sim \begin{bmatrix} 1&0&0 \\ 0&1&0 \end{bmatrix}.$$
From this we conclude that $c_1=c_2=0$ and hence the vector equation is solved by only the trivial solution. So by definition, we know that $\{\vec{p}_1,\vec{p}_2\}$ is an independent set and from earlier we know that $S = \mathrm{span}\{\vec{p}_1,\vec{p}_2\}$. Therefore we have shown that $\mathscr{B}=\{\vec{p}_1,\vec{p}_2\}$ is a basis for $S$.
