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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

$S$ .