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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

$[\vec{p}]_{\mathbb{B}} = \begin{bmatrix} -2 \\ 3 \\ 6 \end{bmatrix}.$