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Section 6.2 #19: Determine whether or not $\mathcal{B}=\left\{ \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}, \begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix}, \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix}, \begin{bmatrix} 1&1 \\ 1&-1 \end{bmatrix} \right\}$ is a basis for $(\mathbb{R}^{2 \times 2},\mathbb{R})$.
Solution: First we must check if $\mathcal{B}$ is linearly independent. Consider the vector equation
$$c_1 \begin{bmatrix} 1&0\\0&1 \end{bmatrix} + c_2 \begin{bmatrix} 0&-1 \\ 1&0 \end{bmatrix} + c_3 \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix} + c_4 \begin{bmatrix} 1&1 \\ 1&-1 \end{bmatrix} = \begin{bmatrix}0&0 \\ 0&0 \end{bmatrix}.$$
Simplify the left-hand side to get
$$\begin{bmatrix} c_1+c_3+c_4 & -c_2+c_3+c_4 \\ c_2+c_3+c_4 & c_1 + c_3 - c_4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$
This yields the system
$$\left\{ \begin{array}{rrrrl}
c_1& & +c_3 & +c_4 &= 0 \\
& -c_2 & +c_3 & +c_4 &= 0 \\
& c_2 &+c_3 & +c_4 &= 0 \\
c_1 & & +c_3 & -c_4 &= 0
\end{array} \right.$$
Solve this sytem by writing an augmented matrix and calculating its reduced echelon form:
$$\begin{bmatrix} 1&0&1&1&0 \\ 0&-1&1&1&0 \\ 0&1&1&1&0 \\ 1&0&1&-1&0 \end{bmatrix} \sim \begin{bmatrix} 1&0&0&0&0 \\ 0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1&0 \end{bmatrix},$$
which shows independence. Now check if $\mathcal{B}$ spans $\mathbb{R}^{2 \times 2}$. Clearly $\mathrm{span}(\mathcal{B}) \subseteq \mathbb{R}^{2 \times 2}$ and so now we must check if $\mathbb{R}^{2 \times 2} \subseteq \mathrm{span}(\mathcal{B})$. Pick an arbitrary element $\begin{bmatrix} a & b \\ c&d \end{bmatrix}$ in $\mathbb{R}^{2 \times 2}$. Consider the equation
$$c_1 \begin{bmatrix} 1&0\\0&1 \end{bmatrix} + c_2 \begin{bmatrix} 0&-1 \\ 1&0 \end{bmatrix} + c_3 \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix} + c_4 \begin{bmatrix} 1&1 \\ 1&-1 \end{bmatrix} = \begin{bmatrix}a&b \\ c&d \end{bmatrix}.$$
This leads to a matrix to put into reduced echelon form:
$$\begin{bmatrix} 1&0&1&1&a \\ 0&-1&1&1&b \\ 0&1&1&1&c \\ 1&0&1&-1&d \end{bmatrix} \sim \begin{bmatrix} 1&0&0&0&\frac{2a-b-c}{2} \\ 0&1&0&0&\frac{c-b}{2} \\ 0&0&1&0&\frac{-a+b+c+d}{2} \\ 0&0&0&1&\frac{a-d}{2} \end{bmatrix},$$
proving that $\mathrm{span}(\mathcal{B}) \subseteq \mathbb{R}^{2\times 2}$. Therefore $\mathcal{B}$ is a basis for $\mathbb{R}^{2 \times 2}$.
Section 6.2 #23: Determine whether or not $\mathcal{B}=\left\{ 1-x, 1-x^2, x-x^2 \right\}$ is a basis of $(\mathscr{P}_2,\mathbb{R})$.
Solution: Consider an arbitrary $ax^2+bx+c \in \mathscr{P}_2$. Consider the vector equation
$$c_1 (1-x) + c_2 (1-x^2) + c_3 (x-x^2) = ax^2+bx+c.$$
Combine like terms on the left to get
$$(-c_2 -c_3)x^2 + (-c_1+c_3)x + (c_1+c_2) = ax^2+bx+c.$$
This yields the following system of equations:
$$\left\{ \begin{array}{rrrl}
&-c_2&-c_3&=a \\
-c_1 & & +c_3 &= b \\
c_1 & +c_2 & &= c
\end{array} \right.$$
Write the augmented matrix and compute
$$\begin{bmatrix} 0&-1&-1&a \\ -1&0&1&b \\ 1&1&0&c \end{bmatrix} \sim \begin{bmatrix} 1&0&-1&0 \\ 0&1&1&0 \\ 0&0&0&1 \end{bmatrix}.$$
This shows that there are no solutions and so $\mathrm{span}(\mathcal{B}) \neq \mathscr{P}_2$.
Section 6.2 #27: Find the coordinate vector of $A = \begin{bmatrix} 1&2 \\ 3&4 \end{bmatrix}$ with respect to the basis $\mathcal{B} = \left\{ \begin{bmatrix} 1 & 0 \\ 0&0 \end{bmatrix}, \begin{bmatrix} 1&1 \\ 0&0 \end{bmatrix}, \begin{bmatrix} 1&1 \\ 1&0 \end{bmatrix}, \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix} \right\}$ of $(\mathbb{R}^{2 \times 2}, \mathbb{R})$.
Solution: Consider the vector equation
$$c_1 \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} + c_2 \begin{bmatrix} 1&1 \\ 0&0 \end{bmatrix} + c_3 \begin{bmatrix} 1&1 \\ 1&0 \end{bmatrix} + c_4 \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix} = \begin{bmatrix} 1&2 \\ 3&4 \end{bmatrix}.$$
Combining the vectors on the left yields
$$\begin{bmatrix} c_1+c_2+c_3+c_4 & c_2+c_3+c_4 \\ c_3+c_4 & c_4 \end{bmatrix} = \begin{bmatrix} 1&2 \\ 3&4 \end{bmatrix}.$$
This yields the following system of equations:
$$\left\{ \begin{array}{rrrrll}
c_1 &+c_2 &+c_3 & +c_4 &= 1 & (i) \\
&c_2 &+c_3 &+c_4 &= 2 & (ii) \\
&&c_3 &+c_4 &= 3 & (iii) \\
&&&c_4 &= 4 & (iv)
\end{array} \right.$$
Plugging $(iv)$ into $(iii)$ yields $c_3=-1$. Plugging the known values into $(ii)$ yields $c_2 = -1$. Plugging the known values into $(i)$ yields $c_1 = -1$. Therefore we have found the coordinate vector:
$$\left[ \begin{bmatrix} 1&2 \\ 3&4 \end{bmatrix} \right]_{\mathcal{B}} = \begin{bmatrix} -1 \\ -1 \\ -1 \\ 4 \end{bmatrix}.$$
Section 6.2 #28: Find the coordinate vector of $p(x)=1+2x+3x^2$ with respect to the basis $\mathcal{B}=\{1+x,1-x,x^2\}$ of $(\mathcal{P}_2,\mathbb{R})$.
Solution: Set up the vector equation
$$c_1(1+x) + c_2(1-x) + c_3 (x^2) = 3x^2+2x+1.$$
Combine like-terms on the left to get $(c_3)x^2+(c_1-c_2)x+(c_1+c_2)=3x^2+2x+1$. Equating coefficients on each side yields the following system of equations:
$$\left\{ \begin{array}{rrrll}
&&c_3 &= 3 & (i) \\
c_1&-c_2 & &=2 & (ii) \\
c_1 &+c_2 & &=1 & (iii)
\end{array} \right.$$
Writing it as an augmented matrix and putting it into reduced echelon form yields
$$\begin{bmatrix} 0&0&1&3 \\ 1&-1&0&2 \\ 1&1&0&1 \end{bmatrix} \sim \begin{bmatrix} 1&0&0&\frac{3}{2} \\ 0&1&0& -\frac{1}{2}\\ 0&0&1&3 \end{bmatrix}.$$
Hence we have solution $c_1=\dfrac{3}{2}, c_2=-\dfrac{1}{2}$, and $c_3=3$. Therefore we have found the $\mathcal{B}$-coordinates of $3x^2+2x+1$ to be
$$[3x^2+2x+1]_{\mathcal{B}} = \begin{bmatrix} \frac{3}{2} \\ -\frac{1}{2} \\ 3 \end{bmatrix}.$$
Section 6.2 #34: Find the dimension of the vector space $(V,\mathbb{R})$ where $V=\{p(x) \in \mathscr{P}_2 \colon p(0)=0 \}$ and give a basis for $(V,\mathbb{R})$.
Solution: An arbitrary $p(x) \in \mathcal{P}_2$ is of the form $p(x)=ax^2+bx+c$. A $p$ lying in $V$ obeys $p(0)=0$, i.e. $c=0$. Hence any $p \in V$ is of the form $ax^2+bx$. Therefore a basis for $V$ is the set $\{x,x^2\}$ because it spans $V$ and is independent (we always know that the monomial $x^m$ is independent from the monomial $x^n$ whenever $n \neq m$). Therefore $\mathrm{dim} (V,\mathbb{R})=2$.
Section 6.2 #35: Find the dimension of the vector space $(V,\mathbb{R})$ where $V=\{p(x) \in \mathscr{P}_2 \colon p(1)=0\}$ and give a basis for $(V,\mathbb{R})$.
Solution: An arbitrary $p(x) \in \mathcal{P}_2$ is of the from $p(x)=ax^2+bx+c$. A $p$ lying in $V$ obeys $p(1)=0$, i.e. $a+b+c=0$. Solve for one of the variables, say $a$: $a=-b-c$. This means that polynomials $p \in V$ are of the form $(-b-c)x^2+bx+c$ (i.e. it has two free variables) which could be written as $(-x^2+x)b + (-x^2+1)c$. This indicates we can express $V$ as the span of the set $\{-x^2+x, -x^2+1\}$. It is easy to check that this is an independent set of vectors and hence is a basis for $(V,\mathbb{R})$. Therefore $\mathrm{dim} (V,\mathbb{R})=2$.
