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Section 6.2: #5 Is $\{x,1+x\}$ a linearly independent subset of $(\mathscr{P}_1,\mathbb{R})$? If not, express one of the polynomials as a linear combination of the others.
Solution: We consider the vector equation
$$c_1 x + c_2(1+x) = 0.$$
Algebra yields
$$(c_1+c_2)x + c_2 = 0,$$
from which we conclude $c_2=0$ and $c_1 = -c_2 = 0$. Therefore the solution is the trivial solution $\begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ and we conclude that the set is linearly independent.
Section 6.2: #7 Is $\{x, 2x-x^2, 3x+2x^2\}$ a linearly independent subset of $(\mathscr{P}_2,\mathbb{R})$? IF not, express one of the polynomials as a linear combination of the others.
Solution: We consider the vector equation
$$(*) \hspace{35pt} c_1 x + c_2(2x-x^2) + c_3 (3x+2x^2)=0.$$
Algebra yields
$$(-c_2+2c_3)x^2 + (c_1+2c_2+3c_3)x = 0.$$
In other words, we want to solve the following system of equations:
$$\left\{ \begin{array}{llll}
& -c_2 & +2c_3 & =0 \\
c_1 &+2c_2 & +3c_3 &= 0
\end{array} \right.$$
Therefore we set up an augmented matrix and
$$\begin{bmatrix}
0 & -1 & 2 & 0 \\
1&2&3&0
\end{bmatrix} \sim \ldots \sim \begin{bmatrix} 1&0&7&0 \\ 0&1&-2&0 \end{bmatrix}$$
This means there is a nontrivial solution because the reduced echelon form encodes the system
$$\left\{ \begin{bmatrix}
c_1 & & +7c_3 &=0 \\
& c_2 &-2c_3 &= 0
\end{bmatrix} \right.$$
giving us solution vector (with free variable $c_3$)
$$\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} -7c_3 \\ 2c_3 \\ c_3 \end{bmatrix} = c_3 \begin{bmatrix} -7 \\ 2 \\ 0 \end{bmatrix}.$$
Choose, say, $c_3=1$ and we arrive at $c_1=-7$ and $c_2=2$. Therefore $(*)$ becomes
$$-7x + 2(2x-x^2) + (3x+2x^2)=0.$$
Which we may rearrange to say, for instance,
$$-7x = -2(2x-x^2) - (3x+2x^2).$$
Section 6.2: #11 Is $\{1,\sin^2(x),\cos^2(x)\}$ a linearly independent subset of $(\mathscr{F},\mathbb{R})$? If not, express one of the functions as a linear combination of the others.
Solution: No it is not independent because of the Pythagorean identity of trigonomery:
$$\cos^2(x)+\sin^2(x)=1.$$
That equation itself writes the vector $1$ in terms of the vectors $\cos^2(x)$ and $\sin^2(x)$.
Section 6.2: #18 Determine if
$$\mathscr{B} = \left\{ \begin{bmatrix} 1&0\\0&1 \end{bmatrix}, \begin{bmatrix} 1&1& \\ 1&0 \end{bmatrix}, \begin{bmatrix} 1&-1 \\ -1&1 \end{bmatrix} \right\}$$
is a basis for the vector space $(V,\mathbb{F})=(\mathbb{R}^{2 \times 2},\mathbb{R})$.
Solution: We must determine whether or not $\mathscr{B}$ is linearly independent and also whether or not $\mathrm{span}(\mathscr{B})=\mathbb{R}^{2 \times 2}$. To check independence, consider the vector equation
$$c_1 \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} + c_2 \begin{bmatrix} 1&1 \\ 1&0 \end{bmatrix} + c_3 \begin{bmatrix} 1&-1 \\ -1&1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0&0 \end{bmatrix}.$$
Algebra yields
$$\begin{bmatrix} c_1+c_2+c_3 & c_2-c_3 \\ c_2 - c_3 & c_1+c_3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0&0 \end{bmatrix}.$$
This encodes the system of equations
$$\left\{ \begin{array}{llll}
c_1&+c_2&+c_3&=0 \\
&c_2&-c_3&=0\\
&c_2&-c_3&=0 \\
c_1 & &+c_3&=0.
\end{array} \right.$$
Writing an augemented matrix and putting it into reduced echelon form yields
$$\begin{bmatrix} 1&1&1&0 \\ 0&1&-1&0 \\ 0&1&-1&0 \\ 1&0&1&0 \end{bmatrix} \sim \ldots \sim \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&0 \end{bmatrix},$$
which shows that the solution vector is $\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$, the trivial solution. From this we conclude that the set $\mathscr{B}$ is independent. Now we must decide whether or not $\mathrm{span}(\mathscr{B})=\mathbb{R}^{2 \times 2}$. Obviously
$$\mathrm{span}(\mathscr{B}) \subset \mathbb{R}^{2 \times 2}.$$
To check the other implication, consider an arbitrary $\begin{bmatrix} a & b \\ c & d \end{bmatrix} \in \mathbb{R}^{2 \times 2}$. We must check whether or not there is a solution to the following vector equation:
$$c_1 \begin{bmatrix} 1 & 0 \\ 0&1 \end{bmatrix} + c_2 \begin{bmatrix} 1&1 \\ 1&0 \end{bmatrix} + c_3 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} a & b \\ c& d \end{bmatrix}.$$
As before, algebra yields
$$\begin{bmatrix} c_1+c_2+c_3 & c_2-c_3 \\ c_2 - c_3 & c_1+c_3 \end{bmatrix} = \begin{bmatrix} a & b \\ c&d \end{bmatrix}.$$
Setting up an augmented matrix and putting it into reduced echelon form yields
$$\begin{bmatrix} 1&1&1&a \\ 0&1&-1&b \\ 0&1&-1&c \\ 1&0&1&d \end{bmatrix} \sim \ldots \sim \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}.$$
The last line of the reduced echelon form matrix shows that there is no solution to this vector equation. Therefore $\mathrm{span}(\mathscr{B}) \neq \mathbb{R}^{2 \times 2}$ and therefore the set $\mathscr{B}$ is not a basis of $\mathbb{R}^{2 \times 2}$.
Section 6.2 #22: Determine if $\mathscr{B}=\{x, 1+x, x-x^2 \}$ is a basis for $(\mathscr{P}_2,\mathbb{R})$.
Solution: We must check whether or not $\mathscr{B}$ is a linearly independent set and we must check whether or not $\mathscr{B}$ spans $\mathscr{P}_2$. To check independence, consider the vector equation
$$c_1 x + c_2 (1+x) + c_3(x-x^2)=0.$$
Algebra yields
$$-c_3 x^2 + (c_1+c_2+c_3)x + c_2 =0.$$
We see immediately that $c_3=0$ and $c_2=0$ and hence $c_1+c_2+c_3=c_1$ also equals zero. Therefore the set of vectors is an independent set (note: this can be solved with an augmented matrix, but it's straightforward without one). Obviously $\mathrm{span}\{x,1+x,x-x^2\} \subset \mathscr{P}_2$. It remains to check whether or not $\mathscr{P}_2 \subset \mathrm{span}\{x,1+x,x-x^2\}$. Let $ax^2+bx+c \in \mathscr{P}_2$ be an arbitrary polynomial. Consider the following vector equation:
$$c_1 x + c_2(1+x) + c_3(x-x^2)=ax^2+bx+c.$$
As before, algebra yields
$$-c_3 x^2 + (c_1+c_2+c_3)x + c_2 = ax^2+bx+c.$$
We immediately observe that $-c_3=a$ (hence $c_3=-a$) and $c_2=c$. From these values we see that $c_1+c-a=b$, or in other words $c_1=a+b-c$. Therefore we have found a solution to the vector equation, meaning that $\mathscr{P}_2 \subset \mathrm{span}\{x,1+x,x-x^2\}$. Thus $\mathscr{P}_2 = \mathrm{span}\{x,1+x,x-x^2\}$.
We have shown that both $\mathscr{B}$ is linearly independent and that $\mathrm{span}(\mathscr{B})=\mathscr{P}_2$. Therefore $\mathscr{B}$ is a basis for $\mathscr{P}_2$.
