1. **Determine whether each set is a subspace of $P_2$** (the space of all polynomials of degree at most 2). Recall: A subset $W$ of a vector space $V$ is a subspace if it satisfies three conditions: - The zero vector of $V$ is in $W$. - $W$ is closed under vector addition. - $W$ is closed under scalar multiplication. **Set 1:** $\{ p : p(x) = ax^2, a \in \mathbb{R} \}$ - Zero vector: $p(x) = 0x^2 = 0$ is in the set. - Closure under addition: If $p_1(x) = a_1x^2$ and $p_2(x) = a_2x^2$, then $p_1 + p_2 = (a_1 + a_2)x^2$ is in the set. - Closure under scalar multiplication: For scalar $c$, $c p_1 = c a_1 x^2$ is in the set. **Conclusion:** This set is a subspace. **Set 2:** $\{ p : p(x) = a + x^2, a \in \mathbb{R} \}$ - Zero vector: The zero polynomial is $0$, but here $p(x) = a + x^2$ always has $x^2$ term with coefficient 1, so zero polynomial is not in the set. **Conclusion:** This set is not a subspace. **Set 3:** $\{ p : p(x) = a + bx + cx^2, a,b,c \in \mathbb{Z} \}$ - Zero vector: $a=b=c=0$ gives zero polynomial, so zero vector is in the set. - Closure under addition: Sum of two such polynomials has integer coefficients, so closed. - Closure under scalar multiplication: For scalar $c \in \mathbb{R}$, $c p$ may not have integer coefficients unless $c$ is integer. **Conclusion:** Not closed under scalar multiplication by arbitrary real scalars, so not a subspace. 2. **Write three different linear combinations of the vectors** $$A=\begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix}, B=\begin{bmatrix}2 & 1 \\ -1 & -1\end{bmatrix}, C=\begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix}$$ A linear combination is $xA + yB + zC$ for scalars $x,y,z \in \mathbb{R}$. Examples: 1. $1 \cdot A + 0 \cdot B + 0 \cdot C = A$ 2. $0 \cdot A + 1 \cdot B + 1 \cdot C = B + C = \begin{bmatrix}2+1 & 1+2 \\ -1+0 & -1+1\end{bmatrix} = \begin{bmatrix}3 & 3 \\ -1 & 0\end{bmatrix}$ 3. $2 \cdot A -1 \cdot B + 3 \cdot C = 2A - B + 3C = \begin{bmatrix}2 & 2 \\ 0 & 0\end{bmatrix} - \begin{bmatrix}2 & 1 \\ -1 & -1\end{bmatrix} + \begin{bmatrix}3 & 6 \\ 0 & 3\end{bmatrix} = \begin{bmatrix}2-2+3 & 2-1+6 \\ 0+1+0 & 0+1+3\end{bmatrix} = \begin{bmatrix}3 & 7 \\ 1 & 4\end{bmatrix}$ 3. **For each span, write an arbitrary element and check if $v$ is in the span.** (a) Span $\{A, B, C\}$ with $v = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$. Write $xA + yB + zC = v$: $$x \begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix} + y \begin{bmatrix}2 & 1 \\ -1 & -1\end{bmatrix} + z \begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$ Equate entries: - Top-left: $x + 2y + z = 1$ - Top-right: $x + y + 2z = 0$ - Bottom-left: $0x - y + 0z = 0 \Rightarrow -y = 0 \Rightarrow y=0$ - Bottom-right: $0x - y + z = 1 \Rightarrow -0 + z = 1 \Rightarrow z=1$ Substitute $y=0$, $z=1$ into first two: - $x + 0 + 1 = 1 \Rightarrow x=0$ - $x + 0 + 2(1) = 0 \Rightarrow 0 + 2 = 0$ contradiction. **Conclusion:** No solution, so $v$ is not in the span. (b) Span $\{p_1, p_2, p_3\}$ with $p_1(x) = -x^2 - x -1$, $p_2(x) = x^2 + 3x$, $p_3(x) = 2x -1$, and $v(x) = 2x^2 - 3x + 1$. Write $a p_1 + b p_2 + c p_3 = v$: $$a(-x^2 - x -1) + b(x^2 + 3x) + c(2x -1) = 2x^2 - 3x + 1$$ Group terms: - $x^2$: $-a + b$ - $x$: $-a + 3b + 2c$ - Constant: $-a - c$ Set equal to $v$ coefficients: $$\begin{cases} -a + b = 2 \\ -a + 3b + 2c = -3 \\ -a - c = 1 \end{cases}$$ From third: $-a - c = 1 \Rightarrow c = -a -1$ Substitute $c$ into second: $$-a + 3b + 2(-a -1) = -3 \Rightarrow -a + 3b - 2a - 2 = -3 \Rightarrow -3a + 3b = -1$$ From first: $-a + b = 2 \Rightarrow b = a + 2$ Substitute $b$ into above: $$-3a + 3(a + 2) = -1 \Rightarrow -3a + 3a + 6 = -1 \Rightarrow 6 = -1$$ Contradiction. **Conclusion:** No solution, so $v$ is not in the span. 4. **Given $H = \{ p \in P_2 : p'(2) = 0, \int_0^1 p(x) dx = 0 \}$ is a subspace.** (No question asked here, so no further work.)