1. **Problem 5(a): Show that $M_{2\times 2}(\mathbb{R})$ forms a vector space under standard matrix addition and scalar multiplication.** - The set $M_{2\times 2}(\mathbb{R})$ consists of all $2 \times 2$ matrices with real entries. - **Key axioms:** - **Closure under addition:** If $A, B \in M_{2\times 2}(\mathbb{R})$, then $A + B$ is also a $2 \times 2$ matrix with real entries, so $A + B \in M_{2\times 2}(\mathbb{R})$. - **Closure under scalar multiplication:** For any scalar $c \in \mathbb{R}$ and matrix $A \in M_{2\times 2}(\mathbb{R})$, $cA$ is also a $2 \times 2$ matrix with real entries, so $cA \in M_{2\times 2}(\mathbb{R})$. - **Additive identity:** The zero matrix $0$ with all entries zero is in $M_{2\times 2}(\mathbb{R})$ and acts as the additive identity. - **Additive inverses:** For any $A \in M_{2\times 2}(\mathbb{R})$, the matrix $-A$ is also in $M_{2\times 2}(\mathbb{R})$ and satisfies $A + (-A) = 0$. These key properties, along with others like associativity and distributivity, confirm $M_{2\times 2}(\mathbb{R})$ is a vector space. 2. **Problem 5(b): Another example of a vector space model other than $\mathbb{R}^n$.** - The set of all polynomials with real coefficients, denoted $\mathbb{P}$, forms a vector space under polynomial addition and scalar multiplication. 3. **Problem 6(a): Show that $W = \{(a,b,a+b) \mid a,b \in \mathbb{R}\}$ is a subspace of $\mathbb{R}^3$.** - **Non-empty:** The zero vector $(0,0,0)$ is in $W$ since $0 + 0 = 0$. - **Closed under addition:** For $(a,b,a+b)$ and $(c,d,c+d)$ in $W$, their sum is $(a+c, b+d, (a+b)+(c+d)) = (a+c, b+d, (a+c)+(b+d))$, which is in $W$. - **Closed under scalar multiplication:** For scalar $k$ and $(a,b,a+b)$ in $W$, $k(a,b,a+b) = (ka, kb, k(a+b)) = (ka, kb, ka + kb)$, which is in $W$. Thus, $W$ is a subspace. 4. **Problem 6(b): Express $(5,2,7)$ as a linear combination of generators of $W$.** - Generators can be $(1,0,1)$ and $(0,1,1)$. - Find scalars $x,y$ such that $x(1,0,1) + y(0,1,1) = (5,2,7)$. - This gives the system: - $x = 5$ - $y = 2$ - $x + y = 7$ - Since $5 + 2 = 7$, the system is consistent. - So, $(5,2,7) = 5(1,0,1) + 2(0,1,1)$. 5. **Problem 7: Determine if $S = \{(1,2,3), (0,1,2), (2,5,8)\}$ is linearly dependent or independent.** - Check if scalars $c_1, c_2, c_3$ exist, not all zero, such that: $$c_1(1,2,3) + c_2(0,1,2) + c_3(2,5,8) = (0,0,0)$$ - This gives the system: - $c_1 + 0 + 2c_3 = 0$ - $2c_1 + c_2 + 5c_3 = 0$ - $3c_1 + 2c_2 + 8c_3 = 0$ - From the first equation: $c_1 = -2c_3$. - Substitute into second: $2(-2c_3) + c_2 + 5c_3 = 0 \Rightarrow -4c_3 + c_2 + 5c_3 = 0 \Rightarrow c_2 + c_3 = 0 \Rightarrow c_2 = -c_3$. - Substitute $c_1$ and $c_2$ into third: $$3(-2c_3) + 2(-c_3) + 8c_3 = -6c_3 - 2c_3 + 8c_3 = 0$$ - This is true for all $c_3$. - Since $c_3$ can be any scalar, there exist nontrivial solutions. Therefore, the set $S$ is **linearly dependent**. **Final answers:** - 5(a): $M_{2\times 2}(\mathbb{R})$ is a vector space. - 5(b): Example vector space: polynomials $\mathbb{P}$. - 6(a): $W$ is a subspace of $\mathbb{R}^3$. - 6(b): $(5,2,7) = 5(1,0,1) + 2(0,1,1)$. - 7: $S$ is linearly dependent.