Let's solve the requested problems step by step: --- ### **14. S = {v, w}** **Given:** S = {v, w} **Question:** Either show that Sp(S) = R³ or give an algebraic specification for Sp(S). If Sp(S) ≠ R³, then give a geometric description of Sp(S). #### **Solution:** - Sp(S) is the span of two vectors in R³. - Unless v and w are linearly independent and a third vector is added, two vectors can only span a plane in R³. **Algebraic specification:** Sp(S) = {a*v + b*w | a, b ∈ R} **Geometric description:** Sp(S) is a plane through the origin in R³ (unless v or w is zero or they are multiples, in which case it's a line). **Final Answer:** - Sp(S) = {a*v + b*w | a, b ∈ R} - Geometrically, this is a plane through the origin in R³. --- ### **20. Let S be as in Exercise 14. For each vector below, determine whether the vector is in Sp(S). Express those in Sp(S) as a linear combination of v and w.** From Exercise 14: Let S = {v, w} Vectors to check: a) [1, 1, 2]ᵗ b) [1, -1, 2]ᵗ c) [2, 1, 3]ᵗ d) [2, 1, 3]ᵗ e) [1, 2, 4]ᵗ f) [1, 2, 3]ᵗ **But v and w are not specified!** Usually, v and w are the standard basis vectors or arbitrary vectors, but since the specific vectors are not given in the image, we can't do the calculations without that info. **So, the answer depends on the actual v and w. Please provide the vectors v and w for a complete answer.** --- ### **28. A = \(\begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix}\)** **Question:** Give an algebraic specification for the null space and the range of the matrix \(A\). #### **Null Space:** Solve \(A\mathbf{x} = 0\): \[ \begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] Gives: 1. \(x_1 - x_2 = 0 \implies x_1 = x_2\) 2. \(2x_1 + x_2 = 0\) But plug \(x_1 = x_2\) into second equation: \(2x_2 + x_2 = 0 \implies 3x_2 = 0 \implies x_2 = 0\), so \(x_1 = 0\). **Null space:** Only the zero vector. \[ N(A) = \{\mathbf{0}\} \] #### **Range (Column space):** The columns are \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\) and \(\begin{bmatrix} -1 \\ 1 \end{bmatrix}\). Are they linearly independent? Calculate determinant: \[ \det\begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix} = 1*1 - (-1)*2 = 1 + 2 = 3 \neq 0 \] So, columns are linearly independent. **Range:** All of \(\mathbb{R}^2\). **Final Answers:** - Null space: \(\{\mathbf{0}\}\) - Range: \(\mathbb{R}^2\) --- ### **30. A = \(\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}\)** #### **Null Space:** Solve \(A\mathbf{x} = 0\): \[ \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] Equations: 1. \(x_1 + 2x_2 = 0\) 2. \(2x_1 + 4x_2 = 0\) Second is just twice the first. So only one independent equation: \(x_1 + 2x_2 = 0 \implies x_1 = -2x_2\) **Null space:** All scalar multiples of \(\begin{bmatrix} -2 \\ 1 \end{bmatrix}\). \[ N(A) = \{t \begin{bmatrix} -2 \\ 1 \end{bmatrix} : t \in \mathbb{R}\} \] #### **Range (Column space):** Columns: \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\), \(\begin{bmatrix} 2 \\ 4 \end{bmatrix}\) Second is just twice the first. So column space is all multiples of \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\). **Range:** All scalar multiples of \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\). \[ R(A) = \{t \begin{bmatrix} 1 \\ 2 \end{bmatrix} : t \in \mathbb{R}\} \] --- ### **32. A = \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix}\)** This is a \(2 \times 3\) matrix. #### **Null Space:** Solve \(A\mathbf{x} = 0\): \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] Equations: 1. \(x_1 + 2x_2 + 3x_3 = 0\) 2. \(x_2 + 2x_3 = 0 \implies x_2 = -2x_3\) Substitute \(x_2 = -2x_3\) into first equation: \(x_1 + 2(-2x_3) + 3x_3 = 0\) \(x_1 - 4x_3 + 3x_3 = 0\) \(x_1 - x_3 = 0 \implies x_1 = x_3\) Let \(x_3 = t\), then \(x_2 = -2t\), \(x_1 = t\). So, \[ N(A) = \{ t \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} : t \in \mathbb{R}\} \] #### **Range (Column space):** The two rows are linearly independent, so the range is all of \(\mathbb{R}^2\). **Final Answers:** - Null space: \(\{ t \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} : t \in \mathbb{R}\}\) - Range: \(\mathbb{R}^2\) --- ## **Summary Table** | Problem | Null Space | Range (Column Space) | |---------|------------|---------------------| | 14 | See above | See above | | 20 | Need v, w | Need v, w | | 28 | {0} | \(\mathbb{R}^2\) | | 30 | \(t\begin{bmatrix} -2 \\ 1 \end{bmatrix}\) | \(t\begin{bmatrix} 1 \\ 2 \end{bmatrix}\) | | 32 | \(t\begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}\) | \(\mathbb{R}^2\) | --- If you can provide the vectors \(v\) and \(w\) for problems 14 and 20, I can solve those parts in detail!

Certainly! To solve **Problem 20** in detail, I need to clarify the vectors \( v \) and \( w \) from Exercise 14, since the question asks to determine whether certain vectors are in the span of \( S = \{v, w\} \). --- ### Step 1: Clarify the vectors \( v \) and \( w \) **In Exercise 14**, the set \( S = \{v, w\} \) is given, but the actual vectors \( v \) and \( w \) are not explicitly provided in the text you shared. **Assumption:** Based on common problems, if these vectors are not specified, they are often standard basis vectors or provided in previous exercises. **From the image** (which mentions vectors \( v \) and \( w \)), the vectors are likely: \[ v = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \quad \text{and} \quad w = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \] or some similar pair. **But** to be accurate, **please confirm or provide the specific vectors \( v \) and \( w \).** --- ### For demonstration, let's assume: \[ v = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad w = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \] which span the xy-plane in \( \mathbb{R}^3 \). --- ### Step 2: Express the problem Given the vectors: \[ b_1 = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \] \[ b_2 = \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \] \[ b_3 = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \] \[ b_4 = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \] \[ b_5 = \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix} \] \[ b_6 = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \] The task: **Determine whether each \( b_i \) is in the span of \( v \) and \( w \).** --- ### Step 3: Check if \( b_i \) is in the span of \( v \) and \( w \) Since \( v = [1, 0, 0]^T \) and \( w = [0, 1, 0]^T \), the span of \( v \) and \( w \) is all vectors in \( \mathbb{R}^3 \) with zero in the third component: \[ \text{Span}\{v, w\} = \left\{ \begin{bmatrix} a \\ b \\ 0 \end{bmatrix} : a, b \in \mathbb{R} \right\} \] **Therefore:** - A vector \( b = [b_1, b_2, b_3]^T \) **is in** the span of \( v \) and \( w \) **if and only if** \( b_3 = 0 \). --- ### Step 4: Check each \( b_i \) | \( b_i \) | Components | Is in span? | Explanation | |------------|--------------|--------------|----------------| | \( b_1 = [1, 1, 2]^T \) | \( 2 \neq 0 \) | **No** | Third component not zero | | \( b_2 = [1, -1, 2]^T \) | \( 2 \neq 0 \) | **No** | Third component not zero | | \( b_3 = [2, 1, 3]^T \) | \( 3 \neq 0 \) | **No** | Third component not zero | | \( b_4 = [2, 1, 3]^T \) | \( 3 \neq 0 \) | **No** | Same as above | | \( b_5 = [1, 2, 4]^T \) | \( 4 \neq 0 \) | **No** | Third component not zero | | \( b_6 = [1, 2, 3]^T \) | \( 3 \neq 0 \) | **No** | Third component not zero | --- ### **Final conclusion:** **All these vectors are *not* in the span of \( v \) and \( w \) (assuming \( v, w \) span only the xy-plane) because their third components are non-zero.** --- ### **Summary:** | Vector \( b \) | Is \( b \in \text{Sp}(v,w) \)? | Expression as linear combination if yes | |----------------|------------------------------|-------------------------------------| | \([1,1,2]^T\) | **No** | N/A | | \([1,-1,2]^T\) | **No** | N/A | | \([2,1,3]^T\) | **No** | N/A | | \([2,1,3]^T\) | **No** | N/A | | \([1,2,4]^T\) | **No** | N/A | | \([1,2,3]^T\) | **No** | N/A | --- ### **Note:** **If** the actual vectors \( v \) and \( w \) or their span differ, please provide them, and I will redo the calculations accordingly. --- ### **Final answer:** > **All the given vectors are not in the span of \( v \) and \( w \) under the assumption that \( v \) and \( w \) span only the xy-plane (i.e., have zero in the third component).** --- **Let me know if you'd like me to proceed with different specific vectors for \( v \) and \( w \), or if you have more details!**

Certainly! Let's proceed with the detailed calculations for **Problem 20**, assuming: \[ u = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \quad v = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \] and the set \( S = \{u, v\} \). --- ### **Understanding the problem** - The set \( S = \{u, v\} \) spans a subspace of \( \mathbb{R}^3 \). - To determine whether each given vector \( b \) is in \( \text{Sp}(u, v) \), we need to check if \( b \) can be written as a linear combination of \( u \) and \( v \): \[ b = a u + b v \] for some scalars \( a, b \in \mathbb{R} \). --- ### **Step 1: Write the general linear combination** \[ b = a \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} + c \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} c \\ 0 \\ a + c \end{bmatrix} \] This gives the system: \[ \begin{cases} b_1 = c \\ b_2 = 0 \\ b_3 = a + c \end{cases} \] --- ### **Step 2: Conditions for \( b \) to be in the span** - For \( b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} \), the following must hold: \[ b_2 = 0 \] and \[ a = b_3 - c = b_3 - b_1 \] - Also, since \( c = b_1 \), the vector \( b \) is in the span **if and only if**: \[ b_2 = 0 \] --- ### **Step 3: Check each vector** Given vectors: | \(b\) | \(b_1, b_2, b_3\) | Condition for membership | Is in span? | Linear combination \(a, c\) if yes | |--------|-------------------|----------------------------|--------------|-------------------------------------| | \(b^{(1)} = [1, 1, 2]^T\) | (1, 1, 2) | \(b_2 = 1 \neq 0\) | **No** | N/A | | \(b^{(2)} = [1, -1, 2]^T\) | (1, -1, 2) | \(b_2 = -1 \neq 0\) | **No** | N/A | | \(b^{(3)} = [2, 1, 3]^T\) | (2, 1, 3) | \(b_2 = 1 \neq 0\) | **No** | N/A | | \(b^{(4)} = [2, 1, 3]^T\) | same as above | \(b_2 = 1 \neq 0\) | **No** | N/A | | \(b^{(5)} = [1, 2, 4]^T\) | (1, 2, 4) | \(b_2 = 2 \neq 0\) | **No** | N/A | | \(b^{(6)} = [1, 2, 3]^T\) | (1, 2, 3) | \(b_2=2 \neq 0\) | **No** | N/A | --- ### **Final conclusion:** All these vectors have \( b_2 \neq 0 \), so **none** are in the span of \( u \) and \( v \). --- ### **Summary:** | Vector \(b\) | Is \(b \in \text{Sp}(u,v)\)? | Explanation | |--------------|------------------------------|--------------| | \([1, 1, 2]^T\) | **No** | \(b_2 \neq 0\) | | \([1, -1, 2]^T\) | **No** | \(b_2 \neq 0\) | | \([2, 1, 3]^T\) | **No** | \(b_2 \neq 0\) | | \([2, 1, 3]^T\) | **No** | same as above | | \([1, 2, 4]^T\) | **No** | \(b_2 \neq 0\) | | \([1, 2, 3]^T\) | **No** | \(b_2 \neq 0\) | --- ### **Expressing in terms of \( u \) and \( v \):** - Since \( b_1 = c \) and \( b_2 = 0 \), \( b \) is in the span **only if** \( b_2 = 0 \). - When \( b_2 = 0 \), the linear combination is: \[ b = c u + a v \] with \( c = b_1 \), \( a = b_3 - b_1 \). **But since all given vectors have \( b_2 \neq 0 \), none can be expressed as such.** --- ### **Final note:** If you have other vectors with \( b_2 = 0 \), they can be expressed as: \[ b = c u + (b_3 - b_1) v \] with \( c = b_1 \). --- **Let me know if you'd like me to verify with different vectors or with different basis vectors!**

Certainly! Here is a detailed step-by-step solution for **Problem 20**, assuming the vectors \( u = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \) and \( v = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \). --- ### **Understanding the problem** - The set \( S = \{u, v\} \) spans a subspace of \( \mathbb{R}^3 \). - To check if a vector \( b \) is in this span, we express \( b \) as a linear combination: \[ b = a u + c v \] which expands to: \[ b = a \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} + c \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} c \\ 0 \\ a + c \end{bmatrix} \] --- ### **Step 1: Set up the equations** Given a vector: \[ b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} \] It is in the span **if and only if**: \[ b_1 = c \quad (1) \] \[ b_2 = 0 \quad (2) \] \[ b_3 = a + c \quad (3) \] From (2): **For \( b \) to be in the span, its second component must be zero.** --- ### **Step 2: Determine whether each vector is in the span** Since the second component \( b_2 \) must be zero, **any vector with \( b_2 \neq 0 \) is not in the span.** Now, check each vector: - \( b^{(1)} = [1, 1, 2]^T \): \( b_2 = 1 \neq 0 \) → **Not in the span** - \( b^{(2)} = [1, -1, 2]^T \): \( b_2 = -1 \neq 0 \) → **Not in the span** - \( b^{(3)} = [2, 1, 3]^T \): \( b_2 = 1 \neq 0 \) → **Not in the span** - \( b^{(4)} = [2, 1, 3]^T \): same as above → **Not in the span** - \( b^{(5)} = [1, 2, 4]^T \): \( b_2 = 2 \neq 0 \) → **Not in the span** - \( b^{(6)} = [1, 2, 3]^T \): \( b_2 = 2 \neq 0 \) → **Not in the span** --- ### **Final conclusion** **All the given vectors have a non-zero second component, so none of them lie in the span of \( u \) and \( v \).** ### **Summary** - The span of \( u \) and \( v \) consists of all vectors where the second component is zero. - Since all provided vectors have \( b_2 \neq 0 \), they are not in the span. --- ### **Expressing in terms of basis vectors** - For a vector \( b = [b_1, 0, b_3]^T \), it can be written as: \[ b = c u + (b_3 - c) v \] where \( c = b_1 \), provided \( b_2 = 0 \). But since none of the given vectors satisfy this, they cannot be expressed in this form. --- **If you have other vectors with \( b_2 = 0 \), I can show how to express them explicitly.**

Certainly! Here are the final answers for each problem: --- **Problem 14:** **Sp(S) = {a*v + b*w | a, b ∈ R}** Geometrically, the span of two vectors in R³ is a **plane through the origin** (unless they are linearly dependent). --- **Problem 20:** **Given** vectors \( u = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \), \( v = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \) **Final conclusions:** - None of the provided vectors (e.g., [1, 1, 2], [1, -1, 2], etc.) are in the span of \( u \) and \( v \) because their second component is non-zero. - The span consists only of vectors where the second component is zero. --- **Problem 28:** - Null space of \( A = \begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix} \): **{0}** (only the zero vector). - Range of \( A \): **\(\mathbb{R}^2\)** (columns are linearly independent). --- **Problem 30:** - Null space of \( A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \): **all scalar multiples of \(\begin{bmatrix} -2 \\ 1 \end{bmatrix}\).** - Range of \( A \): **all scalar multiples of \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\).** --- **Problem 32:** - Null space of \( A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix} \): **all scalar multiples of \(\begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}\).** - Range of \( A \): **\(\mathbb{R}^2\)** (the row vectors are linearly independent). --- Let me know if you'd like further clarification!