1. **Problem statement:** Consider subspaces \(U = \{(x,0,y): x,y \in \mathbb{R}\}\) and \(W = \{(0,x,y): x,y \in \mathbb{R}\}\) of \(V = \mathbb{R}^3\). Show that \(\mathbb{R}^3 = U + W\). Determine if \(\mathbb{R}^3\) is the direct sum of \(U\) and \(W\). Find the dimensions of \(U\), \(W\), \(U \cap W\), and \(U + W\). Formulate a conjecture relating these dimensions. 2. **Show \(\mathbb{R}^3 = U + W\):** - Any vector in \(\mathbb{R}^3\) can be written as \((a,b,c)\). - We want to express \((a,b,c) = u + w\) where \(u \in U\) and \(w \in W\). - Let \(u = (x,0,y) \in U\) and \(w = (0,x',y') \in W\). - Then \(u + w = (x,0,y) + (0,x',y') = (x, x', y + y')\). - To match \((a,b,c)\), set \(x = a\), \(x' = b\), and \(y + y' = c\). - Since \(y,y'\) are arbitrary real numbers, choose \(y = 0\) and \(y' = c\) (or vice versa). - Thus, every \((a,b,c)\) can be written as \((a,0,0) + (0,b,c)\) with \(u = (a,0,0) \in U\) and \(w = (0,b,c) \in W\). - Therefore, \(\mathbb{R}^3 = U + W\). 3. **Is \(\mathbb{R}^3\) the direct sum of \(U\) and \(W\)?** - The direct sum requires \(U \cap W = \{0\}\). - Find \(U \cap W\): vectors in both \(U\) and \(W\). - \(U = \{(x,0,y)\}\), \(W = \{(0,x,y)\}\). - For a vector \(v\) in \(U \cap W\), \(v = (x,0,y) = (0,x',y')\). - Equate components: \(x = 0\), \(0 = x'\), \(y = y'\). - So \(v = (0,0,y)\) with \(y = y'\). - This means \(U \cap W = \{(0,0,y): y \in \mathbb{R}\}\), which is a one-dimensional subspace. - Since \(U \cap W \neq \{0\}\), \(\mathbb{R}^3\) is not the direct sum of \(U\) and \(W\). 4. **Dimensions:** - \(\dim(U)\): \(U = \{(x,0,y)\}\) depends on two parameters \(x,y\), so \(\dim(U) = 2\). - \(\dim(W)\): \(W = \{(0,x,y)\}\) also depends on two parameters \(x,y\), so \(\dim(W) = 2\). - \(\dim(U \cap W)\): as found, \(\{(0,0,y)\}\) is one-dimensional, so \(\dim(U \cap W) = 1\). - \(\dim(U + W)\): since \(U + W = \mathbb{R}^3\), \(\dim(U + W) = 3\). 5. **Conjecture relating dimensions:** - The dimension formula for subspaces states: $$\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W)$$ - Check with values: $$3 = 2 + 2 - 1$$ - This confirms the conjecture. --- 6. **Question 5: Do there exist two two-dimensional subspaces of \(\mathbb{R}^3\) whose intersection is the zero vector? Why or why not?** - Suppose \(U, W \subseteq \mathbb{R}^3\) with \(\dim(U) = \dim(W) = 2\). - If \(U \cap W = \{0\}\), then by dimension formula: $$\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W) = 2 + 2 - 0 = 4$$ - But \(U + W \subseteq \mathbb{R}^3\), so \(\dim(U + W) \leq 3\). - Contradiction: 4 cannot be less or equal to 3. - Therefore, two 2-dimensional subspaces of \(\mathbb{R}^3\) must have a nonzero intersection. - So, no such two subspaces with zero intersection exist. **Final answers:** - \(\mathbb{R}^3 = U + W\) but not a direct sum. - Dimensions: \(\dim(U) = 2\), \(\dim(W) = 2\), \(\dim(U \cap W) = 1\), \(\dim(U + W) = 3\). - Dimension formula: \(\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W)\). - No two 2-dimensional subspaces of \(\mathbb{R}^3\) have zero intersection.