This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
To prove W = \(x, y, 0) x, y R\ is a subspace of V = R^3, we must verify three conditions: Step 1: Check if the zero vector is in W. The zero vector in R^3 is (0, 0, 0). We can express this vector in the form (x, y, 0) by choosing x=0 and y=0. Since (0, 0, 0) W, W is non-empty. Step 2: Check closure under vector addition. Let u = (x_1, y_1, 0) and v = (x_2, y_2, 0) be two arbitrary vectors in W. Their sum is: u + v = (x_1, y_1, 0) + (x_2, y_2, 0) = (x_1 + x_2, y_1 + y_2, 0) Since x_1, x_2, y_1, y_2 R, it follows that x_1 + x_2 R and y_1 + y_2 R. The resulting vector (x_1 + x_2, y_1 + y_2, 0) is of the form (x, y, 0), so it belongs to W. Thus, W is closed under vector addition. Step 3: Check closure under scalar multiplication. Let u = (x_1, y_1, 0) be a vector in W and c be an arbitrary scalar in R. Their scalar product is: cu = c(x_1, y_1, 0) = (cx_1, cy_1, 0) Since c, x_1, y_1 R, it follows that cx_1 R and cy_1 R. The resulting vector (cx_1, cy_1, 0) is of the form (x, y, 0), so it belongs to W. Thus, W is closed under scalar multiplication. Since all three conditions are satisfied, W is a subspace of V = R^3. W is a subspace of R^3 That's 2 down. 3 left today — send the next one.