The zero vector (0, 0, 0, 0) is in W.
W is closed under addition: if u, v are in W, then u + v is in W.
W is closed under scalar multiplication: if u is in W and c is a scalar, then cu is in W.
Therefore, W is a subspace of V. W is a subspace of V .
Explanation
Understanding the Problem We are given V = R 4 and W = {( x 1 , x 2 , x 3 , 0 ) : x 1 , x 2 , x 3 ∈ R } . We need to determine if W is a subspace of V . To do this, we need to check if W satisfies the following conditions:
The zero vector is in W .
W is closed under addition.
W is closed under scalar multiplication.
Checking the Conditions
Zero Vector: The zero vector in R 4 is ( 0 , 0 , 0 , 0 ) . Since W = {( x 1 , x 2 , x 3 , 0 ) : x 1 , x 2 , x 3 ∈ R } , we can see that ( 0 , 0 , 0 , 0 ) is in W because we can choose x 1 = 0 , x 2 = 0 , and x 3 = 0 .
Closed under Addition: Let u = ( x 1 , x 2 , x 3 , 0 ) and v = ( y 1 , y 2 , y 3 , 0 ) be two vectors in W . Then their sum is u + v = ( x 1 + y 1 , x 2 + y 2 , x 3 + y 3 , 0 ) . Since x 1 , x 2 , x 3 , y 1 , y 2 , y 3 are real numbers, x 1 + y 1 , x 2 + y 2 , and x 3 + y 3 are also real numbers. Therefore, u + v is in W .
Closed under Scalar Multiplication: Let u = ( x 1 , x 2 , x 3 , 0 ) be a vector in W and c be a scalar (a real number). Then c u = ( c x 1 , c x 2 , c x 3 , 0 ) . Since x 1 , x 2 , x 3 are real numbers and c is a real number, c x 1 , c x 2 , and c x 3 are also real numbers. Therefore, c u is in W .
Conclusion Since W satisfies all three conditions, W is a subspace of V .
Examples
Understanding subspaces is crucial in fields like computer graphics, where transformations (e.g., rotations, scaling) are applied to objects represented as vectors. If these transformations maintain the object within a specific subspace, it simplifies computations and ensures certain properties are preserved. For instance, if an object is constrained to a 2D plane within a 3D space, transformations that keep it within that plane (a subspace) are more efficient to calculate and guarantee the object remains 2D.