Every subset of a vector space V that contains the zero vector in V is a subspace of V
f
To prove that every subset of a vector space V that contains the zero vector in V is a subspace of V, we need to show that it satisfies the following properties of a subspace:
1. Closure under addition: For any two vectors u and v in the subset, u+v is also in the subset.
2. Closure under scalar multiplication: For any scalar c and vector u in the subset, cu is also in the subset.
Let W be any subset of V that contains the zero vector.
To prove the first property, we need to show that for any two vectors u and v in W, their sum u + v is also in W.
Since W contains the zero vector, we have 0 ∈ W. Therefore, we can write
u = u + 0
v = v + 0
Now, using the fact that W is a subset of V, we know that V is closed under vector addition. Hence,
u + v = (u + 0) + (v + 0) ∈ W
Thus, W is closed under addition.
To prove the second property, we need to show that for any scalar c and vector u in W, the scalar product cu is also in W.
Again, since W is a subset of V, we know that V is closed under scalar multiplication. Hence,
c * u = c * u + 0 * u = (c * 1) * u + (0) * u
= c * u + 0 * u
= (c + 0) * u
= c * u
Therefore, cu is in W, and W is closed under scalar multiplication.
Since W satisfies both of the above properties, we can conclude that W is a subspace of V.
More Answers:
Inner Products: Linearity, Conjugate Symmetry, And Positive Definiteness In Vector SpacesInfinite Inner Products On R^N: Debunking The Common Misconception – A Comprehensive Guide On Varying Definitions Of Inner Products In Vector Spaces
Proving The Kernel Of A Matrix Transformation As A Subspace Of Rm In Math