Answer
See explanation
Work Step by Step
To show that 𝐻+𝐾 is a subspace of 𝑉, we need to verify the three conditions for a subset to be a subspace:
1) The set 𝐻+𝐾 contains the zero vector.
2) 𝐻+𝐾 is closed under vector addition.
3) 𝐻+𝐾 is closed under scalar multiplication.
Proof:
1. 𝐻+𝐾 contains the zero vector:
Since 𝐻 and 𝐾 are subspaces of V, they both contain the zero vector (0∈ and (0∈𝐾).
By the definition of 𝐻+𝐾, we can write:
𝑤=𝑢+𝑣, where 𝑢∈𝐻 and 𝑣∈𝐾.
Let 𝑢=0 (from 𝐻) and 𝑣=0 (from 𝐾). Then:
𝑤=0+0=0.
Thus, 0∈𝐻+𝐾.
2. 𝐻+𝐾 is closed under addition:
Let 𝑤1,𝑤2∈𝐻+𝐾. By definition, there exist 𝑢1,𝑢2∈𝐻 and 𝑣1,𝑣2∈𝐾 such that:
𝑤1=𝑢1+𝑣1 and 𝑤2=𝑢2+𝑣2.
Now, adding 𝑤1 and 𝑤2:
𝑤1+𝑤2=(𝑢1+𝑣1)+(𝑢2+𝑣2).
Rearranging terms:
𝑤1+𝑤2=(𝑢1+𝑢2)+(𝑣1+𝑣2).
Since 𝐻 and 𝐾 are subspaces, 𝑢1+𝑢2∈𝐻 and 𝑣1+𝑣2∈𝐾.
Therefore:
𝑤1+𝑤2∈𝐻+𝐾.
Hence, 𝐻+𝐾 is closed under addition.
3. 𝐻+𝐾 is closed under scalar multiplication:
Let 𝑤∈𝐻+𝐾 and 𝑐∈𝑅 (or 𝐹, depending on the field). By definition, there exist 𝑢∈𝐻 and 𝑣∈𝐾 such that:
𝑤=𝑢+𝑣.
Now, multiplying 𝑤 by 𝑐:
𝑐𝑤=𝑐(𝑢+𝑣)=𝑐𝑢+𝑐𝑣.
Since 𝐻 and 𝐾 are subspaces, 𝑐𝑢∈𝐻 and 𝑐𝑣∈𝐾.
Therefore:
𝑐𝑤∈𝐻+𝐾.
Thus, 𝐻+𝐾 is closed under scalar multiplication.
Conclusion:
Since 𝐻+𝐾 satisfies all three conditions, we conclude that 𝐻+𝐾 is a subspace of 𝑉.
