(a) To prove that S ∩ S⊥ = 0, we need to show that the intersection of a set S and its orthogonal complement S⊥ is the zero vector.

Let's assume that there is a vector v in S ∩ S⊥. This means that v is in both S and S⊥. By the definition of the orthogonal complement, if v is in S⊥, then the inner product of v with any vector in S is zero. But v is also in S, so the inner product of v with itself is zero. The only vector that has a zero inner product with itself is the zero vector. Therefore, v must be the zero vector.

So, we have shown that if a vector is in S ∩ S⊥, then it must be the zero vector. Therefore, S ∩ S⊥ = 0.

(b) To prove that if R ⊆ S then S⊥ ⊆ R⊥, we need to show that if a vector is orthogonal to all vectors in S, then it is also orthogonal to all vectors in R.

Let's assume that v is in S⊥. This means that the inner product of v with any vector in S is zero. But R is a subset of S, so every vector in R is also in S. Therefore, the inner product of v with any vector in R is also zero. This means that v is in R⊥.

So, we have shown that if a vector is in S⊥, then it is also in R⊥. Therefore, S⊥ ⊆ R⊥.

(c) To prove that S ⊆ (S⊥)⊥, we need to show that if a vector is in S, then it is also in (S⊥)⊥.

Let's assume that v is in S. We need to show that v is in (S⊥)⊥. By the definition of the orthogonal complement, v is in (S⊥)⊥ if the inner product of v with any vector in S⊥ is zero. But by the definition of S⊥, the inner product of any vector in S with any vector in S⊥ is zero. Therefore, the inner product of v with any vector in S⊥ is zero, so v is in (S⊥)⊥.

So, we have shown that if a vector is in S, then it is also in (S⊥)⊥. Therefore, S ⊆ (S⊥)⊥.

(d) To prove that S⊥ = Span(S)⊥, we need to show that a vector is in S⊥ if and only if it is in Span(S)⊥.

Let's assume that v is in S⊥. This means that the inner product of v with any vector in S is zero. But every vector in Span(S) is a linear combination of vectors in S, so the inner product of v with any vector in Span(S) is also zero. Therefore, v is in Span(S)⊥.

Conversely, let's assume that v is in Span(S)⊥. This means that the inner product of v with any vector in Span(S) is zero. But every vector in Span(S) is a linear combination of vectors in S, so the inner product of v with any vector in S is also zero. Therefore, v is in S⊥.

So, we have shown that a vector is in S⊥ if and only if it is in Span(S)⊥. Therefore, S⊥ = Span(S)⊥.

Question

(a) To prove that S ∩ S⊥ = 0, we need to show that the intersection of a set S and its orthogonal complement S⊥ is the zero vector.

Let's assume that there is a vector v in S ∩ S⊥. This means that v is in both S and S⊥. By the definition of the orthogonal complement, if v is in S⊥, then the inner product of v with any vector in S is zero. But v is also in S, so the inner product of v with itself is zero. The only vector that has a zero inner product with itself is the zero vector. Therefore, v must be the zero vector.

So, we have shown that if a vector is in S ∩ S⊥, then it must be the zero vector. Therefore, S ∩ S⊥ = 0.

(b) To prove that if R ⊆ S then S⊥ ⊆ R⊥, we need to show that if a vector is orthogonal to all vectors in S, then it is also orthogonal to all vectors in R.

Let's assume that v is in S⊥. This means that the inner product of v with any vector in S is zero. But R is a subset of S, so every vector in R is also in S. Therefore, the inner product of v with any vector in R is also zero. This means that v is in R⊥.

So, we have shown that if a vector is in S⊥, then it is also in R⊥. Therefore, S⊥ ⊆ R⊥.

(c) To prove that S ⊆ (S⊥)⊥, we need to show that if a vector is in S, then it is also in (S⊥)⊥.

Let's assume that v is in S. We need to show that v is in (S⊥)⊥. By the definition of the orthogonal complement, v is in (S⊥)⊥ if the inner product of v with any vector in S⊥ is zero. But by the definition of S⊥, the inner product of any vector in S with any vector in S⊥ is zero. Therefore, the inner product of v with any vector in S⊥ is zero, so v is in (S⊥)⊥.

So, we have shown that if a vector is in S, then it is also in (S⊥)⊥. Therefore, S ⊆ (S⊥)⊥.

(d) To prove that S⊥ = Span(S)⊥, we need to show that a vector is in S⊥ if and only if it is in Span(S)⊥.

Let's assume that v is in S⊥. This means that the inner product of v with any vector in S is zero. But every vector in Span(S) is a linear combination of vectors in S, so the inner product of v with any vector in Span(S) is also zero. Therefore, v is in Span(S)⊥.

Conversely, let's assume that v is in Span(S)⊥. This means that the inner product of v with any vector in Span(S) is zero. But every vector in Span(S) is a linear combination of vectors in S, so the inner product of v with any vector in S is also zero. Therefore, v is in S⊥.

So, we have shown that a vector is in S⊥ if and only if it is in Span(S)⊥. Therefore, S⊥ = Span(S)⊥.

Knowee AI · Accepted Answer