If S is the subspace of R^3 containing only the zero vector, what is S perp ? If S is spanned by ( 1 , 1, 1), what is S perp? If S is spanned by ( 2, 0, 0) and ( 0, 0, 3), what is S perp?
To find the orthogonal complement (S⊥) of a subspace S, we need to find all vectors that are orthogonal (perpendicular) to every vector in S.
1. If S is the subspace of R^3 containing only the zero vector, then S⊥ is the entire space R^3. This is because every vector in R^3 is orthogonal to the zero vector.
2. If S is spanned by the vector (1, 1, 1), then S⊥ is the subspace of R^3 containing vectors perpendicular to (1, 1, 1). To determine these vectors, we can find the null space of the matrix A, where A is the row vector [1, 1, 1]:
A = [1, 1, 1]
By solving the equation A * x = 0, we get:
[1, 1, 1] * [x1, x2, x3]^T = 0
x1 + x2 + x3 = 0
Here, x1, x2, and x3 can be any real numbers as long as they satisfy the equation. Thus, the orthogonal complement S⊥ is the subspace of R^3 containing all vectors of the form [x, y, -x-y], where x and y can be any real numbers.
3. If S is spanned by the vectors (2, 0, 0) and (0, 0, 3), then S⊥ is the subspace of R^3 containing vectors perpendicular to both (2, 0, 0) and (0, 0, 3). To find these vectors, we can find the null space of the matrix A, where A is the matrix whose rows are the vectors (2, 0, 0) and (0, 0, 3):
A = [2, 0, 0]
[0, 0, 3]
By solving the equation A * x = 0, we get:
[2, 0, 0] * [x1, x2, x3]^T = 0
[0, 0, 3] * [x1, x2, x3]^T = 0
2x1 = 0 ---> x1 = 0
3x3 = 0 ---> x3 = 0
Here, x1 and x3 can be any real numbers as long as they satisfy the equations. We don't have any restrictions on x2. Thus, the orthogonal complement S⊥ is the subspace of R^3 containing all vectors of the form [0, x2, 0], where x2 can be any real number.
To find the orthogonal complement (S⊥) of a subspace S, we need to determine all vectors that are orthogonal (perpendicular) to all vectors in S. In other words, S⊥ consists of all vectors that are perpendicular to every vector in S.
1. If S is the subspace of ℝ³ containing only the zero vector, then S is the trivial subspace {0}. The orthogonal complement of the zero vector is the entire space ℝ³ since every vector in ℝ³ is orthogonal to the zero vector.
Therefore, S⊥ = ℝ³.
2. If S is spanned by (1, 1, 1), then S consists of all scalar multiples of (1, 1, 1). To find S⊥, we need to find a vector that is orthogonal to (1, 1, 1).
Let's call a vector in S⊥ as (a, b, c). For it to be orthogonal to (1, 1, 1), their dot product should be zero:
(1, 1, 1) ⋅ (a, b, c) = a + b + c = 0
We have one equation and three unknowns, which implies there are infinitely many vectors in S⊥. One possible solution is to set b = a and c = -2a, where a is any real number.
Therefore, S⊥ is spanned by (-1, -1, 2). In other words, S⊥ = span{(-1, -1, 2)}.
3. If S is spanned by (2, 0, 0) and (0, 0, 3), then S consists of all linear combinations of these vectors. To find S⊥, we need to determine a vector that is orthogonal to both (2, 0, 0) and (0, 0, 3).
Let's call a vector in S⊥ as (a, b, c). For it to be orthogonal to both (2, 0, 0) and (0, 0, 3), their dot products should be zero:
(2, 0, 0) ⋅ (a, b, c) = 2a = 0
(0, 0, 3) ⋅ (a, b, c) = 3c = 0
From the first equation, a must be zero, and from the second equation, c must be zero. Setting a = c = 0, we are left with b as a free variable.
Therefore, S⊥ is spanned by (0, 1, 0). In other words, S⊥ = span{(0, 1, 0)}.