H:={(x,y,z)T |x+y=z}

(a) Show that ⃗u := (1,0,1)T and ⃗v := (0,1,1)T belong to H. (b) Show that span(⃗u,⃗v) is contained in H.

(c) Show that span(⃗u,⃗v) = H.

We are given the set \( H := \{(x, y, z)^T \mid x + y = z \} \) and the vectors \( \vec{u} := (1, 0, 1)^T \) and \( \vec{v} := (0, 1, 1)^T \). The tasks are to show that:

(a) \( \vec{u} \) and \( \vec{v} \) belong to \( H \).

(b) The span of \( \vec{u} \) and \( \vec{v} \) is contained in \( H \).

(c) The span of \( \vec{u} \) and \( \vec{v} \) equals \( H \).

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### (a) Show that \( \vec{u} \) and \( \vec{v} \) belong to \( H \).

For a vector \( (x, y, z)^T \) to belong to \( H \), it must satisfy \( x + y = z \).

#### For \( \vec{u} = (1, 0, 1)^T \):
- \( x = 1 \), \( y = 0 \), and \( z = 1 \).
- Check if \( x + y = z \):
  \[
  1 + 0 = 1
  \]
  This is true, so \( \vec{u} \in H \).

#### For \( \vec{v} = (0, 1, 1)^T \):
- \( x = 0 \), \( y = 1 \), and \( z = 1 \).
- Check if \( x + y = z \):
  \[
  0 + 1 = 1
  \]
  This is true, so \( \vec{v} \in H \).

Thus, both \( \vec{u} \) and \( \vec{v} \) belong to \( H \).

---

### (b) Show that \( \text{span}(\vec{u}, \vec{v}) \) is contained in \( H \).

The span of \( \vec{u} \) and \( \vec{v} \) consists of all linear combinations of the form:
\[
\alpha \vec{u} + \beta \vec{v} = \alpha (1, 0, 1)^T + \beta (0, 1, 1)^T = (\alpha, \beta, \alpha + \beta)^T
\]
We need to check if any vector of this form belongs to \( H \), i.e., whether \( x + y = z \). For the vector \( (\alpha, \beta, \alpha + \beta)^T \), we have:
- \( x = \alpha \)
- \( y = \beta \)
- \( z = \alpha + \beta \)

Check if \( x + y = z \):
\[
\alpha + \beta = \alpha + \beta
\]
This is always true, so any vector in \( \text{span}(\vec{u}, \vec{v}) \) satisfies the condition \( x + y = z \), meaning \( \text{span}(\vec{u}, \vec{v}) \subseteq H \).

---

### (c) Show that \( \text{span}(\vec{u}, \vec{v}) = H \).

To prove this, we need to show that any vector in \( H \) can be written as a linear combination of \( \vec{u} \) and \( \vec{v} \). Consider any vector \( (x, y, z)^T \in H \). By the definition of \( H \), we know that \( x + y = z \). We want to express \( (x, y, z)^T \) as a linear combination of \( \vec{u} = (1, 0, 1)^T \) and \( \vec{v} = (0, 1, 1)^T \), i.e.,
\[
(x, y, z)^T = \alpha (1, 0, 1)^T + \beta (0, 1, 1)^T = (\alpha, \beta, \alpha + \beta)^T
\]
This gives the system of equations:
\[
x = \alpha \quad \text{and} \quad y = \beta \quad \text{and} \quad z = \alpha + \beta
\]
Since \( z = x + y \), this system is consistent, and we can choose \( \alpha = x \) and \( \beta = y \). Thus, any vector in \( H \) can be written as a linear combination of \( \vec{u} \) and \( \vec{v} \).

Therefore, \( \text{span}(\vec{u}, \vec{v}) = H \).

---

### Conclusion:
- \( \vec{u} \) and \( \vec{v} \) belong to \( H \).
- \( \text{span}(\vec{u}, \vec{v}) \subseteq H \).
- \( \text{span}(\vec{u}, \vec{v}) = H \).

Would you like further details or have any questions?

Here are 5 related questions:
1. How do you compute the span of more than two vectors?
2. What is the geometric interpretation of the span of two vectors?
3. How can you check if a set of vectors spans a space?
4. Can the span of two vectors be larger than the space they are defined in?
5. What role do linear independence and dependence play in the span of vectors?

**Tip**: The span of vectors represents the smallest subspace that contains all linear combinations of those vectors.

H := {(x, y, z)^T | x + y = z}

(a) Show that u := (1, 0, 1)^T and v := (0, 1, 1)^T belong to H.
(b) Show that span(u, v) is contained in H.
(c) Show that span(u, v) = H.

Explore how to prove that the vectors u = (1, 0, 1)^T and v = (0, 1, 1)^T belong to the subspace H defined by x + y = z. This problem includes proving the containment of the span of these vectors in H and showing that span(u, v) equals H. A detailed solution is provided, suitable for undergraduate-level linear algebra students.

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