Find specific vectors **u** and **v** in W such that 

**u**plus+**v**

 is not in W. This is enough to show that W is *not* a vector space.

Two vectors in​ W, **u** and 

**v**​, 

for which 

**u**plus+**v**

 is not in W are

To show that \( W \) is not a vector space, we need to find two vectors \( \mathbf{u} \) and \( \mathbf{v} \) in \( W \) such that their sum \( \mathbf{u} + \mathbf{v} \) is not in \( W \).

Recall that \( W \) is defined as:

\[
W = \left\{ \begin{pmatrix} x \\ y \end{pmatrix} : xy \geq 0 \right\}
\]

### Step 1: Choose Vectors in \( W \)

Let's choose specific vectors \( \mathbf{u} \) and \( \mathbf{v} \) that satisfy the condition \( xy \geq 0 \):

1. **Vector \( \mathbf{u} \):** Let \( \mathbf{u} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \). 
   - Here, \( x = 1 \) and \( y = 1 \).
   - The product \( xy = 1 \times 1 = 1 \), which is greater than 0, so \( \mathbf{u} \in W \).

2. **Vector \( \mathbf{v} \):** Let \( \mathbf{v} = \begin{pmatrix} -1 \\ -1 \end{pmatrix} \).
   - Here, \( x = -1 \) and \( y = -1 \).
   - The product \( xy = (-1) \times (-1) = 1 \), which is greater than 0, so \( \mathbf{v} \in W \).

### Step 2: Compute the Sum \( \mathbf{u} + \mathbf{v} \)

Now, let's add these two vectors:

\[
\mathbf{u} + \mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} + \begin{pmatrix} -1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 + (-1) \\ 1 + (-1) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]

### Step 3: Check if the Sum is in \( W \)

The resulting vector is \( \mathbf{u} + \mathbf{v} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \).

To check if this vector is in \( W \), we need to evaluate \( xy \):

\[
xy = 0 \times 0 = 0
\]

Since \( xy = 0 \), the vector \( \begin{pmatrix} 0 \\ 0 \end{pmatrix} \) is actually in \( W \). Therefore, this specific choice does not show that \( W \) is not a vector space.

### Step 4: Correct the Choice

Let's try a different set of vectors:

1. **Vector \( \mathbf{u} \):** Let \( \mathbf{u} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \). This is the same as above, so \( \mathbf{u} \in W \).

2. **Vector \( \mathbf{v} \):** Let \( \mathbf{v} = \begin{pmatrix} -1 \\ 1 \end{pmatrix} \).
   - Here, \( x = -1 \) and \( y = 1 \).
   - The product \( xy = (-1) \times 1 = -1 \), which is less than 0, so \( \mathbf{v} \) is **not** in \( W \).

To ensure that the sum falls out of \( W \), we need:

- **Vector \( \mathbf{u} \):** Let \( \mathbf{u} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \).
- **Vector \( \mathbf{v} \):** Let \( \mathbf{v} = \begin{pmatrix} -1 \\ 1 \end{pmatrix} \).

Now, let's try \( \mathbf{v} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \):

This is the correct condition since:
 
- The product is \( 1 \times -1 = -1\) and \( xy \geq 0\) clearly fails.

### Step 5: Correct Result
1. \(u = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\)  
2. \(v = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\)

Clearly \(\mathbf{u} + \mathbf{v} =\begin{pmatrix} 2 \\ 0\end{pmatrix} \notin W\). 

These violate vector addition condition, implying non-closure.

### Questions:

1. How do closure properties generally ensure \(W\) is not a vector space?
2. Can you identify scalar multiplication causing another issue with \(W\)?
3. Which core vector space axioms fail with \(W\)?
4. Could extending \(W\) via additional constraints rectify its structure?
5. Can one redefine \(W\) conditions to ensure a vector space?

**Tip:** Remember, failing to meet just one vector space axiom is sufficient to show a set isn't a vector space!

Explore how to find specific vectors u and v in W such that their sum is not in W, demonstrating that W is not a vector space. This problem involves vector space concepts and closures. Suitable for advanced undergraduate students studying linear algebra.

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