Find the basis of v+w and v intersection w where basis for v and w is given respectively

S={(1,1,0,0),(0,1,1,0),(0,0,1,1)}

P={(1,0,1,0),(0,2,1,1),(1,2,1,2)}

We are tasked with finding the basis for two sets: \(V + W\) and \(V \cap W\), where the basis for \(V\) and \(W\) are given as:

- \(V\) is spanned by \(S = \{(1,1,0,0), (0,1,1,0), (0,0,1,1)\}\)
- \(W\) is spanned by \(P = \{(1,0,1,0), (0,2,1,1), (1,2,1,2)\}\)

### Step 1: Find Basis for \(V + W\)
The space \(V + W\) is the union of the subspaces \(V\) and \(W\), which means it is the span of all the vectors in \(S\) and \(P\). To find the basis of \(V + W\), we need to determine the linear independence of the set of all vectors from both subspaces.

So, we take all vectors from both bases and form the following set:
\[
V + W = \{(1,1,0,0), (0,1,1,0), (0,0,1,1), (1,0,1,0), (0,2,1,1), (1,2,1,2)\}
\]
We now check for linear independence and eliminate any linearly dependent vectors to find the basis.

### Step 2: Find Basis for \(V \cap W\)
The space \(V \cap W\) consists of the vectors that are common to both subspaces. To find the basis of \(V \cap W\), we need to find the set of vectors that are linear combinations of both the basis vectors in \(S\) and \(P\).

This can be done by setting up a system of linear equations where any vector in \(V \cap W\) is a linear combination of both the basis of \(V\) and \(W\). Solving this system will give us the basis for \(V \cap W\).

Let me compute these steps.Let me reprocess the linear dependence and intersection computations to complete both steps clearly. I'll ensure all details are calculated correctly now.It seems there was an issue when calculating the intersection between the column spaces of \(V\) and \(W\). I will refine the calculation to correctly find both the reduced matrix (basis of \(V+W\)) and the intersection vectors. Let me proceed with this refinement.### Results:

1. **Basis for \( V + W \):**
   The rank of the combined set of vectors from \( V \) and \( W \) is 4, indicating that the basis for \( V + W \) consists of 4 linearly independent vectors. Therefore, the dimension of \( V + W \) is 4.

2. **Basis for \( V \cap W \):**
   The basis for the intersection \( V \cap W \) is spanned by the vectors:
   \[
   \left\{ \begin{pmatrix} -0.5 \\ 0.5 \\ -0.5 \\ 0.5 \end{pmatrix}, \begin{pmatrix} -0.5 \\ -0.5 \\ 0.5 \\ 0.5 \end{pmatrix} \right\}
   \]
   This implies that the dimension of \( V \cap W \) is 2, and the above two vectors form a basis for the intersection.

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Would you like to see more details or clarifications? Here are some follow-up questions:

1. How is the rank determined for the span of vectors?
2. What method is used to compute the intersection of two vector spaces?
3. Can we visualize the geometric interpretation of these spaces?
4. How can we determine if a set of vectors is linearly independent?
5. Why is the null space relevant in finding the intersection of subspaces?

**Tip:** The rank of a matrix tells you the number of linearly independent rows (or columns), which directly corresponds to the dimension of the span of those vectors.

Find the basis of V + W and V intersection W where basis for V and W is given respectively: S = {(1,1,0,0),(0,1,1,0),(0,0,1,1)} and P = {(1,0,1,0),(0,2,1,1),(1,2,1,2)}

This problem explores how to find the basis of the sum and intersection of two vector spaces. Using the subspaces S = {(1,1,0,0),(0,1,1,0),(0,0,1,1)} and P = {(1,0,1,0),(0,2,1,1),(1,2,1,2)}, we calculate the linear independence and intersection, applying concepts from linear algebra such as vector spaces, basis, and rank. Suitable for students in linear algebra courses.

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