Consider two vector subspaces 
𝑈
1
U 
1
​
  and 
𝑈
2
U 
2
​
 , where 
𝑈
1
U 
1
​
  is spanned by the columns of matrix 
𝐴
1
A 
1
​
  and 
𝑈
2
U 
2
​
  is spanned by the columns of matrix 
𝐴
2
A 
2
​
  with:

𝐴
1
=
(
1
0
1
1
−
2
−
1
2
1
3
1
0
1
)
and
𝐴
2
=
(
3
−
3
0
1
2
3
7
−
5
2
3
−
1
2
)
A 
1
​
 = 
​
  
1
1
2
1
​
  
0
−2
1
0
​
  
1
−1
3
1
​
  
​
 andA 
2
​
 = 
​
  
3
1
7
3
​
  
−3
2
−5
−1
​
  
0
3
2
2
​
  
​
 
Determine the dimension of 
𝑈
1
U 
1
​
  and 
𝑈
2
U 
2
​
 .
Determine the basis of 
𝑈
1
U 
1
​
  and 
𝑈
2
U 
2
​
 .
Determine the basis of 
𝑈
1
∩
𝑈
2
U 
1
​
 ∩U 
2
​
 .

We are tasked with analyzing the vector subspaces \( U_1 \) and \( U_2 \), each spanned by the columns of matrices \( A_1 \) and \( A_2 \) respectively. To address the problem, we'll proceed step by step:

### Given:
Matrix \( A_1 \):
\[
A_1 = \begin{pmatrix}
1 & 0 & 1 & 1 \\
-2 & -1 & 2 & 1 \\
3 & 1 & 0 & 1
\end{pmatrix}
\]
Matrix \( A_2 \):
\[
A_2 = \begin{pmatrix}
3 & -3 & 0 & 1 \\
2 & 3 & 7 & -5 \\
2 & 3 & -1 & 2
\end{pmatrix}
\]

### Step 1: Dimension and Basis of \( U_1 \)

The dimension of \( U_1 \) is the rank of matrix \( A_1 \), which corresponds to the number of linearly independent columns in \( A_1 \). We will perform Gaussian elimination to determine the rank and extract a basis.

### Step 2: Dimension and Basis of \( U_2 \)

Similarly, the dimension of \( U_2 \) is the rank of matrix \( A_2 \), and we'll find the basis by performing Gaussian elimination on \( A_2 \).

### Step 3: Basis of \( U_1 \cap U_2 \)

The intersection \( U_1 \cap U_2 \) is the set of vectors common to both subspaces. This can be found by solving the homogeneous system formed by combining the matrices \( A_1 \) and \( A_2 \). Specifically, we will check for the null space of the combined system or compute the ranks of their concatenated matrices.

I will perform these computations now.### Results:

1. **Dimension of \( U_1 \)**: The rank of \( A_1 \) is 3, meaning the dimension of \( U_1 \) is 3.
2. **Dimension of \( U_2 \)**: The rank of \( A_2 \) is 3, meaning the dimension of \( U_2 \) is 3.
3. **Intersection \( U_1 \cap U_2 \)**: The rank of the concatenated matrix formed by combining \( A_1 \) and \( A_2 \) is 3. This suggests that the dimension of the intersection \( U_1 \cap U_2 \) is 0, meaning there is no non-trivial intersection between the two subspaces, and the only common vector is the zero vector.

Would you like further details on how the bases of \( U_1 \) and \( U_2 \) were determined, or any other clarifications?

### Related Questions:
1. How are linearly independent vectors used to form a basis of a subspace?
2. How is Gaussian elimination used in determining the rank of a matrix?
3. Can the dimension of the intersection \( U_1 \cap U_2 \) ever be greater than the dimensions of \( U_1 \) or \( U_2 \)?
4. What does it mean geometrically when the intersection of two subspaces is trivial?
5. How can we visualize the span of a matrix's columns in 3D space?

### Tip:
The rank of a matrix can be interpreted as the number of linearly independent directions that the matrix spans in vector space.

Consider two vector subspaces U1 and U2, where U1 is spanned by the columns of matrix A1 and U2 is spanned by the columns of matrix A2 with: A1 = (1 0 1 1 −2 −1 2 1 3 1 0 1) and A2 = (3 −3 0 1 2 3 7 −5 2 3 −1 2). Determine the dimension of U1 and U2. Determine the basis of U1 and U2. Determine the basis of U1 ∩ U2.

Learn how to determine the dimension and basis of subspaces U1 and U2, spanned by matrices A1 and A2, using Gaussian elimination. This detailed solution also covers how to find the intersection of two subspaces. Perfect for undergraduate students studying linear algebra.

Math Problem Statement

𝐴 1

( 1 0 1 1 − 2 − 1 2 1 3 1 0 1 ) and 𝐴 2

( 3 − 3 0 1 2 3 7 − 5 2 3 − 1 2 ) A 1

andA 2

Solution

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