Hey there! Let's work through your linear algebra problem step by step, starting with your core question about using a matrix of basis vectors to find (W_1 \cap W_2).

The direction of your idea is right, but you can't just stack the basis vectors into a matrix directly—you need to set up the right system of linear equations to capture vectors that belong to both subspaces. Let's break down the proper process, along with solving for the bases of (W_1 \cap W_2) and (W_1 + W_2).

Step 1: Formalize the bases of (W_1) and (W_2) #

From your problem statement:

• (W_1) has dimension 2. We can pick a valid basis from its generating set, e.g., (\alpha_1 = (1,2,3,6)) and (\alpha_2 = (4,-1,3,6)) (these two are linearly independent, so they form a basis).
• (W_2) has dimension 2, with a linearly independent basis: (\beta_1 = (1,-1,1,1)) and (\beta_2 = (2,-1,4,5)).

Step 2: Set up the equation for (W_1 \cap W_2) #

Any vector (x \in W_1 \cap W_2) must be expressible as a linear combination of both (W_1)'s basis and (W_2)'s basis. Let:

x = k₁α₁ + k₂α₂ = k₁(1,2,3,6) + k₂(4,-1,3,6)

and also:

x = m₁β₁ + m₂β₂ = m₁(1,-1,1,1) + m₂(2,-1,4,5)

Set the two expressions equal and rearrange to get a homogeneous system of equations:

k₁(1,2,3,6) + k₂(4,-1,3,6) - m₁(1,-1,1,1) - m₂(2,-1,4,5) = (0,0,0,0)

This expands to 4 scalar equations:

• (k₁ + 4k₂ - m₁ - 2m₂ = 0)
• (2k₁ - k₂ + m₁ + m₂ = 0)
• (3k₁ + 3k₂ - m₁ - 4m₂ = 0)
• (6k₁ + 6k₂ - m₁ - 5m₂ = 0)

Step 3: Solve the homogeneous system #

Write the system's coefficient matrix:

[1   4  -1  -2]
[2  -1   1   1]
[3   3  -1  -4]
[6   6  -1  -5]

Perform row elementary operations to reduce it to row-echelon form:

[1  0  0  -1]
[0  1  0  -1]
[0  0  1  -3]
[0  0  0   0]

The solution is parameterized by (t \in \mathbb{R}):

• (k₁ = t), (k₂ = t)
• (m₁ = 3t), (m₂ = t)

Substitute back into the expression for (x):

x = t(1,2,3,6) + t(4,-1,3,6) = t(5,1,6,12)

So the basis for (W_1 \cap W_2) is {(5,1,6,12)}, with dimension 1.

Step 4: Find the basis for (W_1 + W_2) #

Use the dimension formula for subspaces:

dim(W₁+W₂) = dim(W₁) + dim(W₂) - dim(W₁∩W₂) = 2 + 2 - 1 = 3

To find a basis, take the union of (W_1) and (W_2)'s bases ((\alpha_1, \alpha_2, \beta_1, \beta_2)) and find a maximal linearly independent subset. For example:

• (\alpha_1, \alpha_2, \beta_1) are linearly independent (since (\beta_1) can't be expressed as a combination of (\alpha_1) and (\alpha_2)—if it could, (dim(W_1 \cap W_2)) would be larger than 1). This set forms a valid basis for (W_1 + W_2).

Your initial thought to use basis vectors in a matrix is on the right track, but you need to frame it as solving a homogeneous system that enforces the vector belongs to both subspaces. Just stacking the basis vectors won't capture that intersection condition—you have to set up the linear combination equality first.

内容的提问来源于stack exchange，提问作者TheLast Cipher