Question

Consider the following subspaces of \( \mathbb{R}^4 \): \[ V_1 = \left\{ (x, y, z, w) \in \mathbb{R}^4 : x + y + 2w = 0 \right\}, \quad V_2 = \left\{ (x, y, z, w) \in \mathbb{R}^4 : 2y + z + w = 0 \right\}, \quad V_3 = \left\{ (x, y, z, w) \in \mathbb{R}^4 : x + 3y + z + 3w = 0 \right\}. \] Then, the dimension of the subspace \( V_1 \cap V_2 \cap V_3 \) is equal to ............... (rounded off to two decimal places).

Hint:

To find the dimension of an intersection of subspaces, solve the system of equations corresponding to the subspaces. The dimension of the solution space is the dimension of the intersection.

Answer 1

Step 1: Find the equations defining the intersection.
The intersection of the three subspaces \( V_1 \cap V_2 \cap V_3 \) consists of all vectors \( (x, y, z, w) \) that satisfy the following system of equations: \[ x + y + 2w = 0, \quad 2y + z + w = 0, \quad x + 3y + z + 3w = 0. \] Step 2: Solve the system of equations.
We can write the system in matrix form: \[ \begin{pmatrix} 1 & 1 & 0 & 2
0 & 2 & 1 & 1
1 & 3 & 1 & 3 \end{pmatrix} \begin{pmatrix} x
y
z
w \end{pmatrix} = \begin{pmatrix} 0
0
0 \end{pmatrix}. \] Solving this system (using Gaussian elimination or a similar method) gives us the solution space with dimension 1. Final Answer: \[ \boxed{1}. \]