Question

Consider the following two subspaces of \(\mathbb{R}^4\) \[ W_1=\{(x_1,x_2,x_3,x_4):x_1+x_2+x_3+x_4=0\} \] \[ W_2=\{(x_1,x_2,x_3,x_4):x_1+2x_2+3x_3+4x_4=0\}. \] Which of the following statements is correct?

• A. \(\dim(W_1\cap W_2)=0\)
• B. \(\dim(W_1\cap W_2)=1\)
• C. \(\dim(W_1\cap W_2)=2\)
• D. \(\dim(W_1\cap W_2)=3\)

Hint:

Two independent linear equations in 4 unknowns cut down the solution space by 2 dimensions, so the intersection of the two hyperplanes has dimension 4-2=2.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept.
Each of \(W_1\) and \(W_2\) is defined by a single linear equation in \(\mathbb{R}^4\), so each is a hyperplane of dimension \(4-1=3\). The intersection \(W_1\cap W_2\) is the set of vectors satisfying both equations at once, so it is the solution space of a linear system with two equations in four unknowns.

Step 2: Write the system.
\[ \begin{aligned} x_1+x_2+x_3+x_4&=0\\ x_1+2x_2+3x_3+4x_4&=0 \end{aligned} \]

Step 3: Find the rank of the coefficient matrix.
The coefficient matrix is \[ A=\begin{pmatrix}1&1&1&1\\1&2&3&4\end{pmatrix} \] Its two rows, \((1,1,1,1)\) and \((1,2,3,4)\), are not proportional to each other (there is no single constant \(k\) with every entry of the first row times \(k\) matching the second row), so the two rows are linearly independent. Hence \(\operatorname{rank}(A)=2\).

Step 4: Apply the rank-nullity relation.
The solution space of \(Ax=0\) with \(x\in\mathbb{R}^4\) has dimension \[ \dim(\text{solution space})=4-\operatorname{rank}(A)=4-2=2 \] This solution space is exactly \(W_1\cap W_2\), since it consists of all vectors satisfying both defining equations together.

Step 5: Double-check by finding a basis.
Subtracting the first equation from the second: \[ (x_1+2x_2+3x_3+4x_4)-(x_1+x_2+x_3+x_4)=x_2+2x_3+3x_4=0 \] So \(x_2=-2x_3-3x_4\), and from the first equation \(x_1=-x_2-x_3-x_4=(2x_3+3x_4)-x_3-x_4=x_3+2x_4\). So every solution has the form \[ (x_1,x_2,x_3,x_4)=x_3(1,-2,1,0)+x_4(2,-3,0,1) \] with \(x_3,x_4\) free. This gives 2 free parameters, so \(\dim(W_1\cap W_2)=2\), matching Step 4.

Step 6: Rule out the other options.
Option (A) (\(\dim=0\)) and option (B) (\(\dim=1\)) would need rank \(4\) or \(3\) respectively, impossible with only \(2\) rows. Option (D) (\(\dim=3\)) would need the two equations to be the same equation (rank \(1\)), which they are not, since the rows are not proportional.

Final Answer:
\(\dim(W_1\cap W_2)=2\). \[ \boxed{\dim(W_1\cap W_2)=2} \]