Question

Consider the subspace
\[ U = \{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1+x_2+x_4=0,\ x_3=0\}. \]
Let \(U^{\perp}\) be its orthogonal complement. The value of \(\dim(U^{\perp})\) equals ________ (answer in integer).

Hint:

Hint:
For any subspace \(U\) of \(\mathbb{R}^n\), \(\dim(U)+\dim(U^{\perp})=n\). Find \(\dim(U)\) using rank-nullity on the two defining equations.

Answer 1

Correct Answer: 2

Step 1: Understand the subspace U.
The set U is defined inside \(\mathbb{R}^4\) by two conditions: \(x_1+x_2+x_4=0\) and \(x_3=0\). Each condition is a single linear equation, and together they form a system of 2 linear equations in 4 unknowns.

Step 2: Write the equations in matrix form.
We can write U as the null space of the matrix
\[ A=\begin{pmatrix}1 & 1 & 0 & 1\\ 0 & 0 & 1 & 0\end{pmatrix}, \]
so that \(U=\{v \in \mathbb{R}^4 : Av=0\}\).

Step 3: Check that the two rows of A are independent.
The first row is \((1,1,0,1)\) and the second row is \((0,0,1,0)\). Neither row is a scalar multiple of the other, so the rows are linearly independent. This means
\[ \text{rank}(A)=2. \]

Step 4: Find the dimension of U.
By the rank-nullity theorem, for a matrix A with 4 columns,
\[ \dim(U)=\dim(\text{Null}(A))=4-\text{rank}(A)=4-2=2. \]

Step 5: Use the complement dimension formula.
For any subspace U of \(\mathbb{R}^n\), the orthogonal complement satisfies
\[ \dim(U)+\dim(U^{\perp})=n. \]
Here \(n=4\) and \(\dim(U)=2\), so
\[ \dim(U^{\perp})=4-2=2. \]

Final Answer:
The dimension of the orthogonal complement of U is 2. \[ \boxed{2} \]