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).
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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.
Step 1: Recall the row space and null space relation.
For a matrix $A$, the null space of $A$ and the row space of $A$ are orthogonal complements of each other inside $\mathbb{R}^n$. Here $U$ is exactly the null space of the matrix built from the two defining equations.
Step 2: Write down that matrix.
The equations $x_1+x_2+x_4=0$ and $x_3=0$ come from the rows
\[ a_1=(1,1,0,1), \qquad a_2=(0,0,1,0). \]
So $U=\{v: a_1\cdot v=0, a_2\cdot v=0\}$, which is precisely the set of vectors perpendicular to both $a_1$ and $a_2$.
Step 3: Identify U perp directly.
Since $U$ consists of all vectors orthogonal to $a_1$ and $a_2$, the orthogonal complement $U^{\perp}$ is the span of $a_1$ and $a_2$ themselves:
\[ U^{\perp}=\text{span}\{a_1,a_2\}. \]
Step 4: Find the dimension of that span.
$a_1=(1,1,0,1)$ and $a_2=(0,0,1,0)$ are not scalar multiples of each other, so they are linearly independent. A span of 2 independent vectors has dimension 2.
Step 5: Conclude.
\[ \dim(U^{\perp})=2. \]
This matches the count from the rank-nullity route, confirming the same answer without first finding a basis for $U$ itself.
\[ \boxed{2} \]
