Question

Consider \(\mathbb{R}^4\) with the inner product \(\langle x, y \rangle = \sum_{i=1}^4 x_i y_i\), for \(x = (x_1, x_2, x_3, x_4)\) and \(y = (y_1, y_2, y_3, y_4)\).
Let \(M = \{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1 = x_3\}\) and \(M^\perp\) denote the orthogonal complement of M. The dimension of \(M^\perp\) is equal to ................

Hint:

In \(\mathbb{R}^n\), a subspace defined by \(k\) linearly independent linear homogeneous equations has dimension \(n-k\). Its orthogonal complement has dimension \(k\). Here, \(M\) is defined by one equation in \(\mathbb{R}^4\), so \(\dim(M) = 4-1=3\) and \(\dim(M^\perp) = 1\).

Answer 1

Step 1: Understanding the Concept:
The problem asks for the dimension of the orthogonal complement of a given subspace \(M\) in \(\mathbb{R}^4\). For any subspace \(W\) of a finite-dimensional inner product space \(V\), we have the relation \(\dim(V) = \dim(W) + \dim(W^\perp)\).
Step 2: Key Formula or Approach:
1. Determine the dimension of the subspace \(M\).
2. Use the formula \(\dim(M^\perp) = \dim(\mathbb{R}^4) - \dim(M)\) to find the result.
Step 3: Detailed Calculation:
The subspace \(M\) is defined by the single linear constraint \(x_1 = x_3\), which can be written as \(x_1 - x_3 = 0\).
A vector \((x_1, x_2, x_3, x_4)\) is in \(M\) if and only if \(x_1 = x_3\).
We can choose \(x_2, x_3, x_4\) as free variables. Let \(x_2=s, x_3=t, x_4=u\). Then \(x_1=t\).
An arbitrary vector in M can be written as: \[ (t, s, t, u) = t(1, 0, 1, 0) + s(0, 1, 0, 0) + u(0, 0, 0, 1) \] The vectors \(\{(1, 0, 1, 0), (0, 1, 0, 0), (0, 0, 0, 1)\}\) span M. They are also linearly independent. Therefore, they form a basis for M.
The dimension of M is the number of vectors in its basis, so \(\dim(M) = 3\).
Now we use the dimension formula for orthogonal complements: \[ \dim(M^\perp) = \dim(\mathbb{R}^4) - \dim(M) \] \[ \dim(M^\perp) = 4 - 3 = 1 \] Alternative Method:
A vector \(v = (v_1, v_2, v_3, v_4)\) is in \(M^\perp\) if it is orthogonal to every vector in M. It is sufficient to be orthogonal to the basis vectors of M.
Let \(u_1=(1,0,1,0), u_2=(0,1,0,0), u_3=(0,0,0,1)\) be the basis of M.
\(\langle v, u_1 \rangle = v_1 + v_3 = 0 \implies v_3 = -v_1\)
\(\langle v, u_2 \rangle = v_2 = 0\)
\(\langle v, u_3 \rangle = v_4 = 0\)
So, any vector in \(M^\perp\) must be of the form \((v_1, 0, -v_1, 0) = v_1(1, 0, -1, 0)\). The space \(M^\perp\) is spanned by the single vector \((1, 0, -1, 0)\).
Therefore, \(\dim(M^\perp) = 1\).
Step 4: Final Answer:
The dimension of \(M^\perp\) is 1.
Step 5: Why This is Correct:
Both methods show that the dimension is 1. The first method uses the rank-nullity theorem equivalent for subspaces, relating the dimension of a subspace and its orthogonal complement. The second method explicitly constructs a basis for the orthogonal complement.