Question

Let \(M = \left(I_n - \frac{1}{n}\mathbf{1}\mathbf{1}^T\right)\) be a matrix, where \(\mathbf{1} = (1,1,1,\ldots,1)^T \in \mathbb{R}^n\) and \(I_n\) is the identity matrix of order \(n\).

Which of the following options is/are correct?

• A. \(M^T = M\)
• B. \(M^2 = I_n\)
• C. \(\text{Trace}(M) = n\)
• D. \(M\) is a projection matrix

Hint:

Use the identity \(\mathbf{1}^T\mathbf{1}=n\) to expand \(M^2\) and \(\text{Trace}(M)\); a symmetric idempotent matrix is by definition a projection matrix.

Answer 1

The Correct Option is A, D

Step 1: Understand the matrix M.
\(M = I_n - \frac{1}{n}\mathbf{1}\mathbf{1}^T\) is the well known centering matrix. \(\mathbf{1}\mathbf{1}^T\) is an \(n \times n\) matrix of all ones, so \(M\) subtracts the average of the entries of a vector from each of its components when applied. Before checking the options, note the key scalar fact \(\mathbf{1}^T\mathbf{1} = n\), since it is a sum of \(n\) ones.

Step 2: Check option (A), symmetry.
\(I_n\) is symmetric, and \(\mathbf{1}\mathbf{1}^T\) is symmetric because \((\mathbf{1}\mathbf{1}^T)^T = \mathbf{1}\mathbf{1}^T\) (the transpose of an outer product of a vector with itself is itself). A difference of two symmetric matrices is symmetric, so \(M^T = M\). Option (A) is correct.

Step 3: Check option (B), whether M squared is the identity.
Expand \(M^2\):
\[ M^2 = \left(I_n - \frac{1}{n}\mathbf{1}\mathbf{1}^T\right)^2 = I_n - \frac{2}{n}\mathbf{1}\mathbf{1}^T + \frac{1}{n^2}\mathbf{1}\mathbf{1}^T\mathbf{1}\mathbf{1}^T \]
Use \(\mathbf{1}^T\mathbf{1} = n\), so \(\mathbf{1}\mathbf{1}^T\mathbf{1}\mathbf{1}^T = \mathbf{1}(\mathbf{1}^T\mathbf{1})\mathbf{1}^T = n\,\mathbf{1}\mathbf{1}^T\).
\[ M^2 = I_n - \frac{2}{n}\mathbf{1}\mathbf{1}^T + \frac{1}{n^2}\cdot n\,\mathbf{1}\mathbf{1}^T = I_n - \frac{2}{n}\mathbf{1}\mathbf{1}^T + \frac{1}{n}\mathbf{1}\mathbf{1}^T = I_n - \frac{1}{n}\mathbf{1}\mathbf{1}^T = M \]
So \(M^2 = M\), not \(I_n\) (for \(n > 1\), \(M \neq I_n\)). Option (B) is incorrect.

Step 4: Check option (C), the trace.
\[ \text{Trace}(M) = \text{Trace}(I_n) - \frac{1}{n}\text{Trace}(\mathbf{1}\mathbf{1}^T) \]
\(\text{Trace}(I_n) = n\), and \(\text{Trace}(\mathbf{1}\mathbf{1}^T) = \mathbf{1}^T\mathbf{1} = n\) (the trace of an outer product equals the inner product).
\[ \text{Trace}(M) = n - \frac{1}{n}\cdot n = n - 1 \]
This is \(n-1\), not \(n\), so option (C) is incorrect.

Step 5: Check option (D), whether M is a projection matrix.
A matrix is called a projection matrix when it is symmetric and idempotent, meaning \(M^T = M\) and \(M^2 = M\). Step 2 showed \(M^T = M\) and Step 3 showed \(M^2 = M\). Both conditions hold, so \(M\) is indeed a (orthogonal) projection matrix, it projects any vector onto the subspace orthogonal to \(\mathbf{1}\). Option (D) is correct.

Final Answer:
Options (A) and (D) are correct. \[ \boxed{\text{(A), (D)}} \]