Question

Let \( P = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \). Which of the following statements is/are correct?
1. P\(^T\)P = I
2. P is skew-symmetric
3. The value of each eigenvalue of P is 1
4. The trace of P is equal to the sum of its eigenvalues

Hint:

A fundamental theorem in linear algebra states that the trace of a square matrix is always equal to the sum of its eigenvalues. This can be a quick check or sometimes the only correct option in multiple-choice questions.

Answer 1

Step 1: Understanding the Question:
We need to evaluate four statements about the given 3x3 matrix P and determine which are true.
Step 2: Detailed Evaluation of Each Statement:
The given matrix is \( P = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \).
1. P\(^T\)P = I First, find the transpose of P, \(P^T\). \[ P^T = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} = P \] Since \(P^T = P\), the matrix is symmetric. The condition becomes \(P^2 = I\). Let's calculate \(P^2\): \[ P^2 = P \times P = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} = \begin{pmatrix} (1+0+1) & (0+0+0) & (1+0+1) \\ (0+0+0) & (0+1+0) & (0+0+0) \\ (1+0+1) & (0+0+0) & (1+0+1) \end{pmatrix} = \begin{pmatrix} 2 & 0 & 2 \\ 0 & 1 & 0 \\ 2 & 0 & 2 \end{pmatrix} \] Since \(P^2 \neq I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\), statement 1 is false.
2. P is skew-symmetric A matrix M is skew-symmetric if \(M^T = -M\). We already found that \(P^T = P\). Since \(P \neq -P\), the matrix is symmetric, not skew-symmetric. Statement 2 is false.
3. The value of each eigenvalue of P is 1 To find the eigenvalues (\(\lambda\)), we solve the characteristic equation \(det(P - \lambda I) = 0\). \[ \det \begin{pmatrix} 1-\lambda & 0 & 1 \\ 0 & 1-\lambda & 0 \\ 1 & 0 & 1-\lambda \end{pmatrix} = 0 \] Expanding along the second row: \[ (1-\lambda) \left[ (1-\lambda)(1-\lambda) - (1)(1) \right] = 0 \] \[ (1-\lambda) [ (1-\lambda)^2 - 1 ] = 0 \] \[ (1-\lambda) [ \lambda^2 - 2\lambda + 1 - 1 ] = 0 \] \[ (1-\lambda) [ \lambda^2 - 2\lambda ] = 0 \] \[ (1-\lambda) \lambda (\lambda - 2) = 0 \] The eigenvalues are \(\lambda_1 = 0, \lambda_2 = 1, \lambda_3 = 2\). Since the eigenvalues are not all 1, statement 3 is false.
4. The trace of P is equal to the sum of its eigenvalues The trace of a matrix is the sum of its diagonal elements. \[ \text{Tr}(P) = 1 + 1 + 1 = 3 \] The sum of the eigenvalues we found is: \[ \Sigma \lambda = 0 + 1 + 2 = 3 \] Since \(\text{Tr}(P) = \Sigma \lambda\), statement 4 is true. This property is true for all square matrices.
Step 3: Final Answer:
Only statement 4 is correct.