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: Check Statement 4 (General Matrix Property).
For any square matrix, the Trace (sum of diagonal elements) is always equal to the sum of its eigenvalues. Trace of $P = 1 + 1 + 1 = 3$.
Step 2: Calculate Eigenvalues to Verify.
$|P - \lambda I| = 0 \implies (1-\lambda)[(1-\lambda)^2 - 1] = 0$. $(1-\lambda) = 0 \implies \lambda_1 = 1$. $(1-\lambda)^2 = 1 \implies 1-\lambda = \pm 1$. $\lambda_2 = 0, \lambda_3 = 2$. Sum of eigenvalues $= 1 + 0 + 2 = 3$. Statement 4 is correct.
Step 3: Eliminate Others.
$P^T P = P^2 \neq I$ (Not orthogonal). $P^T = P \neq -P$ (Not skew-symmetric). Eigenvalues are $\{0, 1, 2\}$, not all 1.