Question

Matrix \(P\) is given as
\[ P = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \]
The TRUE option is

• A. Trace of \(P\) is equal to the sum of the Eigen values of \(P\).
• B. \(P^TP\) is an identity matrix.
• C. \(P\) is a skew-symmetric matrix.
• D. Absolute magnitude of each Eigen value is 1.

Hint:

Trace always equals the sum of eigenvalues for any square matrix; check the other three options by direct computation.

Answer 1

The Correct Option is A

Step 1: Write down the matrix and recall the key facts.
\[ P = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \]
Two facts decide this question: for any square matrix, the trace (sum of diagonal entries) always equals the sum of its eigenvalues, and a matrix is symmetric if \(P^T=P\) and skew-symmetric if \(P^T=-P\).

Step 2: Find the eigenvalues of \(P\).
The characteristic equation is \(\det(P-\lambda I)=0\):
\[ \det \begin{bmatrix} 1-\lambda & 0 & 1 \\ 0 & 1-\lambda & 0 \\ 1 & 0 & 1-\lambda \end{bmatrix} = 0 \]
Expanding along the second row, which has only one nonzero entry, gives
\[ (1-\lambda)\left[(1-\lambda)^2 - 1\right] = 0 \]
\[ (1-\lambda)(1-\lambda-1)(1-\lambda+1) = 0 \implies (1-\lambda)(-\lambda)(2-\lambda) = 0 \]
So the eigenvalues are \(\lambda = 0, 1, 2\).

Step 3: Check option (A).
Trace of \(P\) = sum of diagonal entries = \(1+1+1=3\). Sum of eigenvalues = \(0+1+2=3\). These are equal, so option (A) is TRUE. This is in fact a general theorem, true for every square matrix, so the calculation above is only a check, not the reason it holds.

Step 4: Check the remaining options and rule them out.
Option (B): since \(P\) is symmetric (\(P^T=P\)), \(P^TP=P^2\). Computing \(P^2\), the \((1,1)\) entry is \(1(1)+0(0)+1(1)=2 \neq 1\), so \(P^2\) is not the identity matrix; (B) is false.
Option (C): a skew-symmetric matrix needs every diagonal entry to be zero, since \(p_{ii}=-p_{ii}\) forces \(p_{ii}=0\), but \(P\) has diagonal entries \(1,1,1\), so \(P\) is symmetric, not skew-symmetric; (C) is false.
Option (D): the eigenvalues are \(0,1,2\), and \(|0| \neq 1\), so not every eigenvalue has magnitude \(1\); (D) is false.

Final Answer:
Option (A) is the only true statement: trace of \(P\) equals the sum of the eigenvalues of \(P\). \[ \boxed{\text{Option (A)}} \]