Question

In a simple graph, which of the following statements is/are true?
(I) Adjacency matrix is symmetric.
(II) Trace of the adjacency matrix is 1.

• A. Only (I)
• B. Only (II)
• C. Both (I) and (II)
• D. Neither (I) nor (II)

Hint:

Simple graphs have no self-loops, so the adjacency matrix diagonal, and hence its trace, is always zero.

Answer 1

The Correct Option is A

Method: Verify both statements using an explicit small example instead of relying on the general definition alone.

Take the simple graph $K_3$, a triangle with vertices $1, 2, 3$, where every pair of vertices is connected by an edge. Its adjacency matrix, ordering rows and columns as $1, 2, 3$, is:

\[ A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} \]

Compare $A$ with its transpose $A^T$. Swapping rows and columns of $A$ reproduces the exact same matrix, since the off-diagonal $1$s sit in mirrored positions ($A_{12} = A_{21} = 1$, $A_{13} = A_{31} = 1$, $A_{23} = A_{32} = 1$). So $A = A^T$, confirming symmetry. This mirrored pattern is not special to $K_3$; it holds for any simple graph because an edge $\{i, j\}$ contributes identically to $A_{ij}$ and $A_{ji}$.

Now read off the diagonal entries: $A_{11} = A_{22} = A_{33} = 0$, because a simple graph forbids loops (a vertex cannot be adjacent to itself). Adding the diagonal entries gives:

\[ \text{trace}(A) = 0 + 0 + 0 = 0 \ne 1 \]

This example confirms the general rule: the adjacency matrix of any simple graph is symmetric, and its trace is always $0$, never $1$.

\[\boxed{\text{Only (I)}}\]