Question

Let \(A\) be the adjacency matrix of the given graph with vertices \(\{1,2,3,4,5\}\). 

Let \(\lambda_1, \lambda_2, \lambda_3, \lambda_4, \lambda_5\) be the eigenvalues of \(A\) (not necessarily distinct). Find: \[ \lambda_1 + \lambda_2 + \lambda_3 + \lambda_4 + \lambda_5 \;=\; \_\_\_\_\_\_ . \]

Hint:

For adjacency matrices, \(\sum \lambda_i=\mathrm{tr}(A)\). If loops are present, many graph-theory conventions count each loop as \(2\) on the diagonal so that row sums equal vertex degrees. Always check the loop convention—here it yields \(A_{33}=A_{44}=2\).

Answer 1

Step 1 (Key fact): For any square matrix \(A\), the sum of its eigenvalues equals \(\mathrm{tr}(A)\), the trace of \(A\). Hence \(\sum_{i=1}^{5}\lambda_i=\mathrm{tr}(A)=\sum_{i=1}^{5}A_{ii}\).
Step 2 (Adjacency diagonal entries): In the standard adjacency-matrix convention used in such problems, each self-loop at a vertex contributes \(2\) on the diagonal (so that the row sum equals the degree counting a loop twice).
Step 3 (Read the graph): The picture shows self-loops at vertices \(3\) and \(4\), and no loops at \(1,2,5\). Therefore, \(A_{33}=2,\; A_{44}=2,\; A_{11}=A_{22}=A_{55}=0\).
Step 4 (Trace and sum of eigenvalues): \(\mathrm{tr}(A)=A_{11}+A_{22}+A_{33}+A_{44}+A_{55}=0+0+2+2+0=4\).
\(\Rightarrow\) \(\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5=4\).
\[ \boxed{4} \]