Prove that \[ \begin{pmatrix} 0&5&1&0\\ 5&0&5&1\\ 1&5&0&5\\ 0&1&5&0 \end{pmatrix} \] has two positive and two negative eigenvalues, counted with multiplicity.
Solution
Let \[ A= \begin{pmatrix} 0&5&1&0\\ 5&0&5&1\\ 1&5&0&5\\ 0&1&5&0 \end{pmatrix}. \]
1. The subspaces\[ V_+\coloneqq \operatorname{span}\{e_1+e_4,e_2+e_3\}, \qquad V_-\coloneqq \operatorname{span}\{e_1-e_4,e_2-e_3\} \]are \(A\)-invariant, and\[ \RR^4=V_+\oplus V_-. \]
Set \[ u_1=e_1+e_4, \qquad u_2=e_2+e_3, \qquad v_1=e_1-e_4, \qquad v_2=e_2-e_3. \] Direct calculation gives \[ Au_1=6u_2, \qquad Au_2=6u_1+5u_2, \] and \[ Av_1=4v_2, \qquad Av_2=4v_1-5v_2. \] Thus both subspaces are invariant. The four displayed vectors are linearly independent, so their two spans form a direct-sum decomposition of \(\RR^4\).
2. Relative to the ordered bases \((u_1,u_2)\) and \((v_1,v_2)\), the tworestrictions of \(A\) are\[ A_+= \begin{pmatrix} 0&6\\ 6&5 \end{pmatrix}, \qquad A_-= \begin{pmatrix} 0&4\\ 4&-5 \end{pmatrix}. \]
This is exactly the coordinate form of the four identities in step 1.
3. Each of \(A_+\) and \(A_-\) has one positive and one negative eigenvalue.
Both matrices are real symmetric, so their eigenvalues are real. Moreover, \[ \det A_+=-36<0, \qquad \det A_-=-16<0. \] For a real \(2\times2\) matrix with real eigenvalues, a negative determinant means that the product of the two eigenvalues is negative. Hence each restriction has one positive and one negative eigenvalue.
4. The matrix \(A\) has exactly two positive and two negative eigenvalues,counted with multiplicity.
By step 1, \(A\) is the direct sum of its restrictions to \(V_+\) and \(V_-\). Thus its multiset of eigenvalues is the union of the eigenvalue multisets of \(A_+\) and \(A_-\). Step 3 contributes one positive and one negative eigenvalue from each summand, giving two of each.
5. Q.E.D.
Step 4 is the required signature statement.