Question

Let \(M\) be any \(3\times3\) symmetric matrix with eigenvalues \(1,2,3\). Let \(N\) be any \(3\times3\) matrix with real eigenvalues such that \(MN+N^{T}M=3I\), where \(I\) is the \(3\times3\) identity matrix. Which of the following cannot be eigenvalue(s) of the matrix \(N\)?

• A. \(\dfrac14\)
• B. \(\dfrac34\)
• C. \(\dfrac12\)
• D. \(\dfrac74\)

Hint:

For equations of the form \(A X + X^{T} A = cI\) with \(A\succ0\), testing a real eigenpair \(Xv=\lambda v\) and forming \(v^{T}(\cdot)v\) often yields tight bounds on \(\lambda\) via Rayleigh quotients.

Answer 1

The Correct Option is A, D

Step 1: Diagonalize \(M\).
Since \(M\) is symmetric, there is an orthogonal \(Q\) with \(Q^{T}MQ=D=\operatorname{diag}(1,2,3)\). Put \(\tilde N=Q^{T}NQ\). The condition becomes \(D\tilde N+\tilde N^{T}D=3I\). Similarity preserves eigenvalues, so \(\tilde N\) and \(N\) have the same eigenvalues.

Step 2: Rayleigh identity for a real eigenpair.
Let \(\tilde N v=\lambda v\) with real \(\lambda\) and \(v\neq0\). Then
\(v^{T}(D\tilde N+\tilde N^{T}D)v=3\,v^{T}v\).
The left side equals \(v^{T}D(\lambda v)+(\lambda v)^{T}Dv=2\lambda\, v^{T}Dv\). Hence
\(2\lambda\, v^{T}Dv=3\,v^{T}v \implies \lambda=\dfrac{3}{2}\dfrac{v^{T}v}{v^{T}Dv}\).

Step 3: Bound the quotient.
For \(D=\operatorname{diag}(1,2,3)\), its eigenvalues satisfy \(1\le \dfrac{v^{T}Dv}{v^{T}v}\le 3\). Therefore
\(\dfrac{1}{3}\le \dfrac{v^{T}v}{v^{T}Dv}\le 1 \implies \dfrac12 \le \lambda \le \dfrac32.\)

Step 4: Decide feasible options.
Every real eigenvalue of \(N\) must lie in \([1/2,\,3/2]\). Thus \(\tfrac14\) and \(\tfrac74\) are impossible, while \(\tfrac12\) and \(\tfrac34\) are attainable (e.g., by taking \(v\) along the \(D\)-eigenvectors with eigenvalues \(3\) and \(2\), respectively).