Suppose I have a special block, Hermitian matrix $F(t):= \left[ {\begin{array}{cc} H & t \bar S \\ tS & \bar H \\ \end{array} } \right] $ where '-' denotes complex conjugate. The blocks H and S are themselves Hermitian and Complex Symmetric respectively in this case and $t \in \mathbb R .$ $E_k(H)$ stands for the eigenspace belonging to $\lambda_k(H)$. Here we consider the eigenvalues in decreasing order. For a unit vector $v$ in $\mathbb C^n$. $$\mathcal{U_v:= {\ \left[ {\begin{array}{cc} z_1v \\ \bar z_{2}\bar v\\ \end{array} } \right]}; z_1,z_2 \in \mathbb C} $$ Let $v_1,v_2,\ldots,v_d$ be a basis of the the eigenspace $E_k(H)$.

I am unable to prove that $E_{2k-1}(F(0))=E_{2k}(F(0))= \bigoplus \limits_{j=1}^d \mathcal U_{v_{j}} .$ Where $\bigoplus$ denotes the direct sum.

Can anyone help me in this regard.

  • $\begingroup$ What does the subscript $k$ mean? Is $\lambda_k$ the $k$'th largest eigenvalue? Also obviously two different eigenspaces can not be the same space, unless your counting of eigenvalues allows $\lambda_{2k-1} = \lambda_{2k}$. Do you perhaps mean that the the two spaces have equal dimension? $\endgroup$ – Vincent Jul 20 '17 at 12:53
  • $\begingroup$ Here we consider the eigenvalues in decreasing order. $\endgroup$ – baam Jul 20 '17 at 13:58

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