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Consider a complex $N-1 \times 1$ vector $b$ and a complex constant c. Let $I$ denote the $N-1 \times N-1$ identity matrix. Then what can we say about the eigenvalues of the matrix

\begin{align} \begin{bmatrix} I & b \\ b^H & c \end{bmatrix} \end{align}

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Apart from Alexander Gruber's answer, we may also determine the eigenvalues in two other ways.

Call your matrix $A$. As the case $b=0$ is trivial, we assume that $b\not=0$. First, note that if $v\perp b$, then $(v^T, 0)$ will be a left eigenvector of $A$ corresponding to the eigenvalue $1$. Hence the multiplicity of this eigenvalue is at least $n-2$.

Next, if $\lambda\not=1$ is an eigenvalue of $A$, then by a determinant formula for block matrices, we have $0=\det(A-\lambda I)=\det(I-\lambda I)\det[(c-\lambda)-b^T(I-\lambda I)^{-1}b]$ and therefore $(c-\lambda)-\|b\|^2/(1-\lambda)=0$. Hence the remaining two eigenvalues of $A$ are given by the two roots of the quadratic equation $\lambda^2 - (c+1)\lambda + (c-\|b\|^2)=0$.

Alternatively, suppose $w^T=(b^T+v^T, q-1)$ is a left eigenvectors of $A$, where $v\perp b$. Then $w^TA = \left(qb^T + v^T,\ \|b\|^2 + b^Tv + (q-1)c\right)$. In order that $qb^T+v^T$ is a multiple of $b^T+v^T$, we must have $v=0$ and the eigenvalue is $q$. So, by comparing the last elements of $w^T$ and $w^TA$, we must have $\|b\|^2 + (q-1)c = q(q-1)$, i.e. $q^2 - (c+1)q + (c-\|b\|^2)=0$.

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  • $\begingroup$ I suppose that determinant formula you are referring is related to schur complements. The determinant is the product of the block matrix and its schur complement? $\endgroup$ – dineshdileep Dec 17 '12 at 9:16
  • $\begingroup$ elegant answer!!. $\endgroup$ – dineshdileep Dec 17 '12 at 9:18
  • $\begingroup$ @dineshdileep Yes, it's the product of the determinants of a block and its Schur complement. $\endgroup$ – user1551 Dec 17 '12 at 9:39
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Call your matrix $M_b$. Then $$\text{det}M_b-\lambda I_{n}=\left((c-\lambda)(\lambda-1)+\sum_{i=1}^{n-1}b_i^2\right)(\lambda-1)^{n-2},$$ so you're going to have $1$ as an eigenvalue of multiplicity $n-2$ and then $$\frac{1}{2}\left(1+c\pm\sqrt{(c+1)^2+4\left(-c+\sum_{i=1}^{n-1}b_i^2\right)}\right)$$ as your other two eigenvalues.

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  • $\begingroup$ Thanks a lot for the answer!!. $\endgroup$ – dineshdileep Dec 17 '12 at 9:19

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