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Let's have the following $(n+1) \times (n+1)$ matrix with block elements

$\mathbf{Y} = \begin{bmatrix} \mathbf{A} & - \mathbf{w} \\ - \mathbf{w}^{T} & b \end{bmatrix}$

where $\mathbf{A}$ is a $n \times n$ matrix, $\mathbf{w}$ is a $n$-dimensional vector, and $b$ is a scalar.

Is it possible to calculate the eigenvalues of $\mathbf{Y}$ if we already know the eigenvalues of $\mathbf{A}$, as well as the values of $\mathbf{w}$ and $b$?

Thanks in advance!

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2 Answers 2

up vote 4 down vote accepted

No, it is not. Consider the following example in the case $n=2$: $$ Y_1 = \pmatrix{1 & 0 & 1\cr 0 & -1 & 0\cr 1 & 0 & 0\cr},\ Y_2 = \pmatrix{-1 & 0 & 1\cr 0 & 1 & 0\cr 1 & 0 & 0\cr}$$ Here $A_1 = \pmatrix{1 & 0\cr 0 & -1\cr}$ and $A_2 = \pmatrix{-1 & 0\cr 0 & 1\cr}$ have the same eigenvalues, and the $b$ and $w$ are the same. But $Y_1$ and $Y_2$ have different determinants ($+1$ and $-1$ respectively) and therefore different eigenvalues.

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Ok thank you for your answers! –  Ita Atz Dec 20 '12 at 2:35

This is not an answer (as Robert has already provided one), but gives a hint at how the $w$ affects the eigenvalues (and doesn't fit in the comments) in some particular cases.

If $U^T YU=\Lambda$, where $\Lambda$ is diagonal with elements $\lambda_1,...,\lambda_n$, then $\begin{bmatrix} U^T & 0 \\ 0 & I \end{bmatrix} \begin{bmatrix} Y & -w \\ -w^T & b \end{bmatrix} \begin{bmatrix} U & 0 \\ 0 & I \end{bmatrix} = \begin{bmatrix} \Lambda & -Uw \\ -w^T U & b \end{bmatrix}$. Let $\tilde{w} = Uw$ for convenience.

Suppose $x \notin \{\lambda_1,...,\lambda_n\}$, then we have \begin{eqnarray} \det(\begin{bmatrix} xI-\Lambda & \tilde{w} \\ \tilde{w}^T & x-b \end{bmatrix}) &=& \det (xI-\Lambda)(x-b - \tilde{w}^T (xI-\Lambda)^{-1} \tilde{w}) \\ &=& \det (xI-\Lambda)(x-b - \sum_i \frac{\tilde{w}_i^2}{x-\lambda_i}) \\ &=& (x-\lambda_1)\cdots(x-\lambda_n)(x-b)-\sum_i \tilde{w}_i^2 \prod_{j \neq i}(x -\lambda_j) \end{eqnarray} Hence $\det(xI-Y) = (x-\lambda_1)\cdots(x-\lambda_n)(x-b)-\sum_i \tilde{w}_i^2 \prod_{j \neq i}(x -\lambda_j)$.

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