2
$\begingroup$

I have a square real matrix of the following block form:

$$M(k) = \left(\begin{array}{cc} 0 & A\\ kB & C \end{array}\right)$$

where $A$ is $1 \times m$, $B$ is $m \times 1$ and $C$ is $m \times m$, which makes $M$ an $n \times n$ matrix with $n = m+1$. Here $k$ is a scalar.

Is there a way to determine how the eigenvalues of $M$ depend on $k$? In other words, if I compute the eigenvalues of $M$ for $k = 1$, what can I say about the eigenvalues of $M(k)$ for other values of $k$?

  • 0
    Does $A = B^T$ by any chance? Is $C$ a symmetric matrix? – 2017-01-26
  • 0
    @Omnomnomnom No and No. – 2017-01-26
  • 0
    It is notable that $M(k)$ is similar to $$ \pmatrix{0&\sqrt{k}A\\ \sqrt{k}B & C} $$ and that this is a rank-2 update of the matrix $$ \pmatrix{0&0\\0&C} $$ you can use this to get some handy inequalities on the eigenvalues, but I wouldn't expect a nice closed form for the eigenvalues in terms of $k$. – 2017-01-26
  • 0
    Chapter VIII of Bhatia's Matrix Analysis might be a useful resource – 2017-01-26
  • 0
    @Omnomnomnom I am mostly interested in the *signs* of the eigenvalues. Maybe something can be worked out for that. – 2017-01-26
  • 0
    what do you mean by the sign? Are you sure that the eigenvalues will be real? – 2017-01-26
  • 0
    @Omnomnomnom Sorry, I meant the sign of the real parts. – 2017-01-26
  • 0
    Aha, now we're getting somewhere. Is $C$ (or $C + C^T$) positive definite? – 2017-01-26
  • 0
    @Omnomnomnom I don't think it is in general. – 2017-01-27
  • 0
    I'm out of ideas, then – 2017-01-27
  • 0
    @Omnomnomnom Just out of curiosity, if $C$ or $C + C^T$ were positive-definite, what would you have done? – 2017-01-27
  • 0
    if $M + M^T$ is positive definite, then its eigenvalues have positive real part. If $C + C^T$ is positive semidefinite, then so is $M(0) + M(0)^T$. From there, I think we could deduce that $M(k)$ has at most two negative eigenvalues (eigenvalues with negative real part) for $k \neq 0$; I think. Honestly, not quite sure how that would work. Unfortunately, when $k \neq 0$, $M(k)$ can be neither positive nor negative semidefinite. – 2017-01-27

0 Answers 0