Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Denote $ \rho(A)$ to be the spectral radius of a matrix $A,$ that is the maximal eigenvalue of $A.$ We say that a matrix $M$ is positive definite, respectively positive semidefinite, if $x^TMx>0$ and $ x^TMx \geq 0$ respectively for all vectors $x$ with nonzero entries.

I want to show that if $ \rho(A)>1,$ then there exists a real symmetric matrix $B$ that is not positive semidefinite such that $A^TBA - B = -C$ holds for some positive definite matrix $C.$

Any hints or proof would be appreciated.

I've shown the part where, if $ \rho(A)<1$ then for every positive definite matrix $C,$ $ A^TBA - B = -C$ has a unique solution $B$ that is also symmetric and positive definite. ;)

share|cite|improve this question

Presumably $A$ and $B$ are real. Your requirement is possible if and only if $1$ and $-1$ are not eigenvalues of $A$.

Suppose $1$ (or $-1$) is an eigenvalue of $A$ and $x$ is a corresponding eigenvector. Then your requirement implies that $B\succ A^TBA$, which is impossible because $x^TBx=x^TA^TBAx$.

Suppose $\pm1$ are not eigenvalues of $A$. By a change of basis ($B$ is a quadratic form and $A$ is a linear transformation; so they undergo the same change of basis), we may assume that $A$ is already in its real Jordan form. Let $B$ be a block diagonal matrix with block sizes conforming to $A$'s. We can solve the problem blockwise.

For each Jordan block $A_0$ of $A$ with eigenvalue modulus greater than $1$, set the corresponding diagonal subblock of $B$ to the negative definite matrix $B_0$, where $-B_0$ is the positive definite solution to the Lyapunov equation $A_0^{-T}(-B_0)A_0^{-1} - (-B_0) - I = 0$. For each Jordan block $A_0$ with eigenvalue modulus $\le1$ (the modulus can be $1$, it's just that the eigenvalues are a conjugate pair of nonreal eigenvalues on the unit circle), set the corresponding diagonal subblock of $B_0$ to a positive semidefinite solution to $A_0^TB_0A_0 - B_0 = -I$. Since $\rho(A)>1$, $B$ always possesses a negative definite diagonal subblock. Therefore $B$ is not positive semidefinite.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.