2
$\begingroup$

Consider a symmetric positive definite matrix $X \in \mathbb{R}^{n \times n}$. If there is a $Y \in \mathbb{R} ^{n\times n}$ such that $X-Y^TXY$ is symmetric positive definite as well, how can be shown that $\max{|\lambda_i(Y)|<1}$, where $\lambda_i(Y)$'s are the eigenvalues of $Y$? (in other words, the spectral radius of $Y$ is less than 1).

$\endgroup$
3
$\begingroup$

This is actually easier than it looks. $X$ is symmetric positive definite, which means that you can take powers of $X$ that are also s.p.d., including finding $X^{-1/2}$ and $X^{1/2}$ such that $X^{-1/2}X^{1/2} = I$ and $X^{1/2}X^{1/2} = X$. Given any s.p.d. matrix $A$, $X^{-1/2}AX^{-1/2}$ is also s.p.d.

So $$ X^{-1/2}(X - Y^TXY)X^{-1/2} = I - X^{-1/2}Y^TX^{1/2}X^{1/2}YX^{-1/2} = I - Z^TZ, $$ for $Z = X^{1/2}YX^{-1/2}$. Because $I-Z^TZ$ is s.p.d., its eigenvalues are greater than zero, which means that the eigenvalues of $Z^TZ$ are less than one. But $Z^TZ$ is itself symmetric and positive semidefinite, so its eigenvalues are nonnegative. The eigenvalues of $Z^TZ$ are also the squares of the eigenvalues of $Z$. More importantly, the eigenvalues of $Z$ are the same as the eigenvalues of $Y$ since they are similar.

This means that the squares of the eigenvalues of $Y$ are $0\leq \lambda_i^2 < 1$. So $\max_i|\lambda_i(Y)|< 1$.

$\endgroup$
3
2
$\begingroup$

Since $Y$ is real, $Y^T=Y^\ast$. Now, let $(\lambda,v)$ be an eigenpair of $Y$. By the given assumptions, we have $v^\ast Xv>0$ and $(1-|\lambda|^2|)v^\ast Xv=v^\ast(X-Y^\ast XY)v>0$. Therefore $1-|\lambda|^2>0$ for every eigenvalue $\lambda$ of $Y$. In particular, the spectral radius of $Y$ is smaller than $1$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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