# If $X$ and $X-Y^TXY$ are symmetric positive definite, then $\max{|\lambda_i(Y)|<1}$

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).

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 $\max_i|\lambda_i(Y)|< 1$.
• Could you explain more why eigenvalues of $Z$ and $Y$ are the same? May 5 '13 at 20:02
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 $1$.