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Let $M \in \mathbb{R}^{n \times n}$ be a symmetric, positive definite matrix, such that $M \preccurlyeq I$ (i.e., $I-M$ is positive semi-definite). Prove or disprove that the matrix $$ \left[\begin{matrix} I-M & -M \\ I-M & I-M \end{matrix}\right] $$ has all eigenvalues inside the unit circle.

Comments. I can only prove the claim if $n=1$.

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Call your block matrix $A$. Since the blocks commute, its characteristic polynomial is $p(\lambda) = \det(A - \lambda I) = \det(((1-\lambda) I - M)^2 + M(I-M)) = \det((2\lambda - 1)M + (\lambda - 1)^2)$. This is $0$ when $2 \lambda - 1 \ne 0$ and $(\lambda - 1)^2/(1-2\lambda)$ is an eigenvalue of $M$. Now solving the equation $$ \dfrac{(\lambda - 1)^2}{1-2\lambda} = t$$ we get $$\lambda = 1-t \pm \sqrt{t^2-t}$$ For $t \in (0,1)$ (where the eigenvalues of $M$ are), this is in the unit circle because $t^2-t < 0$ and $|\lambda|^2 = (1-t)^2 + t - t^2 = 1 - t < 1$.

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  • $\begingroup$ Thanks a lot your the answer. Can you please further clarity what is meant by "Since the blocks commute"? $\endgroup$ – user693 Jul 10 '15 at 16:11
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Since $M$ is symmetric, it is orthogonally similar to a diagonal matrix: $M=QDQ^T$ where $Q$ is orthogonal and $D$ is diagonal. We have $$ \pmatrix{Q&0\\0&Q}^T\pmatrix{I-M&-M\\I-M&I-M}\pmatrix{Q&0\\0&Q}=\pmatrix{I-D&-D\\I-D&I-D}=:C. $$ Note that $C$ (having the same eigenvalues as the original matrix) is very special, it is a $2\times 2$ block matrix where the blocks are diagonal matrices. It is useful to know that the eigenvalues of such a matrix are given by the eigenvalues of $2\times 2$ matrices composed of the corresponding diagonal entries in the diagonal blocks (this can be shown simply by choosing "special eigenvectors"). Hence the eigenvalues of $C$ are the eigenvalues of matrices $$ \pmatrix{1-d&-d\\1-d&1-d}, $$ (and vice versa) where $d$ are diagonal entries of $D$ (the eigenvalues of $M$). Since you can deal with the case $n=1$, you can now deal with any $n>1$.

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