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Let ${\bf A} \in {\Bbb R}^{n \times n}$ be a symmetric positive definite M-matrix whose non-diagonal entries are non-positive. Let ${\bf B} \in {\Bbb R}^{n \times n}$ be a diagonal matrix, whose diagonal entries are non-negative. Let ${\bf C} \in {\Bbb R}^{n \times n}$ also be a diagonal matrix. Let

$${\bf M} := \begin{bmatrix} \mathbf{A} & \mathbf{0} \newline \mathbf{0} & \mathbf{A} \newline \end{bmatrix} + \begin{bmatrix} -\mathbf{B} & \mathbf{C} \newline \mathbf{C} & \mathbf{B} \newline \end{bmatrix}$$

$${\bf N} := \begin{bmatrix} \mathbf{A} & \mathbf{0} \newline \mathbf{0} & \mathbf{A} \newline \end{bmatrix} + \begin{bmatrix} -\mathbf{B} & |\mathbf{C}| \newline |\mathbf{C}|& \mathbf{B} \newline \end{bmatrix}$$

Compared to $\mathbf{M}$, all the entries of $\mathbf{C}$ in $\mathbf{N}$ take their absolute values. Is the smallest eigenvalue of $\mathbf{N}$ always smaller than the smallest eigenvalue of $\mathbf{M}$?

Rodrigo de Azevedo
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K416
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2 Answers2

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Clearly $\lambda_\min(N)\le\lambda_\min(M)$ if and only if $\lambda_\min(D^{-1}ND+pI)\le\lambda_\min(D^{-1}MD+pI)$ for any invertible matrix $D$ and real number $p$. So, by picking $D=D^{-1}=\operatorname{diag}(I_n,-I_n)$ and a large positive $p$, we may redifine $M$ and $N$ as $$ M=\pmatrix{X&-C\\ -C&Y}\quad\text{and}\quad N=\pmatrix{X&-|C|\\ -|C|&Y} $$ so that $X=A+B+pI$, $Y=A-B+pI$ and $N$ are positive definite M-matrices.

When $p>0$ is sufficiently large, we may further assume that $\|X^{-1}C\|_F$, $\|Y^{-1}C\|_F$, $\|X^{-1}|C|\|_F$ and $\|Y^{-1}|C|\|_F$ are small. Hence \begin{align*} M^{-1} =\pmatrix{I&-X^{-1}C\\ -Y^{-1}C&I}^{-1}\pmatrix{X^{-1}\\ &Y^{-1}} =\sum_{k=0}^\infty\pmatrix{0&X^{-1}C\\ Y^{-1}C&0}^{k}\pmatrix{X^{-1}\\ &Y^{-1}} \end{align*} and there is a similar power series expansion for $N^{-1}$ (with $C$ replaced by $|C|$). It follows from the entrywise non-negativeness of $X^{-1}$ and $Y^{-1}$ that $-N^{-1}\le M^{-1}\le N^{-1}$ entrywise. As $N^{-1}$ is entrywise nonnegative, we may further infer that $-N^{-k}\le M^{-k}\le N^{-k}$ for every natural number $k$. Therefore, by Gelfand’s formula, $$ \rho(M^{-1}) =\lim_{k\to\infty}\|M^{-k}\|_F^{1/k} \le\lim_{k\to\infty}\|N^{-k}\|_F^{1/k} =\rho(N^{-1}). $$ Since the redefined $M$ and $N$ are positive definite, the inequality above is equivalent to $\lambda_\min(M)^{-1}\le\lambda_\min(N)^{-1}$, i.e., $\lambda_\min(N)\le\lambda_\min(M)$. Therefore we have $\lambda_\min(N)\le\lambda_\min(M)$ for the original $M$ and $N$ too.

user1551
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re-write it as
$X = -M +\delta I$ and $Y=-N + \delta I$ for some large enough $\delta \gt 0$. $X = \begin{bmatrix} \delta I + B-A & \mathbf{0} \newline \mathbf{0} & \delta I - B-A \newline \end{bmatrix}+ \begin{bmatrix} \mathbf 0 & -C \newline - C & \mathbf 0 \newline \end{bmatrix}\succ \mathbf 0$

$Y = \begin{bmatrix} \delta I + B-A & \mathbf{0} \newline \mathbf{0} & \delta I - B-A \newline \end{bmatrix}+ \begin{bmatrix} \mathbf 0 & -\vert C\vert \newline - \vert C\vert & \mathbf 0 \newline \end{bmatrix}\succ \mathbf 0$
where the block diagonal matrix is non-negative in the Perron-Frobenius sense. The goal is to prove $\big\Vert X\big\Vert_2=\rho\big(X\big) \leq \rho\big(Y\big) =\big\Vert Y\big\Vert_2$.

$D := \begin{bmatrix} I_n & \mathbf{0} \newline \mathbf{0} & -I_n\newline \end{bmatrix}$
noting $D=D^{-1}$ and $DYD = \vert Y\vert =\vert X\vert = \begin{bmatrix} \delta I + B-A & \vert C\vert \newline \vert C\vert & \delta I - B-A \newline \end{bmatrix}$

$\implies \rho\big(X\big)\leq \rho\big(\vert X\vert\big) =\rho\big(DYD\big)=\rho\big(Y\big)$
per $\rho(B) \leq \rho(|B|)$, where $\rho$ is the spectral radius

user8675309
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