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I would like to find the monotonically increasing range of a function related to matrix norms: $$ f(x)=\left\|I-e^{-Ax}\right\| $$ Where $I$ is the identity matrix, $A\in\mathbb{R}^{n\times n}$ is a real square matrix, $x\in\mathbb{R}_{\ge 0}$ is a nonnegative real number, and $\left\|\cdot\right\|$ represents the spectral norm of a matrix.

I have verified in MATLAB that, when choosing different values for the fixed matrix $A$, if $x$ is not big, for example $x\in\left[0, 1\right]$, the larger $x$ is, the larger the matrix norm will be. So I guess $f(x)$ is a monotonically increasing function in $\left[0, a\right]$, where $a\in\mathbb{R}_{>0}$ is a small real number. However, I cannot verify it through a strict mathematical proof. So the question is how to find the largest $a$ that $f(x)$ keeps monotonically increasing in $\left[0, a\right]$.

Anyone can help? Many Thanks!

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This is not a full answer, just a proof of a special case.

Assume $A$ is diagonal, $A=\text{diag}(a_1,\dots,a_n)$. Then $$I-e^{-Ax} = I-e^{\text{diag}(-a_1x,\dots,-a_nx)} = I-\text{diag}(e^{-a_1x},\dots,e^{-a_n x}) = \text{diag}(1-e^{-a_1x},\dots,1-e^{-a_n x}).$$

Recall that the spectral norm of a matrix is its largest singular value. Now, for any real diagonal matrix, it holds that its largest singular value is the largest absolute value of its eigenvalues, which are just the diagonal entries. Thus, $$||I-e^{-Ax}|| = \max\lbrace |1-e^{-a_1x}|,\dots,|1-e^{-a_nx}|\rbrace.$$ $e^{-a_i x}$ either increases monotonically to infinity or decreases monotonically to $0$, depending on whether $a_i$ is positive or negative. Since it is $1$ for $x=0$, we see that $|1-e^{-a_ix}|$ thus also grows monotonically in $x$, for any $i$. Then so does their maximum, which finishes the proof.

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