Assume that $A \in \mathbb{R}^{N \times N}$ is a positive semi-definite matrix, both $\|A\|_1$ and $\|A^{-1}\|_1$ are uniformly bounded as $N \to \infty.$ Here $\| \cdot \|_1$ is the induced $L_1$ Norm. Prove that there exists a constant $\kappa > 0$ such that $$\|(I_N+A)^{-1}\|_1 \leq \kappa $$ as $N \to \infty.$ Else show a counterexample.

I attempted to demonstrate that the boundedness of $\|(I_N+A)^{-1}\|_1$ can be proven using the definition of $\| \cdot \|_1$ as follows

$$ \|(I_N + A)^{-1} \|_1 = \sup_{x \neq 0} \frac{ \|(I_N + A)^{-1} x\|_1 }{\|x\|_1} = \sup_{z \neq 0} \frac{ \|z\|_1 }{\|(I_N+A)z\|_1} \leq \sup_{z \neq 0} \frac{ \|z\|_1 }{|\|Az\|_1 - \|z\|_1|} = \sup_{z \neq 0} \frac{1}{ \Big| \frac{\|Az\|_1}{\|z\|_1} - 1 \Big| } $$

However, it seems possible that $\inf_{z} \Big|\frac{\|Az\|_1}{\|z\|_1} - 1 \Big|=0$, so I doubt this proof is wrong.

In addition, I also find I cannot use the theorem that $ \|(I_N - B)^{-1}\| \leq \frac{1}{1 - \|B\|} $ if $\|B\| < 1$.

But intuitively, $(I_N + A)^{-1}$ seems to be smaller than $A^{-1}$, and when we consider $L_2$ norm, it is easy to find that $\|(I_N + A)^{-1}\|_2 \leq 1$, so I guess the boundedness still holds when we consider $L_1$ norm.

Any guidance on how to tackle this problem? Thanks a ton!

Assume that $A \in \mathbb{R}^{N \times N}$ is a positive semi-definite matrix, both $\|A\|_1$ and $\|A^{-1}\|_1$ are uniformly bounded as $N \to \infty.$ Here $\| \cdot \|_1$ is the induced $L_1$ Norm. Prove that there exists a constant $\kappa > 0$ such that $$\|(I_N+A)^{-1}\|_1 \leq \kappa $$ as $N \to \infty.$ Else show a counterexample.

I attempted to demonstrate that the boundedness of $\|(I_N+A)^{-1}\|_1$ can be proven using the definition of $\| \cdot \|_1$ as follows

$$ \|(I_N + A)^{-1} \|_1 = \sup_{x \neq 0} \frac{ \|(I_N + A)^{-1} x\|_1 }{\|x\|_1} = \sup_{z \neq 0} \frac{ \|z\|_1 }{\|(I_N+A)z\|_1} \leq \sup_{z \neq 0} \frac{ \|z\|_1 }{|\|Az\|_1 - \|z\|_1|} = \sup_{z \neq 0} \frac{1}{ \Big| \frac{\|Az\|_1}{\|z\|_1} - 1 \Big| } $$

However, it seems possible that $\inf_{z} \Big|\frac{\|Az\|_1}{\|z\|_1} - 1 \Big|=0$, so I doubt this proof is wrong.

But intuitively, $(I_N + A)^{-1}$ seems to be smaller than $A^{-1}$, and when we consider $L_2$ norm, it is easy to find that $\|(I_N + A)^{-1}\|_2 \leq 1$, so I guess the boundedness still holds when we consider $L_1$ norm.

Any guidance on how to tackle this problem? Thanks a ton!

Assume that $A \in \mathbb{R}^{N \times N}$ is a positive semi-definite matrix, both $\|A\|_1$ and $\|A^{-1}\|_1$ are uniformly bounded as $N \to \infty.$ Here $\| \cdot \|_1$ is the induced $L_1$ Norm. Prove that there exists a constant $\kappa > 0$ such that $$\|(I_N+A)^{-1}\|_1 \leq \kappa $$ as $N \to \infty.$ Else show a counterexample.

I attempted to demonstrate that the boundedness of $\|(I_N+A)^{-1}\|_1$ can be proven using the definition of $\| \cdot \|_1$ as follows

$$ \|(I_N + A)^{-1} \|_1 = \sup_{x \neq 0} \frac{ \|(I_N + A)^{-1} x\|_1 }{\|x\|_1} = \sup_{z \neq 0} \frac{ \|z\|_1 }{\|(I_N+A)z\|_1} \leq \sup_{z \neq 0} \frac{ \|z\|_1 }{|\|Az\|_1 - \|z\|_1|} = \sup_{z \neq 0} \frac{1}{ \Big| \frac{\|Az\|_1}{\|z\|_1} - 1 \Big| } $$

However, it seems possible that $\inf_{z} \Big|\frac{\|Az\|_1}{\|z\|_1} - 1 \Big|$ could be equal to $0$, so I doubt this proof is wrong.

In addition, I also find I cannot use the fact that $ \|(I_N - B)^{-1}\| \leq \frac{1}{1 - \|B\|} $ if $\|B\| < 1$.

But intuitively, $(I_N + A)^{-1}$ seems to be smaller than $A^{-1}$, and when we consider $L_2$ norm, it is easy to find that $\|(I_N + A)^{-1}\|_2 \leq 1$, so I guess the boundedness still holds when we consider $L_1$ norm.

Any guidance on how to tackle this problem? Thanks a ton!

Assume that $A \in \mathbb{R}^{N \times N}$ is a positive semi-definite matrix, both $\|A\|_1$ and $\|A^{-1}\|_1$ are uniformly bounded as $N \to \infty.$ Here $\| \cdot \|_1$ is the induced $L_1$ Norm. Prove that there exists a constant $\kappa > 0$ such that $$\|(I_N+A)^{-1}\|_1 \leq \kappa $$ as $N \to \infty.$ Else show a counterexample.

I attempted to demonstrate that the boundedness of $\|(I_N+A)^{-1}\|_1$ can be proven using the definition of $\| \cdot \|_1$ as follows

$$ \|(I_N + A)^{-1} \|_1 = \sup_{x \neq 0} \frac{ \|(I_N + A)^{-1} x\|_1 }{\|x\|_1} = \sup_{z \neq 0} \frac{ \|z\|_1 }{\|(I_N+A)z\|_1} \leq \sup_{z \neq 0} \frac{ \|z\|_1 }{|\|Az\|_1 - \|z\|_1|} = \sup_{z \neq 0} \frac{1}{ \Big| \frac{\|Az\|_1}{\|z\|_1} - 1 \Big| } $$

However, it seems possible that $\inf_{z} \Big|\frac{\|Az\|_1}{\|z\|_1} - 1 \Big|=0$, so I doubt this proof is wrong.

In addition, I also find I cannot use the theorem that $ \|(I_N - B)^{-1}\| \leq \frac{1}{1 - \|B\|} $ if $\|B\| < 1$.

But intuitively, $(I_N + A)^{-1}$ seems to be smaller than $A^{-1}$, and when we consider $L_2$ norm, it is easy to find that $\|(I_N + A)^{-1}\|_2 \leq 1$, so I guess the boundedness still holds when we consider $L_1$ norm.

Any guidance on how to tackle this problem? Thanks a ton!