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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!

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  • $\begingroup$ Can you do this in case $A$ is diagonal? Can the spectral theorem reduce the general case to the diagonal case? $\endgroup$
    – Gerald Edgar
    Commented Apr 8, 2023 at 11:01
  • $\begingroup$ @GeraldEdgar: The diagonal case is trivial ($\|(1+D)^{-1}\|_1\le 1$ when $D_{jj}\ge 0$), but the unitary matrix that diagonalizes $A$ could have large $1$ norm. $\endgroup$
    – Christian Remling
    Commented Apr 8, 2023 at 16:52
  • $\begingroup$ Yes, for example, consider a unitary matrix where the first row is $N^{-1/2}(1,1,...,1)$. So it seems that the eigenvalue decomposition theorem cannot be utilized either. $\endgroup$
    – Golden Silence
    Commented Apr 9, 2023 at 2:14

1 Answer 1

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Take $A=-I$. Then both $\Vert A\Vert_1,\Vert A^{-1}\Vert_1 $ are bounded although $I+A=0$ is not invertible.

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    $\begingroup$ Thanks, but I assume that $A$ is a positive semi-definite matrix so it cannot be $-I$ $\endgroup$
    – Golden Silence
    Commented Apr 10, 2023 at 13:14

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