Trying to understand why the largest singular value of $M_n^{-1}$ is $1.4286$ no matter what $n$ I choose

Question

Let $n \ge 3$ be an odd number. Let $I_n$ be the $n$ dimensional identity matrix and let $A_n$ be the $n\times n$ matrix where every element is zero except the central element which is, say, $0.3$. For example, for $n=3$ and $n=5$ we have $$ \begin{align} I_3 &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}, \\ A_3 &= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0.3 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}, \\ I_5 &= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}, \\ A_5 &= \begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0.3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}. \end{align} $$ Then define the block matrix $M_n$ by $$ \begin{align} M_n &= \begin{bmatrix} I_n & A_n & A_n \\ A_n & I_n & A_n \\ A_n & A_n & I_n \end{bmatrix}. \end{align} $$ No matter what value of $n$ I choose I find that the largest singular value of $M_n^{-1}$ is $1.4286$. How can this be explained?

Here is some Matlab code that demonstrates this phenomenon.

I_3 = [1,0,0;0,1,0;0,0,1];
A_3 = [0,0,0;0,0.3,0;0,0,0];
M_3 = [I_3, A_3, A_3; A_3, I_3, A_3; A_3, A_3, I_3];
M_3_inv = inv(M_3);

I_5 = [1,0,0,0,0;0,1,0,0,0;0,0,1,0,0;0,0,0,1,0;0,0,0,0,1];
A_5 = [0,0,0,0,0;0,0,0,0,0;0,0,0.3,0,0;0,0,0,0,0;0,0,0,0,0;];
M_5 = [I_5, A_5, A_5; A_5, I_5, A_5; A_5, A_5, I_5];
M_5_inv = inv(M_5);

I_7 = [...
    1,0,0,0,0,0,0;...
    0,1,0,0,0,0,0;...
    0,0,1,0,0,0,0;...
    0,0,0,1,0,0,0;...
    0,0,0,0,1,0,0;...
    0,0,0,0,0,1,0;...
    0,0,0,0,0,0,1];
A_7 = [...
    0,0,0,0,0,0,0;...
    0,0,0,0,0,0,0;...
    0,0,0,0,0,0,0;...
    0,0,0,0.3,0,0,0;...
    0,0,0,0,0,0,0;...
    0,0,0,0,0,0,0;...
    0,0,0,0,0,0,0];
M_7 = [I_7, A_7, A_7; A_7, I_7, A_7; A_7, A_7, I_7];
M_7_inv = inv(M_7);

[U_3,S_3,V_3] = svd(M_3_inv);
[U_5,S_5,V_5] = svd(M_5_inv);
[U_7,S_7,V_7] = svd(M_7_inv);

max(diag(S_3))
max(diag(S_5))
max(diag(S_7))

Answer 1

First, note that $M_n$ is symmetric, which means that $M_n^{-1}$ is symmetric, which means that its maximal singular value is equal to its spectral radius (the maximal absolute value among its eigenvalues).

Let $\otimes$ denote the Kronecker product. We can write your matrix as $$ M_n = J \otimes A_n + I_{3n}, $$ where $J$ is the $3 \times 3$ matrix $$ J = \pmatrix{0&1&1\\1&0&1\\1&1&0}. $$ Verify that the eigenvalues of $J$ are $-1,-1,2$. On the other hand, $A_n$ has eigenvalue $0.3$ with multiplicity $1$ and $0$ with multiplicity $n-1$. Correspondingly, the properties of the Kronecker product allow us to find that $J \otimes A_n$ has eigenvalues $2 \times 0.3 = 0.6$ with multiplicity $1$, $-0.3$ with multiplicity $2$, and $0$ with multiplicity $3(n-1)$.

Let $P_n = J \otimes A_n$ so that $M_n^{-1} = (I_{3n} + P_n)^{-1}$. The eigenvalues of $(I_{3n} + P_n)^{-1}$ are equal to $(1 + \lambda)^{-1}$ for every eigenvalue $\lambda$ of $P_n$. With that, we see that the eigenvalues of $M_n^{-1}$ are equal to $$ 1/(1 + 0.6), \quad 1/(1 - 0.3), \quad 1/(1 - 0). $$ Among these, the largest absolute value is $|1/(0-0.3)| = 1.4286$. So, this is the maximal singular value of $M_n.$

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In addition to facilitating our analysis, the Kronecker product also gives us a nice way to express this matrix in Matlab:

A_7 = zeros(7)
A_7(4,4) = 0.3
J = ~eye(3)
M = kron(J,A_7) + eye(3*7)