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Let $A$ be a non-symmetric positive definite matrix with the spectral radius $\rho(A) = \max_i|\lambda_i|$ (note $\lambda_i$ can be complex). By positive definite, I mean $x^\top A x > 0$ for all possible $x$.

Let $B$ be a symmetric PSD matrix with the same shape as $A$.

My question is: would the spectral radius $\rho(A+B)$ be larger than (or equal to) $\rho(A)$? This looks intuitive but I have difficulty in proving it, any hints?

For general case, I have found a counter-example. But what if I know some structures of A. Say A has the following form: $$A = \begin{bmatrix} A_{11} & A_{12} \\ -A_{12}^\top & A_{22}\end{bmatrix}$$ where $A_{11}$ and $A_{22}$ are symmetric positive definite.

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  • $\begingroup$ It's now solved with the following counter-example: $A = \begin{bmatrix} 1.01 & 4 & 4 \\ -2 & 1.01 & 4 \\ -2 & -2 & 1.01 \end{bmatrix}$ and B a all-one matrix. $\endgroup$
    – Batman
    Nov 1, 2020 at 22:54
  • $\begingroup$ Note that $\rho(A^n)$ approaches $\|A^n\|$ for large $n$ so you want to prove that $\|(A+B)^n v\|>\|A^n v\|$ for all $v$ and large $n$. But if $A$ rotates your vector more than $\pi/2$ then $\|Av+Bv\|<||Av\|$ $\endgroup$ Nov 2, 2020 at 20:03

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I didn't expect this to be true because to increase $\rho$ you need to rule out rotations, but your special structure doesn't rule them out (ie, skew-symmetric matrix + identity is a rotation)

It's easy to generate counter-examples programmatically. For instance the following two matrices form a counter-example:

$$ A= \left( \begin{array}{cccc} 2.01 & 0.01 & -1.00421 & -0.475834 \\ 0.01 & 2.01 & 1.84741 & -1.73169 \\ 1.00421 & -1.84741 & 2.01 & 0.01 \\ 0.475834 & 1.73169 & 0.01 & 2.01 \\ \end{array} \right) $$

$$ B= \left( \begin{array}{cccc} \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \end{array} \right) $$

Generated using following Mathematica code

rho[mat_] := Max[Abs[Eigenvalues[mat]]];
try[d_] := (
   B = ConstantArray[1, {d, d}]/2;
   ii = IdentityMatrix[d/2];
   AB = RandomReal[{-2, 2}, {d/2, d/2}];
   A = ArrayFlatten[{{2 ii + .01, AB}, {-AB\[Transpose], 2 ii + .01}}];
   result = rho[A + B] - rho[A];
   If[result < 0,
    Print[A];
    Print[B];
    ];
   result
   );
Table[try[4], {20}]
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    $\begingroup$ The interesting part is that for a 2x2 matrix no such example exists (you can show this mathematically). $\endgroup$
    – Alex Botev
    Nov 2, 2020 at 21:25
  • $\begingroup$ true, also using pure rotations for A doesn't work so that's not the whole story $\endgroup$ Nov 2, 2020 at 21:37

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