# Would adding a symmetric positive semi-definite matrix to a non-symmetric positive definite matrix increase the spectral radius?

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.

• 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. Nov 1, 2020 at 22:54
• 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 $\|Av+Bv\|<||Av\|$ Nov 2, 2020 at 20:03

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}]

• The interesting part is that for a 2x2 matrix no such example exists (you can show this mathematically). Nov 2, 2020 at 21:25
• true, also using pure rotations for A doesn't work so that's not the whole story Nov 2, 2020 at 21:37