Spectral radius of products of symmetric positive definite matrices

eigenvalues-eigenvectorslinear algebramatricespositive definitespectral-radius

I have matrices $A, B, C, D \in \mathbb R^{n\times n}$ which are all symmetric positive (semi-)definite.

If I have that

\begin{align}
\rho(AB) &< \gamma^2, \\
\rho(CD) &< \gamma^2, \quad \text{and}\\
\rho(BD) &< \gamma^2,
\end{align}

can I conclude that
$$
\rho(AC) < \gamma^2 \tag{?}
$$

Here, $\rho$ denotes the spectral radius, i.e., the largest in modulo eigenvalue, and $\gamma \in \mathbb R$ is a constant.

Best Answer

No. If $n=1$ and $X \ge 0$, then $ \rho(X)=X.$

Now let $A=1, B=3, C=5, D=1/2$ and $ \gamma =2.$

Then we have

$\rho(AB) < \gamma^2,$ $\rho(CD) <\gamma^2$ and $\rho(BD) < \gamma^2$,

but $\rho(AC) > \gamma^2 .$

Related Solutions

[Math] Spectral radius and positive definite of matrices

Presumably $A$ and $B$ are real. Your requirement is possible if and only if $1$ and $-1$ are not eigenvalues of $A$.

Suppose $1$ (or $-1$) is an eigenvalue of $A$ and $x$ is a corresponding eigenvector. Then your requirement implies that $B\succ A^TBA$, which is impossible because $x^TBx=x^TA^TBAx$.

Suppose $\pm1$ are not eigenvalues of $A$. By a change of basis ($B$ is a quadratic form and $A$ is a linear transformation; so they undergo the same change of basis), we may assume that $A$ is already in its real Jordan form. Let $B$ be a block diagonal matrix with block sizes conforming to $A$'s. We can solve the problem blockwise.

For each Jordan block $A_0$ of $A$ with eigenvalue modulus greater than $1$, set the corresponding diagonal subblock of $B$ to the negative definite matrix $B_0$, where $-B_0$ is the positive definite solution to the Lyapunov equation $A_0^{-T}(-B_0)A_0^{-1} - (-B_0) - I = 0$. For each Jordan block $A_0$ with eigenvalue modulus $\le1$ (the modulus can be $1$, it's just that the eigenvalues are a conjugate pair of nonreal eigenvalues on the unit circle), set the corresponding diagonal subblock of $B_0$ to a positive semidefinite solution to $A_0^TB_0A_0 - B_0 = -I$. Since $\rho(A)>1$, $B$ always possesses a negative definite diagonal subblock. Therefore $B$ is not positive semidefinite.

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

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}]

Best Answer

Related Solutions

[Math] Spectral radius and positive definite of matrices

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

Related Question