[Math] Matrix norm inequality proof: inverse of two p.s.d matrices sum

linear algebramatricesnormed-spaces

I wonder if the following matrix norm inequality holds:
Let $A$ and $B$ are both strictly symmetric positive definite matrix
$\|(A+B)^{-1}\|_2\leq \|A^{-1}\|_2$ ?

Thanks in advance.

Best Answer

If $\|\cdot\|_2$ denotes the maximum singular value, then it can be proved as follows.

The inequality in the original problem is equivalent to showing $$ \sigma_{\max}((A+B)^{-1})\le \sigma_{\max}(B^{-1}) \Leftrightarrow \frac{1}{\sigma_{\min}(A+B)}\le \frac{1}{\sigma_{\min}(B)} \Leftrightarrow \sigma_{\min}(B)\le\sigma_{\min}(A+B) $$ Since $A$ is PD, the last inequality above is easy to prove (consider the eigenvector for the minimum eigenvalue of $A+B$, then $\sigma_{\min}(A+B)=x^T(A+B)x\ge x^TBx\ge \sigma_{\min}(B))$.

Related Solutions

[Math] Norm inequality for sum and difference of positive-definite matrices

This answer only answers the first question. It does so in a way that may be unnecessarily complicated for your taste.

If $A$ and $B$ are Hermitian matrices, let $A\leq B$ mean that $B-A$ is positive semidefinite. A Hermitian matrix is positive semidefinite if and only if all of its eigenvalues are nonnegative, and the spectral norm of a Hermitian matrix is the maximum of the absolute values of its eigenvalues.

Then $$-\|X_2\|I\leq -X_2\leq X_1-X_2\leq X_1\leq \|X_1\|I.$$ This implies that all of the eigenvalues of $X_1-X_2$ lie in the interval $[-\|X_2\|,\|X_1\|]$, which implies that $\|X_1-X_2\|\leq \max\{\|X_1\|,\|X_2\|\}$.

On the other hand, $0\leq X_1\leq X_1+X_2$ implies that $\|X_1\|\leq\|X_1+X_2\|$, and similarly $\|X_2\|\leq \|X_1+X_2\|$, so $\max\{\|X_1\|,\|X_2\|\}\leq \|X_1+X_2\|$.

[Math] Counterexamples to the Matrix norm AM-GM inequality

Consider $$A = \begin{pmatrix} 46 & 7 & 8 \\ 7 & 36 & 16 \\ 8 & 16 & 30\end{pmatrix}\qquad \text{ and }\qquad B=\begin{pmatrix} 34 & 7 & 12 \\ 7 & 48 & 10 \\ 12 & 10 & 44\end{pmatrix}.$$ Then $$\frac{1}{2}\|A+B\|_1 =\frac{1}{2}\|A+B\|_{\infty}=62$$ however $$ \|A^{1/2}B^{1/2}\|_1\approx 62.1527 \qquad \text{ and }\qquad \|A^{1/2}B^{1/2}\|_{\infty}\approx 62.1037.$$

Here is the Matlab code to generate such counter examples:

n = 3;  % choose the dimension of the matrices
stop = 0;  
while stop == 0  
  % Generates a pair of symmetric positive definite matrices
  A = floor(10*rand(n));  
  A=A+A';  
  A = A+10*n*eye(n);  
  B = floor(10*rand(n));  
  B = B+B';  
  B = B+10*n*eye(n);  
  % Computes their square root
  AS = A^(1/2);  
  BS=B^(1/2);
  % Check if the AM-GM is not satisfied
  if norm(AS*BS,1)>norm(A+B,1)/2 && norm(AS*BS,inf)>norm(A+B,inf)/2  
    stop =1;  
  end  
end

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[Math] Norm inequality for sum and difference of positive-definite matrices

[Math] Counterexamples to the Matrix norm AM-GM inequality

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