Norm inequality for matrix and vector $\|ABx\|_p \leq \|A\|_p\|Bx\|_p$

matricesnormed-spaces

When I read a lecture note, it says

$$\|ABx\|_p \leq \|A\|_p\|Bx\|_p$$

$A$, $B$ are matrices with appropriate dimension.
$x$ is a vector

My questions are

Why does this inequality hold? This includes matrix norm and vector norm.
Under what conditions, the following holds? $$\|ABx\|_p = \|A\|_p\|Bx\|_p=\|Bx\|_p$$

I believe $A$ has to be an orthogonal matrix, i.e., $A^TA = AA^T = I$ at least so $\|A\|_p=1$.

Thanks!

Best Answer

For compatible norms, including induced norms (as $\|\cdot\|_p)$ we have $\|Ay\|\leq \|A\|\ \|y\|$ in your case $y=Bx$. This inequality follows from the definition of $\|A\|=\sup\left\lbrace \frac{\|Ay\|}{\|y\|}:y\neq 0\right\rbrace$. For the equality, you need that the vector $y$ maximizes the quantity $\frac{\|Ay\|}{\|y\|}$, which is not always an easy task. Check this question for more information Norm equivalence of a vector norm and its induced matrix norm using compactness argument

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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

Best Answer

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