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Suppose $A,B,C,D$ are matrices with operator norm at most 1. Then is it true that $\|A(B+C) +D(B-C)\|/2$ is at most $1$ (where $\|\cdot \|$ is the operator norm of a matrix, where by operator norm, I mean largest eigenvalue of the matrix)? I have tried many examples and it seems true, but I am not able to prove something in general.

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  • $\begingroup$ The largest eigenvalue of the matrix is not a norm. $\endgroup$ – copper.hat Aug 27 '18 at 14:59
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    $\begingroup$ The induced two norm of a matrix $M$ (also the largest singular value) is the square root of the largest eigenvalue of $M^TM$. Perhaps this is the source of your confusion? $\endgroup$ – copper.hat Aug 27 '18 at 15:01
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$A=I$, $B=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$, $C=D=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. All of these have an induced two norm of one.

Then $A(B+C)+D(B-C) = \begin{bmatrix} 2 & 0 \\ 2 & 0 \end{bmatrix}$ which has an induced norm of $2 \sqrt{2} > 2$.

octave:1> A = [ 1 0 ; 0 1 ] ;
octave:2> B = [ 1 0 ; 0 -1 ] ;
octave:3> C = D = [ 0 -1 ; 1 0] ;
octave:4> 
octave:4> norm(A)
ans =  1
octave:5> norm(B)
ans =  1
octave:6> norm(C)
ans =  1
octave:7> norm(D)
ans =  1
octave:8> 
octave:8> S = A*(B+C)+D*(B-C)
S =

   2   0
   2   0

octave:9> norm(S)
ans =  2.8284
octave:10> 
octave:10> sqrt(eig(S'*S))
ans =

   0.00000
   2.82843
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  • $\begingroup$ I am sorry but why is the spectral norm of the final matrix $2\sqrt{2}$ isn't it just 2 (the eigen values are $\{2,0\}$)? $\endgroup$ – wwjohnsmith Aug 27 '18 at 14:42
  • $\begingroup$ Just compute the norm explicitly, it is not difficult. Try applying the matrix to the vector $(1,0)^T$. $\endgroup$ – copper.hat Aug 27 '18 at 14:47
  • $\begingroup$ Exactly my point, if you try with the vector $(1,0)^T$, the eigenvalue is $2$ and it is a rank 1 matrix, so this is the only non-zero eigen value! $\endgroup$ – wwjohnsmith Aug 27 '18 at 14:49
  • $\begingroup$ What have the eigenvalues got to do with it? Just compute the induced two norm. If you apply the matrix to the vector $(1,0)^T$ you will get the vector $(2,2)^T$ which has norm $2 \sqrt{2}$. $\endgroup$ – copper.hat Aug 27 '18 at 14:54
  • $\begingroup$ The norm is not a function of the eigenvalues of the matrix. $\endgroup$ – copper.hat Aug 27 '18 at 14:54

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