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Kate Juschenko
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by substituting $A$ with $A-B$ and $B$ by $-B$ your condition will be equivalent to the following: $||P_{A+B}x||\leq ||P_Ax||+||P_Bx||.$ the last one is not true in general (for non-selfadjointpositive matrices), for example it does not hold for $A=((0,1),(0,0))$, $B=((1,1),(1,1))$ and $x=\frac{1}{\sqrt{2}}(1,-1)$.

by substituting $A$ with $A-B$ and $B$ by $-B$ your condition will be equivalent to the following: $||P_{A+B}x||\leq ||P_Ax||+||P_Bx||.$ the last one is not true in general (for non-selfadjoint matrices), for example it does not hold for $A=((0,1),(0,0))$, $B=((1,1),(1,1))$ and $x=\frac{1}{\sqrt{2}}(1,-1)$.

by substituting $A$ with $A-B$ and $B$ by $-B$ your condition will be equivalent to the following: $||P_{A+B}x||\leq ||P_Ax||+||P_Bx||.$ the last one is not true in general (for non-positive matrices), for example it does not hold for $A=((0,1),(0,0))$, $B=((1,1),(1,1))$ and $x=\frac{1}{\sqrt{2}}(1,-1)$.

Source Link
Kate Juschenko
  • 4.7k
  • 22
  • 47

by substituting $A$ with $A-B$ and $B$ by $-B$ your condition will be equivalent to the following: $||P_{A+B}x||\leq ||P_Ax||+||P_Bx||.$ the last one is not true in general (for non-selfadjoint matrices), for example it does not hold for $A=((0,1),(0,0))$, $B=((1,1),(1,1))$ and $x=\frac{1}{\sqrt{2}}(1,-1)$.