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Although it's simply stated, this is neither a homework problem or trivial (I think, but I'd be happy to be proven wrong :) ).

Let $A,B$ be matrices and $x$ be a vector. Is it true that $$ \|P_{A+B} x\| \geq \|P_A x\| - \|P_B x\|, $$ where $P_A$ is the projection onto the range space of $A$? (or is it true if you square the norms?)

I'm having difficulty even figuring out how to attack this: every attempt I've made falters on the facts that the range space of $A + B$ is not simply related to those of $A$ and $B$ and that the projection is nonlinear. Random instances haven't yet provided counterexamples to the inequality.

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If I understand your notation, the matrices are square, or might as well be...in that case, if all three of $A,B,A+B$ are full rank, then you are just comparing versions of the norm of $x$ itself, and anything you can thing of is trivially true. In the answers below, not full rank. –  Will Jagy May 2 '11 at 19:59

2 Answers 2

up vote 4 down vote accepted

Let $$A=\pmatrix{1&0\cr 0&0}, B=\pmatrix{0&0\cr 1&0}, x=(x_1,x_2).$$ Then $P_Ax=(x_1,0)$, $P_Bx=(0,x_2)$, $P_{A+B}x=((x_1+x_2)/2,(x_1+x_2)/2)$. Thus you are asking if $$|(x_1+x_2)/\sqrt{2}|\ge |x_1|-|x_2|.$$ Clearly, this is false in general.

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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-positive matrices), for example it does not hold for $A=((0,1),(0,0))$, $B=((1,1),(1,1))$ and $x=\frac{1}{\sqrt{2}}(1,-1)$.

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