# Strict inequality of vector norms

Given a non orthogonal projection $p$ and non zero vector $x$. I am going to prove that $$\|Px\|<c\|x\|$$ for some $c<1$, where $\|\cdot\|$ is the usual Euclidean norm. I can only have the following: $$\|Px\|\le \|P\|\|x\|\le \|x\|$$ which is weaker than what I wanted. Any advice?

• Just a comment on the "weaker statement": If $P$ is a non-orthogonal projection, $\|P\|>1$. – Algebraic Pavel Jun 18 '15 at 6:46
• Do you mean $||P||<1||?$ – Jlamprong Jun 26 '15 at 4:50
• A submultiplicative norm of any nonzero projector is bounded from below by one because it is idempotent. – Algebraic Pavel Jun 26 '15 at 18:04
• @AlgebraicPavel: I understand. So, how would I bound $\|Px\|$? – Jlamprong Jun 28 '15 at 19:20

## 1 Answer

If you have a "shear-projection" in 2-d which is not orthogonal, e.g. has matrix representation $A = (0,0;\alpha,1)$ for some $\alpha \neq 0$, then with the vector $v = (0,1)^T$, you have $Av = v$ so your inequality can't hold. Are you leaving out some conditions to prohibit this case?