# triangle inequality-ish property misunderstood, simple proof

Suppose $A$ is an invertible linear mapping in $\mathbb R^n$, and $B$ is a linear mapping from $\mathbb R^n$ to $\mathbb R^n$.

Suppose that $\|B-A\| < \frac{1}{\|A^{-1}\|}$. Show that $B$ is infact invertible as well.

The proof:

$|Bx| \geq |Ax| - |(B-A)x| \geq |Ax| - \|B-A\||x|$

Since $|x| = |A^{-1}Ax| \leq \|A^{-1}\| |Ax|$, or in other words $\frac{|x|}{\|A^{-1}\|} \leq |Ax|$, we overall have that $|Bx| \geq (\frac{1}{\|A^{-1}\|}-\|B-A\|)|x|= \alpha |x|$ where $\alpha > 0$.

So when $x \neq 0$, we have that $|Bx| > 0$, so $B$ is invertible.

My problem

This proof relies on two statements:

1) $\|A\| |x| \geq |Ax|$ which I've managed to prove on my own.

2) $|Bx| \geq |Ax| - |(B-A)x|$. This seems like a strange version of the triangle inequality I can't wrap my head around. Why is this true?

EDIT

$|(B-A)x| = |(A-B)x|$ from norm properties. If we switch them, the triangle becomes clear. Thanks all.

• Just add zero to $Ax$, then triangle inequality away – Aweygan Mar 8 '18 at 16:58

(1) is more-or-less the definition of $\|A\|$: $\|A\|$ is the least $c$ such that $|Ax|\le c|x|$ for all $x$.
(2) reads $|z|\ge|y|-|z-y|$ for $y=Ax$ and $z=Bx$. This is the triangle inequality applied to $z$ and $y-z$: $$|z|+|y-z|\ge|y|$$ so $$|z|\ge |y|-|y-z|=|y|-|z-y|.$$
• Ah, $|(B-A)x| = |(A-B)x|$...ofcourse. Now the triangle is clear. – Oria Gruber Mar 8 '18 at 17:00
Observe that $$|Ax|=|(B-A)x+Bx|\leq|(B-A)x|+|Bx|$$
• Why is $|Ax| = |2Bx - Ax|$? That's not obvious to me at all. – Oria Gruber Mar 8 '18 at 16:59