Proof for norm property (Linear Algebra)

[ElectricSeal]

Hello,

I'm stuck with the following exercise:
V is a Vectorial Space with dot product (<.>) and norm (||.||).
For two orthogonal vectors v and w, prove that ||v-w|| >= ||v||.

I tried using dot properties like this:
||v|| = sqrt(<v, v>)
||v-w|| = sqrt(<v-w, v-w>) = sqrt(<v, v-w> - <w, v-w>) = sqrt(<v, v> - <v, w> - <w, v> - <w, w>)
<v, w> = <w, v> = 0 because v and w are orthogonal so:
||v-w|| = sqrt(<v, v> - <w, w>)
However, this is <= sqrt(<v, v>)

What am I doing wrong? Is this even the correct path to solve this?
Thanks!

 

[pka]

the following exercise:
V is a Vectorial Space with dot product (<.>) and norm (||.||).
For two orthogonal vectors v and w, prove that ||v-w|| >= ||v||.

You may not like this but the truth is it all depends upon your textbook.

Here is the proof that I give:
\(\displaystyle \begin{align*}\|v-w\|&\ge\left| \|v\| - \|w\|\right| \\ &\ge \|v\|-\|w\|\\&\ge \|v\|\end{align*}\)

 

[Dr.Peterson]

Hello,

I'm stuck with the following exercise:
V is a Vectorial Space with dot product (<.>) and norm (||.||).
For two orthogonal vectors v and w, prove that ||v-w|| >= ||v||.

I tried using dot properties like this:
||v|| = sqrt(<v, v>)
||v-w|| = sqrt(<v-w, v-w>) = sqrt(<v, v-w> - <w, v-w>) = sqrt(<v, v> - <v, w> - <w, v> - <w, w>)
<v, w> = <w, v> = 0 because v and w are orthogonal so:
||v-w|| = sqrt(<v, v> - <w, w>)
However, this is <= sqrt(<v, v>)

What am I doing wrong? Is this even the correct path to solve this?
Thanks!

Check the signs (or parenthesization) in your distribution: <v-w, v-w>) ≠ <v, v> - <v, w> - <w, v> - <w, w> !