Exercise
https://texercises.com/exercise/scalar-product-and-norm/
Question
Solution
Short
Video
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.

Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Let V be a vector space and langle rangle is an inner product on V. Then ||v||:sqrtlangle v v rangle is a norm on V.

Solution:
Proof. We ase V langle rangle is a Hermitian space the case of Euclidean space is very similar. bf Triangle inequality: Let uvin V. ||u+v||^langle u+v u+v rangle langle uu rangle + langle uv rangle + langle vu rangle + langle vv rangle ||u||^+textRelangle uv rangle + ||v||^. * Note that forall zin mathbbC we have |textRez|leq |z|. Indeed if zx+iy xyin mathbbR then |textRez||x|leq sqrtx^+y^|z|. Applying this to our situation we have textRelangle uv rangle leq |textRelangle uv rangle| leq |langle uv rangle|. Going back to * we get ||u+v||^ leq ||u||^+|langle uv rangle|+||v||^ ||u||+||v||^. Since both ||u+v|| and ||u||+||v|| are non-negative the last inequality implies that ||u+v|| leq ||u||+||v||. The other properties positive definite symmetric of being a norm todo marked as exercise
Meta Information
\(\LaTeX\)-Code
Exercise:
Let V be a vector space and langle rangle is an inner product on V. Then ||v||:sqrtlangle v v rangle is a norm on V.

Solution:
Proof. We ase V langle rangle is a Hermitian space the case of Euclidean space is very similar. bf Triangle inequality: Let uvin V. ||u+v||^langle u+v u+v rangle langle uu rangle + langle uv rangle + langle vu rangle + langle vv rangle ||u||^+textRelangle uv rangle + ||v||^. * Note that forall zin mathbbC we have |textRez|leq |z|. Indeed if zx+iy xyin mathbbR then |textRez||x|leq sqrtx^+y^|z|. Applying this to our situation we have textRelangle uv rangle leq |textRelangle uv rangle| leq |langle uv rangle|. Going back to * we get ||u+v||^ leq ||u||^+|langle uv rangle|+||v||^ ||u||+||v||^. Since both ||u+v|| and ||u||+||v|| are non-negative the last inequality implies that ||u+v|| leq ||u||+||v||. The other properties positive definite symmetric of being a norm todo marked as exercise
Contained in these collections:

Attributes & Decorations
Tags
eth, fs23, linalg ii, norm, proof, scalar product
Content image
Difficulty
(3, default)
Points
0 (default)
Language
ENG (English)
Type
Proof
Creator rk
Decoration
File
Link