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Vi let P be a fuction from R^n to R^n that fullfills the following two requirements: 1. P is linear, so for all x,y, includes R^n and all a, b includes R applies P(ax + by) = aP(x) + bP(y) 2. P is a projection, which means P•P = P Show that P is NOT equal to | så P is not an injective.

Vi let P be a fuction from R^n to R^n that fullfills the following two requirements: 1. P is linear, so for all x,y, includes R^n and all a, b includes R applies P(ax + by) = aP(x) + bP(y) 2. P is a projection, which means P•P = P Show that P is NOT equal to | så P is not an injective.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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3. send handwritten solution asap i added orignal pic also 
Vi let P be a fuction from R^n to R^n that fullfills the following two requirements:
1. P is linear, so for all x,y, includes R^n and all a, b includes R applies P(ax + by) = aP(x) + bP(y)

2. P is a projection, which means P•P = P

Show that P is NOT equal to | så P is not an injective.

 

3. Funktioner. Vi låter P vara en funktion från R" till R" som uppfyller följande två krav: 1. P är linjär, det vill säga för alla a, y E R" och alla a, b e R gäller P(ax +by) = aP(x)+bP(y). 2. P är en projektion, det vill säga Po P= P. Visa att om P+i så är P inte injektiv.
Transcribed Image Text:3. Funktioner. Vi låter P vara en funktion från R" till R" som uppfyller följande två krav: 1. P är linjär, det vill säga för alla a, y E R" och alla a, b e R gäller P(ax +by) = aP(x)+bP(y). 2. P är en projektion, det vill säga Po P= P. Visa att om P+i så är P inte injektiv.
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