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Show that $||v||^2 = ||P_0v||^2 + ||v - P_0v||^2$ for orthogoonal projection

1

1

$begingroup$

I'm working on some practice problems from Noble & Daniel's Applied Linear Algebra (3rd), specifically here looking for help with question 5 from section 5.8 on pg. 232.

Suppose that $P_0$ is the orthogonal projection onto $V_0$. Show that for every v in V,

$||v||^2 = ||P_0v||^2 + ||v - P_0v||^2$

This whole section about orthogonal projections and bases has been a bit confusing for me. I'm not sure where to begin on this question. I know that $||v - P_0v||$ is orthogonal to each of the $v_i$ in v but and I thought that might be useful but I'm not sure how to leverage it.

matrices orthogonal-matrices projection-matrices gram-schmidt

share|cite|improve this question

asked Mar 24 at 19:47

cparmstrongcparmstrong

1084

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hey, question states when you use a projection if your vector does not lie completely in the projected space, norm of your vector depends on not only the projected part but also remaining part. Regarding the proof use $||x||^2=x^T x$
  $endgroup$
  – keoxkeox
  Mar 24 at 19:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean open up every term with it
  $endgroup$
  – keoxkeox
  Mar 24 at 19:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm guessing $V_0$ is a subspace of $V$? I'm further guessing that $V$ is a real inner product space, possibly finite-dimensional? Also, are you aware of Pythagoras's theorem for real inner product spaces? It would be good to get some further context here.
  $endgroup$
  – Theo Bendit
  Mar 24 at 19:59
  

  
  
  
  

add a comment | 

1

1

$begingroup$

I'm working on some practice problems from Noble & Daniel's Applied Linear Algebra (3rd), specifically here looking for help with question 5 from section 5.8 on pg. 232.

Suppose that $P_0$ is the orthogonal projection onto $V_0$. Show that for every v in V,

$||v||^2 = ||P_0v||^2 + ||v - P_0v||^2$

This whole section about orthogonal projections and bases has been a bit confusing for me. I'm not sure where to begin on this question. I know that $||v - P_0v||$ is orthogonal to each of the $v_i$ in v but and I thought that might be useful but I'm not sure how to leverage it.

matrices orthogonal-matrices projection-matrices gram-schmidt

share|cite|improve this question

asked Mar 24 at 19:47

cparmstrongcparmstrong

1084

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hey, question states when you use a projection if your vector does not lie completely in the projected space, norm of your vector depends on not only the projected part but also remaining part. Regarding the proof use $||x||^2=x^T x$
  $endgroup$
  – keoxkeox
  Mar 24 at 19:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean open up every term with it
  $endgroup$
  – keoxkeox
  Mar 24 at 19:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm guessing $V_0$ is a subspace of $V$? I'm further guessing that $V$ is a real inner product space, possibly finite-dimensional? Also, are you aware of Pythagoras's theorem for real inner product spaces? It would be good to get some further context here.
  $endgroup$
  – Theo Bendit
  Mar 24 at 19:59
  

  
  
  
  

add a comment | 

$begingroup$

I'm working on some practice problems from Noble & Daniel's Applied Linear Algebra (3rd), specifically here looking for help with question 5 from section 5.8 on pg. 232.

Suppose that $P_0$ is the orthogonal projection onto $V_0$. Show that for every v in V,

$||v||^2 = ||P_0v||^2 + ||v - P_0v||^2$

This whole section about orthogonal projections and bases has been a bit confusing for me. I'm not sure where to begin on this question. I know that $||v - P_0v||$ is orthogonal to each of the $v_i$ in v but and I thought that might be useful but I'm not sure how to leverage it.

matrices orthogonal-matrices projection-matrices gram-schmidt

share|cite|improve this question

asked Mar 24 at 19:47

cparmstrongcparmstrong

1084

$endgroup$

share|cite|improve this question

asked Mar 24 at 19:47

cparmstrongcparmstrong

1084

share|cite|improve this question

asked Mar 24 at 19:47

cparmstrongcparmstrong

1084

asked Mar 24 at 19:47

cparmstrongcparmstrong

1084

asked Mar 24 at 19:47

cparmstrongcparmstrong

1084

2 Answers
2

active

oldest

votes

1

$begingroup$

It's a calculation. Assume a real vector space, and note that $v=P_0v+w$ where $P_0vperp w$ so the Pythagorean theorem applies to show that

$|v|^2=|P_0v+w|^2=|P_0v|^2+|w|^2=|P_0v|^2+|v-P_0v|^2.$

share|cite|improve this answer

answered Mar 24 at 20:15

MatematletaMatematleta

12.2k21020

$endgroup$

add a comment | 

0

$begingroup$

In general, $||v+w||^2 = <v+w,v+w> = ldots = ||v||^2 + ||w||^2 + 2<v,w>$. Now if $v bot w$, you have that $<v,w>=0$, and you are done.

share|cite|improve this answer

answered Mar 24 at 20:19

dcolazindcolazin

58519

$endgroup$

add a comment | 

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1

$begingroup$

It's a calculation. Assume a real vector space, and note that $v=P_0v+w$ where $P_0vperp w$ so the Pythagorean theorem applies to show that

$|v|^2=|P_0v+w|^2=|P_0v|^2+|w|^2=|P_0v|^2+|v-P_0v|^2.$

share|cite|improve this answer

answered Mar 24 at 20:15

MatematletaMatematleta

12.2k21020

$endgroup$

add a comment | 

1

$begingroup$

It's a calculation. Assume a real vector space, and note that $v=P_0v+w$ where $P_0vperp w$ so the Pythagorean theorem applies to show that

$|v|^2=|P_0v+w|^2=|P_0v|^2+|w|^2=|P_0v|^2+|v-P_0v|^2.$

share|cite|improve this answer

answered Mar 24 at 20:15

MatematletaMatematleta

12.2k21020

$endgroup$

add a comment | 

$begingroup$

It's a calculation. Assume a real vector space, and note that $v=P_0v+w$ where $P_0vperp w$ so the Pythagorean theorem applies to show that

$|v|^2=|P_0v+w|^2=|P_0v|^2+|w|^2=|P_0v|^2+|v-P_0v|^2.$

share|cite|improve this answer

answered Mar 24 at 20:15

MatematletaMatematleta

12.2k21020

$endgroup$

share|cite|improve this answer

answered Mar 24 at 20:15

MatematletaMatematleta

12.2k21020

answered Mar 24 at 20:15

MatematletaMatematleta

12.2k21020

answered Mar 24 at 20:15

MatematletaMatematleta

12.2k21020

0

$begingroup$

In general, $||v+w||^2 = <v+w,v+w> = ldots = ||v||^2 + ||w||^2 + 2<v,w>$. Now if $v bot w$, you have that $<v,w>=0$, and you are done.

share|cite|improve this answer

answered Mar 24 at 20:19

dcolazindcolazin

58519

$endgroup$

add a comment | 

0

$begingroup$

In general, $||v+w||^2 = <v+w,v+w> = ldots = ||v||^2 + ||w||^2 + 2<v,w>$. Now if $v bot w$, you have that $<v,w>=0$, and you are done.

share|cite|improve this answer

answered Mar 24 at 20:19

dcolazindcolazin

58519

$endgroup$

add a comment | 

$begingroup$

In general, $||v+w||^2 = <v+w,v+w> = ldots = ||v||^2 + ||w||^2 + 2<v,w>$. Now if $v bot w$, you have that $<v,w>=0$, and you are done.

share|cite|improve this answer

answered Mar 24 at 20:19

dcolazindcolazin

58519

$endgroup$

share|cite|improve this answer

answered Mar 24 at 20:19

dcolazindcolazin

58519

answered Mar 24 at 20:19

dcolazindcolazin

58519

answered Mar 24 at 20:19

dcolazindcolazin

58519