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Symmetry of inner product

1

$begingroup$

Peter J. Cameron's "Notes on Linear Algebra" defines

An inner product on a real vector space $V$ is a function $b: V times V to R$ satisfying

b is bilinear (that is, b is linear in the first variable when the second is kept
constant and vice versa);

b is positive definite, that is, $b(v,v) geq 0$ for all $v in V$, and $b(v,v) = 0$ if and only if $v = 0$.

Using the notation $x cdot y = b(x,y)$, it then expands (without explanation) $(v+xw) cdot (v+xw)$ to $x^2 (w cdot w) + 2x (v cdot w) + v cdot v$.

However when I try to verify this, I get to $x^2(w cdot w) + x(v cdot w) + x(w cdot v) + v cdot v$, which clearly equals the desired expression if b is symmetric. But I don't believe we've established anywhere that it is, so I guess I'm either misunderstanding a definition, or I'm missing a trick when performing the expansion.

vector-spaces inner-product-space bilinear-form

share|cite|improve this question

asked Apr 1 at 0:11

cb7cb7

1476

$endgroup$

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  1
  

  
  
  
  
  
  

  
  $begingroup$
  Symmetry is generally part of the definition...see, e.g., this or this .
  $endgroup$
  – lulu
  Apr 1 at 0:21
  
  

  
  
  
  


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  I believe you are correct and that this is an oversight by the author. I found the book at maths.qmul.ac.uk/~pjc/notes/linalg.pdf and if you are referring to the proof of theorem 6.1 then I agree.
  $endgroup$
  – John Douma
  Apr 1 at 0:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sure Cameron meant to say that $b$ should be symmetric.
  $endgroup$
  – Rob Arthan
  Apr 1 at 0:41
  

  
  
  
  

add a comment | 

1

$begingroup$

Peter J. Cameron's "Notes on Linear Algebra" defines

An inner product on a real vector space $V$ is a function $b: V times V to R$ satisfying

b is bilinear (that is, b is linear in the first variable when the second is kept
constant and vice versa);

b is positive definite, that is, $b(v,v) geq 0$ for all $v in V$, and $b(v,v) = 0$ if and only if $v = 0$.

Using the notation $x cdot y = b(x,y)$, it then expands (without explanation) $(v+xw) cdot (v+xw)$ to $x^2 (w cdot w) + 2x (v cdot w) + v cdot v$.

However when I try to verify this, I get to $x^2(w cdot w) + x(v cdot w) + x(w cdot v) + v cdot v$, which clearly equals the desired expression if b is symmetric. But I don't believe we've established anywhere that it is, so I guess I'm either misunderstanding a definition, or I'm missing a trick when performing the expansion.

vector-spaces inner-product-space bilinear-form

share|cite|improve this question

asked Apr 1 at 0:11

cb7cb7

1476

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Symmetry is generally part of the definition...see, e.g., this or this .
  $endgroup$
  – lulu
  Apr 1 at 0:21
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I believe you are correct and that this is an oversight by the author. I found the book at maths.qmul.ac.uk/~pjc/notes/linalg.pdf and if you are referring to the proof of theorem 6.1 then I agree.
  $endgroup$
  – John Douma
  Apr 1 at 0:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sure Cameron meant to say that $b$ should be symmetric.
  $endgroup$
  – Rob Arthan
  Apr 1 at 0:41
  

  
  
  
  

add a comment | 

$begingroup$

Peter J. Cameron's "Notes on Linear Algebra" defines

An inner product on a real vector space $V$ is a function $b: V times V to R$ satisfying

b is bilinear (that is, b is linear in the first variable when the second is kept
constant and vice versa);

b is positive definite, that is, $b(v,v) geq 0$ for all $v in V$, and $b(v,v) = 0$ if and only if $v = 0$.

Using the notation $x cdot y = b(x,y)$, it then expands (without explanation) $(v+xw) cdot (v+xw)$ to $x^2 (w cdot w) + 2x (v cdot w) + v cdot v$.

However when I try to verify this, I get to $x^2(w cdot w) + x(v cdot w) + x(w cdot v) + v cdot v$, which clearly equals the desired expression if b is symmetric. But I don't believe we've established anywhere that it is, so I guess I'm either misunderstanding a definition, or I'm missing a trick when performing the expansion.

vector-spaces inner-product-space bilinear-form

share|cite|improve this question

asked Apr 1 at 0:11

cb7cb7

1476

$endgroup$

share|cite|improve this question

asked Apr 1 at 0:11

cb7cb7

1476

share|cite|improve this question

asked Apr 1 at 0:11

cb7cb7

1476

asked Apr 1 at 0:11

cb7cb7

1476

asked Apr 1 at 0:11

cb7cb7

1476