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There are 2 equations of 1-D projection, are they the same?

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2

$begingroup$

I am learning 1-D projection, project a 2-D vector $x ∈ R^2$ onto a 1-D subspace U with basis $b, quad where quad b in R$.

this lecture gives this formula

$displaystyle fracmathbfbmathbfb^Tmathbfb x tag 1$

while this lecture gives this formula

$displaystyle fracx^T mathbfbmathbfb mathbfb tag 2$

are the 2 formulas same? how to prove that?

matrices vector-spaces linear-transformations projection

share|cite|improve this question

edited Jul 14 at 4:07

fu DL

asked Jul 13 at 22:21

fu DLfu DL

18177 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  They are the same since $bf b^Tx=x^Tbf b $ (dot product is commutative). Note that $bf b$ is also in $mathbb R^2$
  $endgroup$
  – Andrei
  Jul 13 at 22:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Andrei thanks man! plz mv or cp your comments to answer, I'll accept it.
  $endgroup$
  – fu DL
  Jul 13 at 23:06
  

  
  
  
  

add a comment | 

2

$begingroup$

I am learning 1-D projection, project a 2-D vector $x ∈ R^2$ onto a 1-D subspace U with basis $b, quad where quad b in R$.

this lecture gives this formula

$displaystyle fracmathbfbmathbfb^Tmathbfb x tag 1$

while this lecture gives this formula

$displaystyle fracx^T mathbfbmathbfb mathbfb tag 2$

are the 2 formulas same? how to prove that?

matrices vector-spaces linear-transformations projection

share|cite|improve this question

edited Jul 14 at 4:07

fu DL

asked Jul 13 at 22:21

fu DLfu DL

18177 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  They are the same since $bf b^Tx=x^Tbf b $ (dot product is commutative). Note that $bf b$ is also in $mathbb R^2$
  $endgroup$
  – Andrei
  Jul 13 at 22:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Andrei thanks man! plz mv or cp your comments to answer, I'll accept it.
  $endgroup$
  – fu DL
  Jul 13 at 23:06
  

  
  
  
  

add a comment | 

$begingroup$

I am learning 1-D projection, project a 2-D vector $x ∈ R^2$ onto a 1-D subspace U with basis $b, quad where quad b in R$.

this lecture gives this formula

$displaystyle fracmathbfbmathbfb^Tmathbfb x tag 1$

while this lecture gives this formula

$displaystyle fracx^T mathbfbmathbfb mathbfb tag 2$

are the 2 formulas same? how to prove that?

matrices vector-spaces linear-transformations projection

share|cite|improve this question

edited Jul 14 at 4:07

fu DL

asked Jul 13 at 22:21

fu DLfu DL

18177 bronze badges

$endgroup$

share|cite|improve this question

edited Jul 14 at 4:07

fu DL

asked Jul 13 at 22:21

fu DLfu DL

18177 bronze badges

share|cite|improve this question

edited Jul 14 at 4:07

fu DL

asked Jul 13 at 22:21

fu DLfu DL

18177 bronze badges

asked Jul 13 at 22:21

fu DLfu DL

18177 bronze badges

asked Jul 13 at 22:21

fu DLfu DL

18177 bronze badges

2 Answers
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2

$begingroup$

They are the same since $bf b^Tx=x^Tbf b$ (dot product is commutative). Note that $bf b$ is also in $mathbb R^2$

share|cite|improve this answer

answered Jul 13 at 23:11

AndreiAndrei

15.4k22 gold badges1414 silver badges3030 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

As Andrei already said, the dot product is commutative (1):
$$mathbfb^T mathbfx = mathbfx^T mathbfb$$
Furthermore the dot product yields a scalar, and we can write a scalar after a vector or in front of it (2):
$$c cdotmathbfx = mathbfxcdot c$$

Putting these two identities together, we can write the two formulas as follows:
$$mathbfbb^Tmathbfx = mathbfbleft( mathbfb^Tmathbfx right) = mathbfbleft( mathbfx^Tmathbfb right) = left( mathbfx^Tmathbfbright)mathbfb = mathbfx^Tmathbfbb$$
Thus, both formulas are equal.

Please note that one can not directly evaluate $mathbfbb$ as this expression is not defined.

share|cite|improve this answer

answered Jul 13 at 23:12

ulfgarulfgar

9633 bronze badges

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add a comment | 

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2

$begingroup$

They are the same since $bf b^Tx=x^Tbf b$ (dot product is commutative). Note that $bf b$ is also in $mathbb R^2$

share|cite|improve this answer

answered Jul 13 at 23:11

AndreiAndrei

15.4k22 gold badges1414 silver badges3030 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

They are the same since $bf b^Tx=x^Tbf b$ (dot product is commutative). Note that $bf b$ is also in $mathbb R^2$

share|cite|improve this answer

answered Jul 13 at 23:11

AndreiAndrei

15.4k22 gold badges1414 silver badges3030 bronze badges

$endgroup$

add a comment | 

$begingroup$

They are the same since $bf b^Tx=x^Tbf b$ (dot product is commutative). Note that $bf b$ is also in $mathbb R^2$

share|cite|improve this answer

answered Jul 13 at 23:11

AndreiAndrei

15.4k22 gold badges1414 silver badges3030 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 13 at 23:11

AndreiAndrei

15.4k22 gold badges1414 silver badges3030 bronze badges

answered Jul 13 at 23:11

AndreiAndrei

15.4k22 gold badges1414 silver badges3030 bronze badges

answered Jul 13 at 23:11

AndreiAndrei

15.4k22 gold badges1414 silver badges3030 bronze badges

3

$begingroup$

As Andrei already said, the dot product is commutative (1):
$$mathbfb^T mathbfx = mathbfx^T mathbfb$$
Furthermore the dot product yields a scalar, and we can write a scalar after a vector or in front of it (2):
$$c cdotmathbfx = mathbfxcdot c$$

Putting these two identities together, we can write the two formulas as follows:
$$mathbfbb^Tmathbfx = mathbfbleft( mathbfb^Tmathbfx right) = mathbfbleft( mathbfx^Tmathbfb right) = left( mathbfx^Tmathbfbright)mathbfb = mathbfx^Tmathbfbb$$
Thus, both formulas are equal.

Please note that one can not directly evaluate $mathbfbb$ as this expression is not defined.

share|cite|improve this answer

answered Jul 13 at 23:12

ulfgarulfgar

9633 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

As Andrei already said, the dot product is commutative (1):
$$mathbfb^T mathbfx = mathbfx^T mathbfb$$
Furthermore the dot product yields a scalar, and we can write a scalar after a vector or in front of it (2):
$$c cdotmathbfx = mathbfxcdot c$$

Putting these two identities together, we can write the two formulas as follows:
$$mathbfbb^Tmathbfx = mathbfbleft( mathbfb^Tmathbfx right) = mathbfbleft( mathbfx^Tmathbfb right) = left( mathbfx^Tmathbfbright)mathbfb = mathbfx^Tmathbfbb$$
Thus, both formulas are equal.

Please note that one can not directly evaluate $mathbfbb$ as this expression is not defined.

share|cite|improve this answer

answered Jul 13 at 23:12

ulfgarulfgar

9633 bronze badges

$endgroup$

add a comment | 

$begingroup$

As Andrei already said, the dot product is commutative (1):
$$mathbfb^T mathbfx = mathbfx^T mathbfb$$
Furthermore the dot product yields a scalar, and we can write a scalar after a vector or in front of it (2):
$$c cdotmathbfx = mathbfxcdot c$$

Putting these two identities together, we can write the two formulas as follows:
$$mathbfbb^Tmathbfx = mathbfbleft( mathbfb^Tmathbfx right) = mathbfbleft( mathbfx^Tmathbfb right) = left( mathbfx^Tmathbfbright)mathbfb = mathbfx^Tmathbfbb$$
Thus, both formulas are equal.

Please note that one can not directly evaluate $mathbfbb$ as this expression is not defined.

share|cite|improve this answer

answered Jul 13 at 23:12

ulfgarulfgar

9633 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 13 at 23:12

ulfgarulfgar

9633 bronze badges

answered Jul 13 at 23:12

ulfgarulfgar

9633 bronze badges

answered Jul 13 at 23:12

ulfgarulfgar

9633 bronze badges