Orthogonal decomposition of vector v with respect to span W

0

I need to find the orthogonal decomposition of vector V
with respect to span W

$$
v=
begin{bmatrix}
2 \ -1 \ 5 \ 6 \
end{bmatrix}
$$
$$
w = span
begin{pmatrix}
begin{bmatrix}
1 \ 1 \ 1 \ 0
end{bmatrix},
begin{bmatrix}
1 \ 0 \ -1 \ 1
end{bmatrix}
end{pmatrix}
$$

I know that $ v = y + z $ where $ y in W $ and $ z in text{orthogonal complement of W} $

matrices vectors orthogonality matrix-decomposition

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asked Mar 6 '16 at 22:10

Max FiltenborgMax Filtenborg

2315

add a comment | 

0

I need to find the orthogonal decomposition of vector V
with respect to span W

$$
v=
begin{bmatrix}
2 \ -1 \ 5 \ 6 \
end{bmatrix}
$$
$$
w = span
begin{pmatrix}
begin{bmatrix}
1 \ 1 \ 1 \ 0
end{bmatrix},
begin{bmatrix}
1 \ 0 \ -1 \ 1
end{bmatrix}
end{pmatrix}
$$

I know that $ v = y + z $ where $ y in W $ and $ z in text{orthogonal complement of W} $

matrices vectors orthogonality matrix-decomposition

share|cite|improve this question

asked Mar 6 '16 at 22:10

Max FiltenborgMax Filtenborg

2315

add a comment | 

0

0

0

I need to find the orthogonal decomposition of vector V
with respect to span W

$$
v=
begin{bmatrix}
2 \ -1 \ 5 \ 6 \
end{bmatrix}
$$
$$
w = span
begin{pmatrix}
begin{bmatrix}
1 \ 1 \ 1 \ 0
end{bmatrix},
begin{bmatrix}
1 \ 0 \ -1 \ 1
end{bmatrix}
end{pmatrix}
$$

I know that $ v = y + z $ where $ y in W $ and $ z in text{orthogonal complement of W} $

matrices vectors orthogonality matrix-decomposition

share|cite|improve this question

asked Mar 6 '16 at 22:10

Max FiltenborgMax Filtenborg

2315

I need to find the orthogonal decomposition of vector V
with respect to span W

$$
v=
begin{bmatrix}
2 \ -1 \ 5 \ 6 \
end{bmatrix}
$$
$$
w = span
begin{pmatrix}
begin{bmatrix}
1 \ 1 \ 1 \ 0
end{bmatrix},
begin{bmatrix}
1 \ 0 \ -1 \ 1
end{bmatrix}
end{pmatrix}
$$

I know that $ v = y + z $ where $ y in W $ and $ z in text{orthogonal complement of W} $

matrices vectors orthogonality matrix-decomposition

matrices vectors orthogonality matrix-decomposition

share|cite|improve this question

asked Mar 6 '16 at 22:10

Max FiltenborgMax Filtenborg

2315

share|cite|improve this question

asked Mar 6 '16 at 22:10

Max FiltenborgMax Filtenborg

2315

share|cite|improve this question

share|cite|improve this question

asked Mar 6 '16 at 22:10

Max FiltenborgMax Filtenborg

2315

asked Mar 6 '16 at 22:10

Max FiltenborgMax Filtenborg

2315

asked Mar 6 '16 at 22:10

Max FiltenborgMax Filtenborg

2315

2315

add a comment | 

add a comment | 

1 Answer
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Project $v$ onto $W$:

$$begin{align}
text{proj}_W v &= frac{1}{3} cdot left( begin{bmatrix}2 & -1 & 5 & 6 end{bmatrix}begin{bmatrix}1 \ 1 \ 1 \ 0 end{bmatrix} right ) begin{bmatrix}1 \ 1 \ 1 \ 0 end{bmatrix} + frac{1}{3}left( begin{bmatrix}2 & -1 & 5 & 6 end{bmatrix}begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} right ) begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} \
&= begin{bmatrix}2 \ 2 \ 2 \ 0 end{bmatrix} + begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} \
&= begin{bmatrix}3 \ 2 \ 1 \ 1 end{bmatrix}
end{align}$$

Then, $z$ is given by $z = v - text{proj}_Wv$:
$$begin{align}
z &= v - text{proj}_Wv \
&= begin{bmatrix}2 \ -1 \ 5 \ 6 end{bmatrix} - begin{bmatrix}3 \ 2 \ 1 \ 1 end{bmatrix} \
&= begin{bmatrix}-1 \ -3 \ 4 \ 5 end{bmatrix}
end{align}$$

$z$ is indeed orthogonal to $W$ since its dot product with both vectors in the spanning set for $W$ is 0.

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answered Mar 6 '16 at 22:52

SplitInfinitySplitInfinity

78236

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Isn't z's dot product with the second vector in the span not 0? Thanks for the help though
  – Max Filtenborg
  Mar 7 '16 at 1:37
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MaxFiltenborg $(-1)(1) + (-3)(0) + (4)(-1) + (5)(1) = -1 + 0 -4 + 5 = 0$
  – SplitInfinity
  Mar 7 '16 at 2:50
  

  
  
  
  

add a comment | 

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1 Answer
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0

Project $v$ onto $W$:

$$begin{align}
text{proj}_W v &= frac{1}{3} cdot left( begin{bmatrix}2 & -1 & 5 & 6 end{bmatrix}begin{bmatrix}1 \ 1 \ 1 \ 0 end{bmatrix} right ) begin{bmatrix}1 \ 1 \ 1 \ 0 end{bmatrix} + frac{1}{3}left( begin{bmatrix}2 & -1 & 5 & 6 end{bmatrix}begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} right ) begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} \
&= begin{bmatrix}2 \ 2 \ 2 \ 0 end{bmatrix} + begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} \
&= begin{bmatrix}3 \ 2 \ 1 \ 1 end{bmatrix}
end{align}$$

Then, $z$ is given by $z = v - text{proj}_Wv$:
$$begin{align}
z &= v - text{proj}_Wv \
&= begin{bmatrix}2 \ -1 \ 5 \ 6 end{bmatrix} - begin{bmatrix}3 \ 2 \ 1 \ 1 end{bmatrix} \
&= begin{bmatrix}-1 \ -3 \ 4 \ 5 end{bmatrix}
end{align}$$

$z$ is indeed orthogonal to $W$ since its dot product with both vectors in the spanning set for $W$ is 0.

share|cite|improve this answer

answered Mar 6 '16 at 22:52

SplitInfinitySplitInfinity

78236

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Isn't z's dot product with the second vector in the span not 0? Thanks for the help though
  – Max Filtenborg
  Mar 7 '16 at 1:37
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MaxFiltenborg $(-1)(1) + (-3)(0) + (4)(-1) + (5)(1) = -1 + 0 -4 + 5 = 0$
  – SplitInfinity
  Mar 7 '16 at 2:50
  

  
  
  
  

add a comment | 

0

Project $v$ onto $W$:

$$begin{align}
text{proj}_W v &= frac{1}{3} cdot left( begin{bmatrix}2 & -1 & 5 & 6 end{bmatrix}begin{bmatrix}1 \ 1 \ 1 \ 0 end{bmatrix} right ) begin{bmatrix}1 \ 1 \ 1 \ 0 end{bmatrix} + frac{1}{3}left( begin{bmatrix}2 & -1 & 5 & 6 end{bmatrix}begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} right ) begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} \
&= begin{bmatrix}2 \ 2 \ 2 \ 0 end{bmatrix} + begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} \
&= begin{bmatrix}3 \ 2 \ 1 \ 1 end{bmatrix}
end{align}$$

Then, $z$ is given by $z = v - text{proj}_Wv$:
$$begin{align}
z &= v - text{proj}_Wv \
&= begin{bmatrix}2 \ -1 \ 5 \ 6 end{bmatrix} - begin{bmatrix}3 \ 2 \ 1 \ 1 end{bmatrix} \
&= begin{bmatrix}-1 \ -3 \ 4 \ 5 end{bmatrix}
end{align}$$

$z$ is indeed orthogonal to $W$ since its dot product with both vectors in the spanning set for $W$ is 0.

share|cite|improve this answer

answered Mar 6 '16 at 22:52

SplitInfinitySplitInfinity

78236

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Isn't z's dot product with the second vector in the span not 0? Thanks for the help though
  – Max Filtenborg
  Mar 7 '16 at 1:37
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MaxFiltenborg $(-1)(1) + (-3)(0) + (4)(-1) + (5)(1) = -1 + 0 -4 + 5 = 0$
  – SplitInfinity
  Mar 7 '16 at 2:50
  

  
  
  
  

add a comment | 

0

0

0

Project $v$ onto $W$:

$$begin{align}
text{proj}_W v &= frac{1}{3} cdot left( begin{bmatrix}2 & -1 & 5 & 6 end{bmatrix}begin{bmatrix}1 \ 1 \ 1 \ 0 end{bmatrix} right ) begin{bmatrix}1 \ 1 \ 1 \ 0 end{bmatrix} + frac{1}{3}left( begin{bmatrix}2 & -1 & 5 & 6 end{bmatrix}begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} right ) begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} \
&= begin{bmatrix}2 \ 2 \ 2 \ 0 end{bmatrix} + begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} \
&= begin{bmatrix}3 \ 2 \ 1 \ 1 end{bmatrix}
end{align}$$

Then, $z$ is given by $z = v - text{proj}_Wv$:
$$begin{align}
z &= v - text{proj}_Wv \
&= begin{bmatrix}2 \ -1 \ 5 \ 6 end{bmatrix} - begin{bmatrix}3 \ 2 \ 1 \ 1 end{bmatrix} \
&= begin{bmatrix}-1 \ -3 \ 4 \ 5 end{bmatrix}
end{align}$$

$z$ is indeed orthogonal to $W$ since its dot product with both vectors in the spanning set for $W$ is 0.

share|cite|improve this answer

answered Mar 6 '16 at 22:52

SplitInfinitySplitInfinity

78236

Project $v$ onto $W$:

$$begin{align}
text{proj}_W v &= frac{1}{3} cdot left( begin{bmatrix}2 & -1 & 5 & 6 end{bmatrix}begin{bmatrix}1 \ 1 \ 1 \ 0 end{bmatrix} right ) begin{bmatrix}1 \ 1 \ 1 \ 0 end{bmatrix} + frac{1}{3}left( begin{bmatrix}2 & -1 & 5 & 6 end{bmatrix}begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} right ) begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} \
&= begin{bmatrix}2 \ 2 \ 2 \ 0 end{bmatrix} + begin{bmatrix}1 \ 0 \ -1 \ 1 end{bmatrix} \
&= begin{bmatrix}3 \ 2 \ 1 \ 1 end{bmatrix}
end{align}$$

Then, $z$ is given by $z = v - text{proj}_Wv$:
$$begin{align}
z &= v - text{proj}_Wv \
&= begin{bmatrix}2 \ -1 \ 5 \ 6 end{bmatrix} - begin{bmatrix}3 \ 2 \ 1 \ 1 end{bmatrix} \
&= begin{bmatrix}-1 \ -3 \ 4 \ 5 end{bmatrix}
end{align}$$

$z$ is indeed orthogonal to $W$ since its dot product with both vectors in the spanning set for $W$ is 0.

share|cite|improve this answer

answered Mar 6 '16 at 22:52

SplitInfinitySplitInfinity

78236

share|cite|improve this answer

share|cite|improve this answer

answered Mar 6 '16 at 22:52

SplitInfinitySplitInfinity

78236

answered Mar 6 '16 at 22:52

SplitInfinitySplitInfinity

78236

answered Mar 6 '16 at 22:52

SplitInfinitySplitInfinity

78236

78236

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Isn't z's dot product with the second vector in the span not 0? Thanks for the help though
  – Max Filtenborg
  Mar 7 '16 at 1:37
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MaxFiltenborg $(-1)(1) + (-3)(0) + (4)(-1) + (5)(1) = -1 + 0 -4 + 5 = 0$
  – SplitInfinity
  Mar 7 '16 at 2:50
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Isn't z's dot product with the second vector in the span not 0? Thanks for the help though
  – Max Filtenborg
  Mar 7 '16 at 1:37
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MaxFiltenborg $(-1)(1) + (-3)(0) + (4)(-1) + (5)(1) = -1 + 0 -4 + 5 = 0$
  – SplitInfinity
  Mar 7 '16 at 2:50
  

  
  
  
  

Isn't z's dot product with the second vector in the span not 0? Thanks for the help though
– Max Filtenborg
Mar 7 '16 at 1:37

Isn't z's dot product with the second vector in the span not 0? Thanks for the help though
– Max Filtenborg
Mar 7 '16 at 1:37

@MaxFiltenborg $(-1)(1) + (-3)(0) + (4)(-1) + (5)(1) = -1 + 0 -4 + 5 = 0$
– SplitInfinity
Mar 7 '16 at 2:50

@MaxFiltenborg $(-1)(1) + (-3)(0) + (4)(-1) + (5)(1) = -1 + 0 -4 + 5 = 0$
– SplitInfinity
Mar 7 '16 at 2:50

add a comment | 

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