Mixing
Distance of a vector from a subspace - Linear Algebra
I want to calculate the distance of the vector $x=(1,1,1,1)$ to the subspace ${(1,0,2,0) , (0,1,0,2)}$
I have solved this in 2 ways that I know of but the thing is, the results are different.
For instance when I use $||x-Pr(x)||$ I get $sqrt{2}$, but when I calculate it using the gram determinant (more info here: Distance between a point and a m-dimensional space in n-dimensional space ($m<n$) ) I get $sqrt{frac{10}{25}}$ which is weird because both ways should be equivalent.
So my question is, what am I missing here? Which one of these result was the correct one?
Thank you.
Edit: Solution using $||x-Pr(x)||$ :
Let $v_1$ be $(1,0,2,0)$ and $v_2$ be $(0,1,0,2)$
Both vectors $v_1$ and $v_2$ are orthogonal meaning the inner product of them is $0$. Now we need to make them orthonormal. After doing so we get $e_1=v_1/sqrt{3}$ and $e_2=v_2/sqrt{3}$
Now we calculate $Pr(x)$.
$$Pr(x)= (x,e_1)e_1+(x,e_2)e_2 =ldots= (1,1,2,2)$$
Therefore the distance is $d(x,U)= ||x-Pr(x)||=||(1,1,1,1)-(1,1,2,2)||= ||(0,0,-1,-1)||= sqrt{2}$
linear-algebra
edited Apr 13 '17 at 12:21
Community♦
1
asked Jun 28 '14 at 14:44
Kyle
1314
add a comment |
I want to calculate the distance of the vector $x=(1,1,1,1)$ to the subspace ${(1,0,2,0) , (0,1,0,2)}$
I have solved this in 2 ways that I know of but the thing is, the results are different.
For instance when I use $||x-Pr(x)||$ I get $sqrt{2}$, but when I calculate it using the gram determinant (more info here: Distance between a point and a m-dimensional space in n-dimensional space ($m<n$) ) I get $sqrt{frac{10}{25}}$ which is weird because both ways should be equivalent.
So my question is, what am I missing here? Which one of these result was the correct one?
Thank you.
Edit: Solution using $||x-Pr(x)||$ :
Let $v_1$ be $(1,0,2,0)$ and $v_2$ be $(0,1,0,2)$
Both vectors $v_1$ and $v_2$ are orthogonal meaning the inner product of them is $0$. Now we need to make them orthonormal. After doing so we get $e_1=v_1/sqrt{3}$ and $e_2=v_2/sqrt{3}$
Now we calculate $Pr(x)$.
$$Pr(x)= (x,e_1)e_1+(x,e_2)e_2 =ldots= (1,1,2,2)$$
Therefore the distance is $d(x,U)= ||x-Pr(x)||=||(1,1,1,1)-(1,1,2,2)||= ||(0,0,-1,-1)||= sqrt{2}$
linear-algebra
edited Apr 13 '17 at 12:21
Community♦
1
asked Jun 28 '14 at 14:44
Kyle
1314
@M.Vinay yes that is indeed what I meant.
– Kyle
Jun 28 '14 at 14:48
$sqrt{3}$ is not the modulus of those vectors
– enzotib
Jun 28 '14 at 15:49
Note that $|v_k| = sqrt{5}$.
– copper.hat
Jun 28 '14 at 15:53
I can't believe I made that mistake lol, well it's all clear now, thanks a lot guys.
– Kyle
Jun 28 '14 at 15:57
add a comment |
I want to calculate the distance of the vector $x=(1,1,1,1)$ to the subspace ${(1,0,2,0) , (0,1,0,2)}$
I have solved this in 2 ways that I know of but the thing is, the results are different.
For instance when I use $||x-Pr(x)||$ I get $sqrt{2}$, but when I calculate it using the gram determinant (more info here: Distance between a point and a m-dimensional space in n-dimensional space ($m<n$) ) I get $sqrt{frac{10}{25}}$ which is weird because both ways should be equivalent.
So my question is, what am I missing here? Which one of these result was the correct one?
Thank you.
Edit: Solution using $||x-Pr(x)||$ :
Let $v_1$ be $(1,0,2,0)$ and $v_2$ be $(0,1,0,2)$
Both vectors $v_1$ and $v_2$ are orthogonal meaning the inner product of them is $0$. Now we need to make them orthonormal. After doing so we get $e_1=v_1/sqrt{3}$ and $e_2=v_2/sqrt{3}$
Now we calculate $Pr(x)$.
$$Pr(x)= (x,e_1)e_1+(x,e_2)e_2 =ldots= (1,1,2,2)$$
Therefore the distance is $d(x,U)= ||x-Pr(x)||=||(1,1,1,1)-(1,1,2,2)||= ||(0,0,-1,-1)||= sqrt{2}$
linear-algebra
edited Apr 13 '17 at 12:21
Community♦
1
asked Jun 28 '14 at 14:44
Kyle
1314
I want to calculate the distance of the vector $x=(1,1,1,1)$ to the subspace ${(1,0,2,0) , (0,1,0,2)}$
I have solved this in 2 ways that I know of but the thing is, the results are different.
For instance when I use $||x-Pr(x)||$ I get $sqrt{2}$, but when I calculate it using the gram determinant (more info here: Distance between a point and a m-dimensional space in n-dimensional space ($m<n$) ) I get $sqrt{frac{10}{25}}$ which is weird because both ways should be equivalent.
So my question is, what am I missing here? Which one of these result was the correct one?
Thank you.
Edit: Solution using $||x-Pr(x)||$ :
Let $v_1$ be $(1,0,2,0)$ and $v_2$ be $(0,1,0,2)$
Both vectors $v_1$ and $v_2$ are orthogonal meaning the inner product of them is $0$. Now we need to make them orthonormal. After doing so we get $e_1=v_1/sqrt{3}$ and $e_2=v_2/sqrt{3}$
Now we calculate $Pr(x)$.
$$Pr(x)= (x,e_1)e_1+(x,e_2)e_2 =ldots= (1,1,2,2)$$
Therefore the distance is $d(x,U)= ||x-Pr(x)||=||(1,1,1,1)-(1,1,2,2)||= ||(0,0,-1,-1)||= sqrt{2}$
linear-algebra
linear-algebra
edited Apr 13 '17 at 12:21
Community♦
1
asked Jun 28 '14 at 14:44
Kyle
1314
edited Apr 13 '17 at 12:21
Community♦
1
asked Jun 28 '14 at 14:44
Kyle
1314
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Jun 28 '14 at 14:44
Kyle
1314
asked Jun 28 '14 at 14:44
Kyle
1314
asked Jun 28 '14 at 14:44
Kyle
1314
1314
@M.Vinay yes that is indeed what I meant.
– Kyle
Jun 28 '14 at 14:48
$sqrt{3}$ is not the modulus of those vectors
– enzotib
Jun 28 '14 at 15:49
Note that $|v_k| = sqrt{5}$.
– copper.hat
Jun 28 '14 at 15:53
I can't believe I made that mistake lol, well it's all clear now, thanks a lot guys.
– Kyle
Jun 28 '14 at 15:57
add a comment |
@M.Vinay yes that is indeed what I meant.
– Kyle
Jun 28 '14 at 14:48
$sqrt{3}$ is not the modulus of those vectors
– enzotib
Jun 28 '14 at 15:49
Note that $|v_k| = sqrt{5}$.
– copper.hat
Jun 28 '14 at 15:53
I can't believe I made that mistake lol, well it's all clear now, thanks a lot guys.
– Kyle
Jun 28 '14 at 15:57
@M.Vinay yes that is indeed what I meant.
– Kyle
Jun 28 '14 at 14:48
@M.Vinay yes that is indeed what I meant.
– Kyle
Jun 28 '14 at 14:48
$sqrt{3}$ is not the modulus of those vectors
– enzotib
Jun 28 '14 at 15:49
$sqrt{3}$ is not the modulus of those vectors
– enzotib
Jun 28 '14 at 15:49
Note that $|v_k| = sqrt{5}$.
– copper.hat
Jun 28 '14 at 15:53
Note that $|v_k| = sqrt{5}$.
– copper.hat
Jun 28 '14 at 15:53
I can't believe I made that mistake lol, well it's all clear now, thanks a lot guys.
– Kyle
Jun 28 '14 at 15:57
I can't believe I made that mistake lol, well it's all clear now, thanks a lot guys.
– Kyle
Jun 28 '14 at 15:57
add a comment |
2 Answers
2
active
oldest
votes
I'll offer the third method:
Let the point in that subspace be given by $x (0,1,0,2)+y(1,0,2,0)$. Then the square of the distance to the point $(1, 1, 1, 1)$ can be written as,
$$(1-y)^2+(1-x)^2+(1-2y)^2+(1-2x)^2$$
We need to minimize the above expression with respect to $x$ and $y$. Due to symetry, it's sufficient to consider only the minimization
$$min_x ( (1-x)^2 +(1-2x)^2) =min_x( 5x^2-6x+2).$$
The minimum is attained at $x=frac 35$ and the minimal value is $frac 15$. Therefore, the square of the minimum distance is $frac 25$ (i.e. the same as you obtained via Gram's determinant).
It seems that you made some mistakes while calculating $|x-mathrm{Pr}, x|$.
edited Nov 20 '18 at 11:35
ijuneja
1498
answered Jun 28 '14 at 15:00
TZakrevskiy
20.1k12354
Interesting, although I asked a few of my friends who solved this question and they all told me the answer was sqrt(2) and they used ||x-Pr(x)|| to calculate it, I can't find a mistake I made in my solution using ||x-Pr(x)||, really weird..
– Kyle
Jun 28 '14 at 15:22
@Kyle If you write the details on how you calculate the projector, we can look into details and find where the error comes from.
– TZakrevskiy
Jun 28 '14 at 15:28
I've edited the OP with the solution using ||x-Pr(x)||
– Kyle
Jun 28 '14 at 15:39
add a comment |
This is a geometric view of TZakrevskiy's answer.
Let $V=operatorname{sp} {v_1,v_2}$. Note that $v_1 bot v_2$ in this problem.
If $p in V$ minimizes the distance between $x$ and $V$, we have
$(x-p) bot v_k $ for $k=1,2$.
Writing $p=sum_k alpha_k v_k$ and using orthogonality, we obtain
$alpha_k = { langle x , v_k rangle over |v_k|^2} = { 3 over 5}$.
Substituting gives $|x-p| = sqrt{2 over 5}$
answered Jun 28 '14 at 15:01
copper.hat
126k559159
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
oldest
votes
active
oldest
votes
I'll offer the third method:
Let the point in that subspace be given by $x (0,1,0,2)+y(1,0,2,0)$. Then the square of the distance to the point $(1, 1, 1, 1)$ can be written as,
$$(1-y)^2+(1-x)^2+(1-2y)^2+(1-2x)^2$$
We need to minimize the above expression with respect to $x$ and $y$. Due to symetry, it's sufficient to consider only the minimization
$$min_x ( (1-x)^2 +(1-2x)^2) =min_x( 5x^2-6x+2).$$
The minimum is attained at $x=frac 35$ and the minimal value is $frac 15$. Therefore, the square of the minimum distance is $frac 25$ (i.e. the same as you obtained via Gram's determinant).
It seems that you made some mistakes while calculating $|x-mathrm{Pr}, x|$.
edited Nov 20 '18 at 11:35
ijuneja
1498
answered Jun 28 '14 at 15:00
TZakrevskiy
20.1k12354
Interesting, although I asked a few of my friends who solved this question and they all told me the answer was sqrt(2) and they used ||x-Pr(x)|| to calculate it, I can't find a mistake I made in my solution using ||x-Pr(x)||, really weird..
– Kyle
Jun 28 '14 at 15:22
@Kyle If you write the details on how you calculate the projector, we can look into details and find where the error comes from.
– TZakrevskiy
Jun 28 '14 at 15:28
I've edited the OP with the solution using ||x-Pr(x)||
– Kyle
Jun 28 '14 at 15:39
add a comment |
I'll offer the third method:
Let the point in that subspace be given by $x (0,1,0,2)+y(1,0,2,0)$. Then the square of the distance to the point $(1, 1, 1, 1)$ can be written as,
$$(1-y)^2+(1-x)^2+(1-2y)^2+(1-2x)^2$$
We need to minimize the above expression with respect to $x$ and $y$. Due to symetry, it's sufficient to consider only the minimization
$$min_x ( (1-x)^2 +(1-2x)^2) =min_x( 5x^2-6x+2).$$
The minimum is attained at $x=frac 35$ and the minimal value is $frac 15$. Therefore, the square of the minimum distance is $frac 25$ (i.e. the same as you obtained via Gram's determinant).
It seems that you made some mistakes while calculating $|x-mathrm{Pr}, x|$.
edited Nov 20 '18 at 11:35
ijuneja
1498
answered Jun 28 '14 at 15:00
TZakrevskiy
20.1k12354
Interesting, although I asked a few of my friends who solved this question and they all told me the answer was sqrt(2) and they used ||x-Pr(x)|| to calculate it, I can't find a mistake I made in my solution using ||x-Pr(x)||, really weird..
– Kyle
Jun 28 '14 at 15:22
@Kyle If you write the details on how you calculate the projector, we can look into details and find where the error comes from.
– TZakrevskiy
Jun 28 '14 at 15:28
I've edited the OP with the solution using ||x-Pr(x)||
– Kyle
Jun 28 '14 at 15:39
add a comment |
I'll offer the third method:
Let the point in that subspace be given by $x (0,1,0,2)+y(1,0,2,0)$. Then the square of the distance to the point $(1, 1, 1, 1)$ can be written as,
$$(1-y)^2+(1-x)^2+(1-2y)^2+(1-2x)^2$$
We need to minimize the above expression with respect to $x$ and $y$. Due to symetry, it's sufficient to consider only the minimization
$$min_x ( (1-x)^2 +(1-2x)^2) =min_x( 5x^2-6x+2).$$
The minimum is attained at $x=frac 35$ and the minimal value is $frac 15$. Therefore, the square of the minimum distance is $frac 25$ (i.e. the same as you obtained via Gram's determinant).
It seems that you made some mistakes while calculating $|x-mathrm{Pr}, x|$.
edited Nov 20 '18 at 11:35
ijuneja
1498
answered Jun 28 '14 at 15:00
TZakrevskiy
20.1k12354
I'll offer the third method:
Let the point in that subspace be given by $x (0,1,0,2)+y(1,0,2,0)$. Then the square of the distance to the point $(1, 1, 1, 1)$ can be written as,
$$(1-y)^2+(1-x)^2+(1-2y)^2+(1-2x)^2$$
We need to minimize the above expression with respect to $x$ and $y$. Due to symetry, it's sufficient to consider only the minimization
$$min_x ( (1-x)^2 +(1-2x)^2) =min_x( 5x^2-6x+2).$$
The minimum is attained at $x=frac 35$ and the minimal value is $frac 15$. Therefore, the square of the minimum distance is $frac 25$ (i.e. the same as you obtained via Gram's determinant).
It seems that you made some mistakes while calculating $|x-mathrm{Pr}, x|$.
edited Nov 20 '18 at 11:35
ijuneja
1498
answered Jun 28 '14 at 15:00
TZakrevskiy
20.1k12354
edited Nov 20 '18 at 11:35
ijuneja
1498
edited Nov 20 '18 at 11:35
ijuneja
1498
edited Nov 20 '18 at 11:35
ijuneja
1498
1498
answered Jun 28 '14 at 15:00
TZakrevskiy
20.1k12354
answered Jun 28 '14 at 15:00
TZakrevskiy
20.1k12354
answered Jun 28 '14 at 15:00
TZakrevskiy
20.1k12354
20.1k12354
Interesting, although I asked a few of my friends who solved this question and they all told me the answer was sqrt(2) and they used ||x-Pr(x)|| to calculate it, I can't find a mistake I made in my solution using ||x-Pr(x)||, really weird..
– Kyle
Jun 28 '14 at 15:22
@Kyle If you write the details on how you calculate the projector, we can look into details and find where the error comes from.
– TZakrevskiy
Jun 28 '14 at 15:28
I've edited the OP with the solution using ||x-Pr(x)||
– Kyle
Jun 28 '14 at 15:39
add a comment |
Interesting, although I asked a few of my friends who solved this question and they all told me the answer was sqrt(2) and they used ||x-Pr(x)|| to calculate it, I can't find a mistake I made in my solution using ||x-Pr(x)||, really weird..
– Kyle
Jun 28 '14 at 15:22
@Kyle If you write the details on how you calculate the projector, we can look into details and find where the error comes from.
– TZakrevskiy
Jun 28 '14 at 15:28
I've edited the OP with the solution using ||x-Pr(x)||
– Kyle
Jun 28 '14 at 15:39
Interesting, although I asked a few of my friends who solved this question and they all told me the answer was sqrt(2) and they used ||x-Pr(x)|| to calculate it, I can't find a mistake I made in my solution using ||x-Pr(x)||, really weird..
– Kyle
Jun 28 '14 at 15:22
Interesting, although I asked a few of my friends who solved this question and they all told me the answer was sqrt(2) and they used ||x-Pr(x)|| to calculate it, I can't find a mistake I made in my solution using ||x-Pr(x)||, really weird..
– Kyle
Jun 28 '14 at 15:22
@Kyle If you write the details on how you calculate the projector, we can look into details and find where the error comes from.
– TZakrevskiy
Jun 28 '14 at 15:28
@Kyle If you write the details on how you calculate the projector, we can look into details and find where the error comes from.
– TZakrevskiy
Jun 28 '14 at 15:28
I've edited the OP with the solution using ||x-Pr(x)||
– Kyle
Jun 28 '14 at 15:39
I've edited the OP with the solution using ||x-Pr(x)||
– Kyle
Jun 28 '14 at 15:39
add a comment |
This is a geometric view of TZakrevskiy's answer.
Let $V=operatorname{sp} {v_1,v_2}$. Note that $v_1 bot v_2$ in this problem.
If $p in V$ minimizes the distance between $x$ and $V$, we have
$(x-p) bot v_k $ for $k=1,2$.
Writing $p=sum_k alpha_k v_k$ and using orthogonality, we obtain
$alpha_k = { langle x , v_k rangle over |v_k|^2} = { 3 over 5}$.
Substituting gives $|x-p| = sqrt{2 over 5}$
answered Jun 28 '14 at 15:01
copper.hat
126k559159
add a comment |
This is a geometric view of TZakrevskiy's answer.
Let $V=operatorname{sp} {v_1,v_2}$. Note that $v_1 bot v_2$ in this problem.
If $p in V$ minimizes the distance between $x$ and $V$, we have
$(x-p) bot v_k $ for $k=1,2$.
Writing $p=sum_k alpha_k v_k$ and using orthogonality, we obtain
$alpha_k = { langle x , v_k rangle over |v_k|^2} = { 3 over 5}$.
Substituting gives $|x-p| = sqrt{2 over 5}$
answered Jun 28 '14 at 15:01
copper.hat
126k559159
add a comment |
This is a geometric view of TZakrevskiy's answer.
Let $V=operatorname{sp} {v_1,v_2}$. Note that $v_1 bot v_2$ in this problem.
If $p in V$ minimizes the distance between $x$ and $V$, we have
$(x-p) bot v_k $ for $k=1,2$.
Writing $p=sum_k alpha_k v_k$ and using orthogonality, we obtain
$alpha_k = { langle x , v_k rangle over |v_k|^2} = { 3 over 5}$.
Substituting gives $|x-p| = sqrt{2 over 5}$
answered Jun 28 '14 at 15:01
copper.hat
126k559159
This is a geometric view of TZakrevskiy's answer.
Let $V=operatorname{sp} {v_1,v_2}$. Note that $v_1 bot v_2$ in this problem.
If $p in V$ minimizes the distance between $x$ and $V$, we have
$(x-p) bot v_k $ for $k=1,2$.
Writing $p=sum_k alpha_k v_k$ and using orthogonality, we obtain
$alpha_k = { langle x , v_k rangle over |v_k|^2} = { 3 over 5}$.
Substituting gives $|x-p| = sqrt{2 over 5}$
answered Jun 28 '14 at 15:01
copper.hat
126k559159
edited Jun 28 '14 at 15:49
answered Jun 28 '14 at 15:01
copper.hat
126k559159
answered Jun 28 '14 at 15:01
copper.hat
126k559159
answered Jun 28 '14 at 15:01
copper.hat
126k559159
126k559159
add a comment |
add a comment |
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@M.Vinay yes that is indeed what I meant.
– Kyle
Jun 28 '14 at 14:48
$sqrt{3}$ is not the modulus of those vectors
– enzotib
Jun 28 '14 at 15:49
Note that $|v_k| = sqrt{5}$.
– copper.hat
Jun 28 '14 at 15:53
I can't believe I made that mistake lol, well it's all clear now, thanks a lot guys.
– Kyle
Jun 28 '14 at 15:57