Mixing
minimization of $|v-b|$ where $bin ker A$.
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
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Let $$A=beginpmatrix5&-10&-15\ -1&2&3\2&-4&-6endpmatrixin mathbb R^3times 3.$$
Fix $vinmathbb R^3$. How can I find $$min_bin ker A|b-v| ?$$
I know that I can find an orthonormal basis of $ker A$, complete in a orthonormal basis of $mathbb R^3$ and then take the orthogonal projection of $v$ onto $ker A$. But I would like to use the mean square method. But I know how to use it only if I have to minimize $min_uinmathbb R^3|Au-c|$ when column of $A$ are linearly independent.
linear-algebra optimization convex-optimization least-squares
edited Jul 27 at 15:10
Royi
2,94012046
asked Jul 27 at 14:29
MSE
1,471315
add a comment |Â
up vote
3
down vote
favorite
Let $$A=beginpmatrix5&-10&-15\ -1&2&3\2&-4&-6endpmatrixin mathbb R^3times 3.$$
Fix $vinmathbb R^3$. How can I find $$min_bin ker A|b-v| ?$$
I know that I can find an orthonormal basis of $ker A$, complete in a orthonormal basis of $mathbb R^3$ and then take the orthogonal projection of $v$ onto $ker A$. But I would like to use the mean square method. But I know how to use it only if I have to minimize $min_uinmathbb R^3|Au-c|$ when column of $A$ are linearly independent.
linear-algebra optimization convex-optimization least-squares
edited Jul 27 at 15:10
Royi
2,94012046
asked Jul 27 at 14:29
MSE
1,471315
If you want to apply the unconstrained least squares, then you need to parameterise the domain so that it is essentially unconstrained. In particular, this means you need to find a basis of $ker A$. As an aside, note that $A= (5,-1,2)^T (1,2,3)$, so finding such a basis is fairly straightforward.
– copper.hat
Jul 27 at 16:11
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $$A=beginpmatrix5&-10&-15\ -1&2&3\2&-4&-6endpmatrixin mathbb R^3times 3.$$
Fix $vinmathbb R^3$. How can I find $$min_bin ker A|b-v| ?$$
I know that I can find an orthonormal basis of $ker A$, complete in a orthonormal basis of $mathbb R^3$ and then take the orthogonal projection of $v$ onto $ker A$. But I would like to use the mean square method. But I know how to use it only if I have to minimize $min_uinmathbb R^3|Au-c|$ when column of $A$ are linearly independent.
linear-algebra optimization convex-optimization least-squares
edited Jul 27 at 15:10
Royi
2,94012046
asked Jul 27 at 14:29
MSE
1,471315
Let $$A=beginpmatrix5&-10&-15\ -1&2&3\2&-4&-6endpmatrixin mathbb R^3times 3.$$
Fix $vinmathbb R^3$. How can I find $$min_bin ker A|b-v| ?$$
I know that I can find an orthonormal basis of $ker A$, complete in a orthonormal basis of $mathbb R^3$ and then take the orthogonal projection of $v$ onto $ker A$. But I would like to use the mean square method. But I know how to use it only if I have to minimize $min_uinmathbb R^3|Au-c|$ when column of $A$ are linearly independent.
linear-algebra optimization convex-optimization least-squares
edited Jul 27 at 15:10
Royi
2,94012046
asked Jul 27 at 14:29
MSE
1,471315
edited Jul 27 at 15:10
Royi
2,94012046
edited Jul 27 at 15:10
Royi
2,94012046
edited Jul 27 at 15:10
Royi
2,94012046
2,94012046
asked Jul 27 at 14:29
MSE
1,471315
asked Jul 27 at 14:29
MSE
1,471315
asked Jul 27 at 14:29
MSE
1,471315
1,471315
If you want to apply the unconstrained least squares, then you need to parameterise the domain so that it is essentially unconstrained. In particular, this means you need to find a basis of $ker A$. As an aside, note that $A= (5,-1,2)^T (1,2,3)$, so finding such a basis is fairly straightforward.
– copper.hat
Jul 27 at 16:11
add a comment |Â
If you want to apply the unconstrained least squares, then you need to parameterise the domain so that it is essentially unconstrained. In particular, this means you need to find a basis of $ker A$. As an aside, note that $A= (5,-1,2)^T (1,2,3)$, so finding such a basis is fairly straightforward.
– copper.hat
Jul 27 at 16:11
If you want to apply the unconstrained least squares, then you need to parameterise the domain so that it is essentially unconstrained. In particular, this means you need to find a basis of $ker A$. As an aside, note that $A= (5,-1,2)^T (1,2,3)$, so finding such a basis is fairly straightforward.
– copper.hat
Jul 27 at 16:11
If you want to apply the unconstrained least squares, then you need to parameterise the domain so that it is essentially unconstrained. In particular, this means you need to find a basis of $ker A$. As an aside, note that $A= (5,-1,2)^T (1,2,3)$, so finding such a basis is fairly straightforward.
– copper.hat
Jul 27 at 16:11
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
The problem is given by:
$$beginalign*
arg min_x quad & frac12 x - y right_2^2 \
textsubject to quad & A x = boldsymbol0
endalign*$$
The Lagrangian is given by:
$$ L left( x, mu right) = frac12 x - y right_2^2 + mu^T A x $$
The KKT System is given by:
beginalign*
nabla_x L left( x, mu right) = x - y + A^T mu & = 0 & text(1) \
mu^T A x & = 0 & text(2) \
endalign*
Taking (1) and multiplying from left by $ A $ and remembering $ A x = 0 $:
$$ A x - A y + A A^T mu = 0 undersetA x = 0Rightarrow A A^T mu = A y Rightarrow mu = left( A A^T right)^dagger A y $$
Plugging the result into (1) yields:
$$ x - y + A^T left( A A^T right)^dagger A y = 0 Rightarrow x = y - A^T left( A A^T right)^dagger A y $$
Plug in your matrix and you'll get the answer.
answered Jul 27 at 15:10
Royi
2,94012046
1
Your expression $(3)$ is incorrect: for equality constraints, your multiplier does not need to satisfy any positivity condition. Also, notice that $AA^T$ is not full rank, so you can't define $(AA^T)^-1$ on the second to last line. (In fact, if $A$ is full rank, then you would get $x = 0$, consistent with the fact that this is the only point in the feasible region).
– Strants
Jul 27 at 15:18
My mistake, I used template for inequality. Fixed it. Regarding, the Inverse, you're correct. It should be replaced with the SVD based inverse. +1 your comment. Thank You.
– Royi
Jul 27 at 15:33
Looks good! Thanks for fixing it.
– Strants
Jul 27 at 15:42
@Strants, It you who pointed me. So it is all you.
– Royi
Jul 27 at 15:49
add a comment |Â
up vote
1
down vote
$DeclareMathOperatorrankRank DeclareMathOperatorimageImage DeclareMathOperatorvspanSpan$Here's a geometric approach that exploits the fact that $rank A = 1$.
Since $rank A = 1$, so $dim ker A = 2$. Also recall that $(ker A)^perp = image A^T$; that is, the kernal of $A$ is perpendicular to the rowspace of $A$. This means that for our minimizer $b^*$, we have $v = b^* + w$, where $w$ is in the row-space of $A$ and $Av = Aw$.
Since our matrix $A$ has a one-dimensional row-space $image A^T = vspan(1, -2, -3)$, we get the explicit formula
$$w = fraclVert A v rVertlVert A(1,-2,-3)rVert (1,-2,-3)$$
and so
$$min_b in ker A lVert b - v rVert = lVert w rVert = lVert Av rVert fraclVert (1, -2, -3) rVertlVert A(1, -2, -3) rVert = fraclVert Av rVert7sqrt10.$$
The method in this anwser can be generalized to argue that
$$min_b in ker A lVert b - vrVert = min_Aw = Av lVert w rVert.$$
Since the matrix $tilde A$ that we get from expressing $A$ as a map from $image A^T$ to $image A$ is invertible, our problem reduces to finding and inverting $tilde A$. In this problem, $tilde A$ is a $1times 1$ matrix, so it is very easy to find the answer.
answered Jul 27 at 15:43
Strants
5,07421636
add a comment |Â
up vote
0
down vote
Hint
Compute $ker(A)$. Let say it's $Spanv_1,v_2$. Then Apply the method you know with the matrix $$M=(v_1 v_2)in mathcal M_3times 2(mathbb R).$$
answered Jul 27 at 14:39
Surb
36.3k84274
Could you please develop ?
– MSE
Jul 27 at 14:51
If you can solve $min_u|Au-b|$ you should be able to solve $min_uinmathbb R^2|Mu-v|$ no ?
– Surb
Jul 27 at 14:53
add a comment |Â
up vote
0
down vote
$$ ker (A) = (x,y,z); x=2y+3z;y,zin mathbbR $$
$$= y(2,1,0)+z(3,0,1);y,zin mathbbR$$
Say $v=(a,b,c)$ then with use of Cauchy-Schwarz inequality we have:
$$begineqnarray||b-v||^2 &=& (a-2y-3z)^2+(b-y)^2+(c-z)^2\
&=& 1over 14(1+4+9)Big((2y+3z-a)^2+(b-y)^2+(c-z)^2Big)\
&geq & 1over 14Big((2y+3z-a)+(2b-2y)+(3c-3z)Big)^2\
&= &1over 14(-a+2b+3c)^2\
endeqnarray
$$
With equality $$2y+3z-aover 1 = b-yover 2 = c-zover 3$$
and this is when $$y=a+5b-3cover 7;;;rm and;;;z =3a-6b+5cover 14$$
So $$min_bin ker A|b-v| = 1over sqrt14|-a+2b+3c|$$
answered Jul 27 at 14:58
greedoid
26.1k93473
@MSE I gave you a full solution to question you ask. And I spend some time to write it down. You don't like it? You don't understand it?
– greedoid
Jul 31 at 14:23
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The problem is given by:
$$beginalign*
arg min_x quad & frac12 x - y right_2^2 \
textsubject to quad & A x = boldsymbol0
endalign*$$
The Lagrangian is given by:
$$ L left( x, mu right) = frac12 x - y right_2^2 + mu^T A x $$
The KKT System is given by:
beginalign*
nabla_x L left( x, mu right) = x - y + A^T mu & = 0 & text(1) \
mu^T A x & = 0 & text(2) \
endalign*
Taking (1) and multiplying from left by $ A $ and remembering $ A x = 0 $:
$$ A x - A y + A A^T mu = 0 undersetA x = 0Rightarrow A A^T mu = A y Rightarrow mu = left( A A^T right)^dagger A y $$
Plugging the result into (1) yields:
$$ x - y + A^T left( A A^T right)^dagger A y = 0 Rightarrow x = y - A^T left( A A^T right)^dagger A y $$
Plug in your matrix and you'll get the answer.
answered Jul 27 at 15:10
Royi
2,94012046
1
Your expression $(3)$ is incorrect: for equality constraints, your multiplier does not need to satisfy any positivity condition. Also, notice that $AA^T$ is not full rank, so you can't define $(AA^T)^-1$ on the second to last line. (In fact, if $A$ is full rank, then you would get $x = 0$, consistent with the fact that this is the only point in the feasible region).
– Strants
Jul 27 at 15:18
My mistake, I used template for inequality. Fixed it. Regarding, the Inverse, you're correct. It should be replaced with the SVD based inverse. +1 your comment. Thank You.
– Royi
Jul 27 at 15:33
Looks good! Thanks for fixing it.
– Strants
Jul 27 at 15:42
@Strants, It you who pointed me. So it is all you.
– Royi
Jul 27 at 15:49
add a comment |Â
up vote
1
down vote
The problem is given by:
$$beginalign*
arg min_x quad & frac12 x - y right_2^2 \
textsubject to quad & A x = boldsymbol0
endalign*$$
The Lagrangian is given by:
$$ L left( x, mu right) = frac12 x - y right_2^2 + mu^T A x $$
The KKT System is given by:
beginalign*
nabla_x L left( x, mu right) = x - y + A^T mu & = 0 & text(1) \
mu^T A x & = 0 & text(2) \
endalign*
Taking (1) and multiplying from left by $ A $ and remembering $ A x = 0 $:
$$ A x - A y + A A^T mu = 0 undersetA x = 0Rightarrow A A^T mu = A y Rightarrow mu = left( A A^T right)^dagger A y $$
Plugging the result into (1) yields:
$$ x - y + A^T left( A A^T right)^dagger A y = 0 Rightarrow x = y - A^T left( A A^T right)^dagger A y $$
Plug in your matrix and you'll get the answer.
answered Jul 27 at 15:10
Royi
2,94012046
1
Your expression $(3)$ is incorrect: for equality constraints, your multiplier does not need to satisfy any positivity condition. Also, notice that $AA^T$ is not full rank, so you can't define $(AA^T)^-1$ on the second to last line. (In fact, if $A$ is full rank, then you would get $x = 0$, consistent with the fact that this is the only point in the feasible region).
– Strants
Jul 27 at 15:18
My mistake, I used template for inequality. Fixed it. Regarding, the Inverse, you're correct. It should be replaced with the SVD based inverse. +1 your comment. Thank You.
– Royi
Jul 27 at 15:33
Looks good! Thanks for fixing it.
– Strants
Jul 27 at 15:42
@Strants, It you who pointed me. So it is all you.
– Royi
Jul 27 at 15:49
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The problem is given by:
$$beginalign*
arg min_x quad & frac12 x - y right_2^2 \
textsubject to quad & A x = boldsymbol0
endalign*$$
The Lagrangian is given by:
$$ L left( x, mu right) = frac12 x - y right_2^2 + mu^T A x $$
The KKT System is given by:
beginalign*
nabla_x L left( x, mu right) = x - y + A^T mu & = 0 & text(1) \
mu^T A x & = 0 & text(2) \
endalign*
Taking (1) and multiplying from left by $ A $ and remembering $ A x = 0 $:
$$ A x - A y + A A^T mu = 0 undersetA x = 0Rightarrow A A^T mu = A y Rightarrow mu = left( A A^T right)^dagger A y $$
Plugging the result into (1) yields:
$$ x - y + A^T left( A A^T right)^dagger A y = 0 Rightarrow x = y - A^T left( A A^T right)^dagger A y $$
Plug in your matrix and you'll get the answer.
answered Jul 27 at 15:10
Royi
2,94012046
The problem is given by:
$$beginalign*
arg min_x quad & frac12 x - y right_2^2 \
textsubject to quad & A x = boldsymbol0
endalign*$$
The Lagrangian is given by:
$$ L left( x, mu right) = frac12 x - y right_2^2 + mu^T A x $$
The KKT System is given by:
beginalign*
nabla_x L left( x, mu right) = x - y + A^T mu & = 0 & text(1) \
mu^T A x & = 0 & text(2) \
endalign*
Taking (1) and multiplying from left by $ A $ and remembering $ A x = 0 $:
$$ A x - A y + A A^T mu = 0 undersetA x = 0Rightarrow A A^T mu = A y Rightarrow mu = left( A A^T right)^dagger A y $$
Plugging the result into (1) yields:
$$ x - y + A^T left( A A^T right)^dagger A y = 0 Rightarrow x = y - A^T left( A A^T right)^dagger A y $$
Plug in your matrix and you'll get the answer.
answered Jul 27 at 15:10
Royi
2,94012046
edited Jul 27 at 15:36
answered Jul 27 at 15:10
Royi
2,94012046
answered Jul 27 at 15:10
Royi
2,94012046
answered Jul 27 at 15:10
Royi
2,94012046
2,94012046
1
Your expression $(3)$ is incorrect: for equality constraints, your multiplier does not need to satisfy any positivity condition. Also, notice that $AA^T$ is not full rank, so you can't define $(AA^T)^-1$ on the second to last line. (In fact, if $A$ is full rank, then you would get $x = 0$, consistent with the fact that this is the only point in the feasible region).
– Strants
Jul 27 at 15:18
My mistake, I used template for inequality. Fixed it. Regarding, the Inverse, you're correct. It should be replaced with the SVD based inverse. +1 your comment. Thank You.
– Royi
Jul 27 at 15:33
Looks good! Thanks for fixing it.
– Strants
Jul 27 at 15:42
@Strants, It you who pointed me. So it is all you.
– Royi
Jul 27 at 15:49
add a comment |Â
1
Your expression $(3)$ is incorrect: for equality constraints, your multiplier does not need to satisfy any positivity condition. Also, notice that $AA^T$ is not full rank, so you can't define $(AA^T)^-1$ on the second to last line. (In fact, if $A$ is full rank, then you would get $x = 0$, consistent with the fact that this is the only point in the feasible region).
– Strants
Jul 27 at 15:18
My mistake, I used template for inequality. Fixed it. Regarding, the Inverse, you're correct. It should be replaced with the SVD based inverse. +1 your comment. Thank You.
– Royi
Jul 27 at 15:33
Looks good! Thanks for fixing it.
– Strants
Jul 27 at 15:42
@Strants, It you who pointed me. So it is all you.
– Royi
Jul 27 at 15:49
1
1
Your expression $(3)$ is incorrect: for equality constraints, your multiplier does not need to satisfy any positivity condition. Also, notice that $AA^T$ is not full rank, so you can't define $(AA^T)^-1$ on the second to last line. (In fact, if $A$ is full rank, then you would get $x = 0$, consistent with the fact that this is the only point in the feasible region).
– Strants
Jul 27 at 15:18
Your expression $(3)$ is incorrect: for equality constraints, your multiplier does not need to satisfy any positivity condition. Also, notice that $AA^T$ is not full rank, so you can't define $(AA^T)^-1$ on the second to last line. (In fact, if $A$ is full rank, then you would get $x = 0$, consistent with the fact that this is the only point in the feasible region).
– Strants
Jul 27 at 15:18
My mistake, I used template for inequality. Fixed it. Regarding, the Inverse, you're correct. It should be replaced with the SVD based inverse. +1 your comment. Thank You.
– Royi
Jul 27 at 15:33
My mistake, I used template for inequality. Fixed it. Regarding, the Inverse, you're correct. It should be replaced with the SVD based inverse. +1 your comment. Thank You.
– Royi
Jul 27 at 15:33
Looks good! Thanks for fixing it.
– Strants
Jul 27 at 15:42
Looks good! Thanks for fixing it.
– Strants
Jul 27 at 15:42
@Strants, It you who pointed me. So it is all you.
– Royi
Jul 27 at 15:49
@Strants, It you who pointed me. So it is all you.
– Royi
Jul 27 at 15:49
add a comment |Â
up vote
1
down vote
$DeclareMathOperatorrankRank DeclareMathOperatorimageImage DeclareMathOperatorvspanSpan$Here's a geometric approach that exploits the fact that $rank A = 1$.
Since $rank A = 1$, so $dim ker A = 2$. Also recall that $(ker A)^perp = image A^T$; that is, the kernal of $A$ is perpendicular to the rowspace of $A$. This means that for our minimizer $b^*$, we have $v = b^* + w$, where $w$ is in the row-space of $A$ and $Av = Aw$.
Since our matrix $A$ has a one-dimensional row-space $image A^T = vspan(1, -2, -3)$, we get the explicit formula
$$w = fraclVert A v rVertlVert A(1,-2,-3)rVert (1,-2,-3)$$
and so
$$min_b in ker A lVert b - v rVert = lVert w rVert = lVert Av rVert fraclVert (1, -2, -3) rVertlVert A(1, -2, -3) rVert = fraclVert Av rVert7sqrt10.$$
The method in this anwser can be generalized to argue that
$$min_b in ker A lVert b - vrVert = min_Aw = Av lVert w rVert.$$
Since the matrix $tilde A$ that we get from expressing $A$ as a map from $image A^T$ to $image A$ is invertible, our problem reduces to finding and inverting $tilde A$. In this problem, $tilde A$ is a $1times 1$ matrix, so it is very easy to find the answer.
answered Jul 27 at 15:43
Strants
5,07421636
add a comment |Â
up vote
1
down vote
$DeclareMathOperatorrankRank DeclareMathOperatorimageImage DeclareMathOperatorvspanSpan$Here's a geometric approach that exploits the fact that $rank A = 1$.
Since $rank A = 1$, so $dim ker A = 2$. Also recall that $(ker A)^perp = image A^T$; that is, the kernal of $A$ is perpendicular to the rowspace of $A$. This means that for our minimizer $b^*$, we have $v = b^* + w$, where $w$ is in the row-space of $A$ and $Av = Aw$.
Since our matrix $A$ has a one-dimensional row-space $image A^T = vspan(1, -2, -3)$, we get the explicit formula
$$w = fraclVert A v rVertlVert A(1,-2,-3)rVert (1,-2,-3)$$
and so
$$min_b in ker A lVert b - v rVert = lVert w rVert = lVert Av rVert fraclVert (1, -2, -3) rVertlVert A(1, -2, -3) rVert = fraclVert Av rVert7sqrt10.$$
The method in this anwser can be generalized to argue that
$$min_b in ker A lVert b - vrVert = min_Aw = Av lVert w rVert.$$
Since the matrix $tilde A$ that we get from expressing $A$ as a map from $image A^T$ to $image A$ is invertible, our problem reduces to finding and inverting $tilde A$. In this problem, $tilde A$ is a $1times 1$ matrix, so it is very easy to find the answer.
answered Jul 27 at 15:43
Strants
5,07421636
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$DeclareMathOperatorrankRank DeclareMathOperatorimageImage DeclareMathOperatorvspanSpan$Here's a geometric approach that exploits the fact that $rank A = 1$.
Since $rank A = 1$, so $dim ker A = 2$. Also recall that $(ker A)^perp = image A^T$; that is, the kernal of $A$ is perpendicular to the rowspace of $A$. This means that for our minimizer $b^*$, we have $v = b^* + w$, where $w$ is in the row-space of $A$ and $Av = Aw$.
Since our matrix $A$ has a one-dimensional row-space $image A^T = vspan(1, -2, -3)$, we get the explicit formula
$$w = fraclVert A v rVertlVert A(1,-2,-3)rVert (1,-2,-3)$$
and so
$$min_b in ker A lVert b - v rVert = lVert w rVert = lVert Av rVert fraclVert (1, -2, -3) rVertlVert A(1, -2, -3) rVert = fraclVert Av rVert7sqrt10.$$
The method in this anwser can be generalized to argue that
$$min_b in ker A lVert b - vrVert = min_Aw = Av lVert w rVert.$$
Since the matrix $tilde A$ that we get from expressing $A$ as a map from $image A^T$ to $image A$ is invertible, our problem reduces to finding and inverting $tilde A$. In this problem, $tilde A$ is a $1times 1$ matrix, so it is very easy to find the answer.
answered Jul 27 at 15:43
Strants
5,07421636
$DeclareMathOperatorrankRank DeclareMathOperatorimageImage DeclareMathOperatorvspanSpan$Here's a geometric approach that exploits the fact that $rank A = 1$.
Since $rank A = 1$, so $dim ker A = 2$. Also recall that $(ker A)^perp = image A^T$; that is, the kernal of $A$ is perpendicular to the rowspace of $A$. This means that for our minimizer $b^*$, we have $v = b^* + w$, where $w$ is in the row-space of $A$ and $Av = Aw$.
Since our matrix $A$ has a one-dimensional row-space $image A^T = vspan(1, -2, -3)$, we get the explicit formula
$$w = fraclVert A v rVertlVert A(1,-2,-3)rVert (1,-2,-3)$$
and so
$$min_b in ker A lVert b - v rVert = lVert w rVert = lVert Av rVert fraclVert (1, -2, -3) rVertlVert A(1, -2, -3) rVert = fraclVert Av rVert7sqrt10.$$
The method in this anwser can be generalized to argue that
$$min_b in ker A lVert b - vrVert = min_Aw = Av lVert w rVert.$$
Since the matrix $tilde A$ that we get from expressing $A$ as a map from $image A^T$ to $image A$ is invertible, our problem reduces to finding and inverting $tilde A$. In this problem, $tilde A$ is a $1times 1$ matrix, so it is very easy to find the answer.
answered Jul 27 at 15:43
Strants
5,07421636
edited Jul 27 at 15:58
answered Jul 27 at 15:43
Strants
5,07421636
answered Jul 27 at 15:43
Strants
5,07421636
answered Jul 27 at 15:43
Strants
5,07421636
5,07421636
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint
Compute $ker(A)$. Let say it's $Spanv_1,v_2$. Then Apply the method you know with the matrix $$M=(v_1 v_2)in mathcal M_3times 2(mathbb R).$$
answered Jul 27 at 14:39
Surb
36.3k84274
Could you please develop ?
– MSE
Jul 27 at 14:51
If you can solve $min_u|Au-b|$ you should be able to solve $min_uinmathbb R^2|Mu-v|$ no ?
– Surb
Jul 27 at 14:53
add a comment |Â
up vote
0
down vote
Hint
Compute $ker(A)$. Let say it's $Spanv_1,v_2$. Then Apply the method you know with the matrix $$M=(v_1 v_2)in mathcal M_3times 2(mathbb R).$$
answered Jul 27 at 14:39
Surb
36.3k84274
Could you please develop ?
– MSE
Jul 27 at 14:51
If you can solve $min_u|Au-b|$ you should be able to solve $min_uinmathbb R^2|Mu-v|$ no ?
– Surb
Jul 27 at 14:53
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint
Compute $ker(A)$. Let say it's $Spanv_1,v_2$. Then Apply the method you know with the matrix $$M=(v_1 v_2)in mathcal M_3times 2(mathbb R).$$
answered Jul 27 at 14:39
Surb
36.3k84274
Hint
Compute $ker(A)$. Let say it's $Spanv_1,v_2$. Then Apply the method you know with the matrix $$M=(v_1 v_2)in mathcal M_3times 2(mathbb R).$$
answered Jul 27 at 14:39
Surb
36.3k84274
answered Jul 27 at 14:39
Surb
36.3k84274
answered Jul 27 at 14:39
Surb
36.3k84274
answered Jul 27 at 14:39
Surb
36.3k84274
36.3k84274
Could you please develop ?
– MSE
Jul 27 at 14:51
If you can solve $min_u|Au-b|$ you should be able to solve $min_uinmathbb R^2|Mu-v|$ no ?
– Surb
Jul 27 at 14:53
add a comment |Â
Could you please develop ?
– MSE
Jul 27 at 14:51
If you can solve $min_u|Au-b|$ you should be able to solve $min_uinmathbb R^2|Mu-v|$ no ?
– Surb
Jul 27 at 14:53
Could you please develop ?
– MSE
Jul 27 at 14:51
Could you please develop ?
– MSE
Jul 27 at 14:51
If you can solve $min_u|Au-b|$ you should be able to solve $min_uinmathbb R^2|Mu-v|$ no ?
– Surb
Jul 27 at 14:53
If you can solve $min_u|Au-b|$ you should be able to solve $min_uinmathbb R^2|Mu-v|$ no ?
– Surb
Jul 27 at 14:53
add a comment |Â
up vote
0
down vote
$$ ker (A) = (x,y,z); x=2y+3z;y,zin mathbbR $$
$$= y(2,1,0)+z(3,0,1);y,zin mathbbR$$
Say $v=(a,b,c)$ then with use of Cauchy-Schwarz inequality we have:
$$begineqnarray||b-v||^2 &=& (a-2y-3z)^2+(b-y)^2+(c-z)^2\
&=& 1over 14(1+4+9)Big((2y+3z-a)^2+(b-y)^2+(c-z)^2Big)\
&geq & 1over 14Big((2y+3z-a)+(2b-2y)+(3c-3z)Big)^2\
&= &1over 14(-a+2b+3c)^2\
endeqnarray
$$
With equality $$2y+3z-aover 1 = b-yover 2 = c-zover 3$$
and this is when $$y=a+5b-3cover 7;;;rm and;;;z =3a-6b+5cover 14$$
So $$min_bin ker A|b-v| = 1over sqrt14|-a+2b+3c|$$
answered Jul 27 at 14:58
greedoid
26.1k93473
@MSE I gave you a full solution to question you ask. And I spend some time to write it down. You don't like it? You don't understand it?
– greedoid
Jul 31 at 14:23
add a comment |Â
up vote
0
down vote
$$ ker (A) = (x,y,z); x=2y+3z;y,zin mathbbR $$
$$= y(2,1,0)+z(3,0,1);y,zin mathbbR$$
Say $v=(a,b,c)$ then with use of Cauchy-Schwarz inequality we have:
$$begineqnarray||b-v||^2 &=& (a-2y-3z)^2+(b-y)^2+(c-z)^2\
&=& 1over 14(1+4+9)Big((2y+3z-a)^2+(b-y)^2+(c-z)^2Big)\
&geq & 1over 14Big((2y+3z-a)+(2b-2y)+(3c-3z)Big)^2\
&= &1over 14(-a+2b+3c)^2\
endeqnarray
$$
With equality $$2y+3z-aover 1 = b-yover 2 = c-zover 3$$
and this is when $$y=a+5b-3cover 7;;;rm and;;;z =3a-6b+5cover 14$$
So $$min_bin ker A|b-v| = 1over sqrt14|-a+2b+3c|$$
answered Jul 27 at 14:58
greedoid
26.1k93473
@MSE I gave you a full solution to question you ask. And I spend some time to write it down. You don't like it? You don't understand it?
– greedoid
Jul 31 at 14:23
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$ ker (A) = (x,y,z); x=2y+3z;y,zin mathbbR $$
$$= y(2,1,0)+z(3,0,1);y,zin mathbbR$$
Say $v=(a,b,c)$ then with use of Cauchy-Schwarz inequality we have:
$$begineqnarray||b-v||^2 &=& (a-2y-3z)^2+(b-y)^2+(c-z)^2\
&=& 1over 14(1+4+9)Big((2y+3z-a)^2+(b-y)^2+(c-z)^2Big)\
&geq & 1over 14Big((2y+3z-a)+(2b-2y)+(3c-3z)Big)^2\
&= &1over 14(-a+2b+3c)^2\
endeqnarray
$$
With equality $$2y+3z-aover 1 = b-yover 2 = c-zover 3$$
and this is when $$y=a+5b-3cover 7;;;rm and;;;z =3a-6b+5cover 14$$
So $$min_bin ker A|b-v| = 1over sqrt14|-a+2b+3c|$$
answered Jul 27 at 14:58
greedoid
26.1k93473
$$ ker (A) = (x,y,z); x=2y+3z;y,zin mathbbR $$
$$= y(2,1,0)+z(3,0,1);y,zin mathbbR$$
Say $v=(a,b,c)$ then with use of Cauchy-Schwarz inequality we have:
$$begineqnarray||b-v||^2 &=& (a-2y-3z)^2+(b-y)^2+(c-z)^2\
&=& 1over 14(1+4+9)Big((2y+3z-a)^2+(b-y)^2+(c-z)^2Big)\
&geq & 1over 14Big((2y+3z-a)+(2b-2y)+(3c-3z)Big)^2\
&= &1over 14(-a+2b+3c)^2\
endeqnarray
$$
With equality $$2y+3z-aover 1 = b-yover 2 = c-zover 3$$
and this is when $$y=a+5b-3cover 7;;;rm and;;;z =3a-6b+5cover 14$$
So $$min_bin ker A|b-v| = 1over sqrt14|-a+2b+3c|$$
answered Jul 27 at 14:58
greedoid
26.1k93473
edited Jul 27 at 15:13
answered Jul 27 at 14:58
greedoid
26.1k93473
answered Jul 27 at 14:58
greedoid
26.1k93473
answered Jul 27 at 14:58
greedoid
26.1k93473
26.1k93473
@MSE I gave you a full solution to question you ask. And I spend some time to write it down. You don't like it? You don't understand it?
– greedoid
Jul 31 at 14:23
add a comment |Â
@MSE I gave you a full solution to question you ask. And I spend some time to write it down. You don't like it? You don't understand it?
– greedoid
Jul 31 at 14:23
@MSE I gave you a full solution to question you ask. And I spend some time to write it down. You don't like it? You don't understand it?
– greedoid
Jul 31 at 14:23
@MSE I gave you a full solution to question you ask. And I spend some time to write it down. You don't like it? You don't understand it?
– greedoid
Jul 31 at 14:23
add a comment |Â
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If you want to apply the unconstrained least squares, then you need to parameterise the domain so that it is essentially unconstrained. In particular, this means you need to find a basis of $ker A$. As an aside, note that $A= (5,-1,2)^T (1,2,3)$, so finding such a basis is fairly straightforward.
– copper.hat
Jul 27 at 16:11