Computing least square solution when eigenvalue and eigenvectors are known.












1












$begingroup$


Suppose a matrix $A$ has eigenvalues 0, 3, 7 and eigenvectors $mathbf{u, v, w,}$ respectively. Find the least square minimum length solution for $Amathbf{x} = mathbf{u+v+w}$.



This was on our engineering math final exam last year and we've tried some techniques about Moore-Penrose pseudoinverse, which didn't seem to work. Can someone help?






linear-algebra







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  • $begingroup$
    What size is $A $?
    $endgroup$
    – AnyAD
    Dec 16 '18 at 6:42










  • $begingroup$
    A is 3 by 3 matrix. All of eigenvalues and eigenvectors of A is known as above.
    $endgroup$
    – Gratus
    Dec 16 '18 at 6:46
















1












$begingroup$


Suppose a matrix $A$ has eigenvalues 0, 3, 7 and eigenvectors $mathbf{u, v, w,}$ respectively. Find the least square minimum length solution for $Amathbf{x} = mathbf{u+v+w}$.



This was on our engineering math final exam last year and we've tried some techniques about Moore-Penrose pseudoinverse, which didn't seem to work. Can someone help?






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    What size is $A $?
    $endgroup$
    – AnyAD
    Dec 16 '18 at 6:42










  • $begingroup$
    A is 3 by 3 matrix. All of eigenvalues and eigenvectors of A is known as above.
    $endgroup$
    – Gratus
    Dec 16 '18 at 6:46














1












1








1


1



$begingroup$


Suppose a matrix $A$ has eigenvalues 0, 3, 7 and eigenvectors $mathbf{u, v, w,}$ respectively. Find the least square minimum length solution for $Amathbf{x} = mathbf{u+v+w}$.



This was on our engineering math final exam last year and we've tried some techniques about Moore-Penrose pseudoinverse, which didn't seem to work. Can someone help?






linear-algebra







share|cite|improve this question









$endgroup$




Suppose a matrix $A$ has eigenvalues 0, 3, 7 and eigenvectors $mathbf{u, v, w,}$ respectively. Find the least square minimum length solution for $Amathbf{x} = mathbf{u+v+w}$.



This was on our engineering math final exam last year and we've tried some techniques about Moore-Penrose pseudoinverse, which didn't seem to work. Can someone help?






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 16 '18 at 6:37









GratusGratus

104




104












  • $begingroup$
    What size is $A $?
    $endgroup$
    – AnyAD
    Dec 16 '18 at 6:42










  • $begingroup$
    A is 3 by 3 matrix. All of eigenvalues and eigenvectors of A is known as above.
    $endgroup$
    – Gratus
    Dec 16 '18 at 6:46


















  • $begingroup$
    What size is $A $?
    $endgroup$
    – AnyAD
    Dec 16 '18 at 6:42










  • $begingroup$
    A is 3 by 3 matrix. All of eigenvalues and eigenvectors of A is known as above.
    $endgroup$
    – Gratus
    Dec 16 '18 at 6:46
















$begingroup$
What size is $A $?
$endgroup$
– AnyAD
Dec 16 '18 at 6:42




$begingroup$
What size is $A $?
$endgroup$
– AnyAD
Dec 16 '18 at 6:42












$begingroup$
A is 3 by 3 matrix. All of eigenvalues and eigenvectors of A is known as above.
$endgroup$
– Gratus
Dec 16 '18 at 6:46




$begingroup$
A is 3 by 3 matrix. All of eigenvalues and eigenvectors of A is known as above.
$endgroup$
– Gratus
Dec 16 '18 at 6:46










3 Answers
3






active

oldest

votes


















0












$begingroup$

The problem here is that there is no such vector $x$ that satisfies
$$
Ax = u+v+w.$$ We can see this easily from the fact that
$$operatorname{ran} A =operatorname{span}{v,w}.$$



$textbf{EDIT:}$ Our problem can be written as follows. First, find $y in operatorname{ran}A$ (that is, $y=Ax$ for some $x$)
such that
$$
lVert y-(u+v+w)rVert leq lVert z-(u+v+w)rVert,quadforall zinoperatorname{ran}A.
$$ Then find $xinmathbb{R}^3$ such that $Ax=y$ and
$$
lVert xrVertleq lVert x'rVert, quadforall x':Ax'=y.
$$



Firstly, the solution $y$ is given by the orthogonal projection $P(u+v+w)$ of $u+v+w$ onto $operatorname{ran} A$. What we need to do is to compute $Pu$ since $P(v+w) = v+w$. To do this, let
$$
Pu = alpha v +beta w.$$ It should be that
$$
u-Pu = u-(alpha v +beta w) perp v,w.
$$ This gives a system of equations about $alpha,beta$:
$$
langle u,vrangle=alpha langle v,v rangle +beta langle w,v rangle,
$$ and
$$
langle u,wrangle= alphalangle v,w rangle+ langle w,w rangle.
$$ (If $A$ is normal, then we have $langle u,vrangle=langle v,wrangle=langle w,vrangle=0$. But this is not true in general.)



Solving this equation gives $y =(alpha+1)v + (beta+1)w$. Finally, note that
$$
A^{-1}({y}) = {frac{(alpha+1)}{3}v+frac{(beta+1)}{7}w+gamma u;|;gammainmathbb{R}}.$$
To minimize $lVert x rVert$ over $xin A^{-1}({y})$, it must be that
$$
x =frac{(alpha+1)}{3}v+frac{(beta+1)}{7}w+gamma u perp u.
$$
This gives us
$$
x = frac{(alpha+1)}{3}left(v-frac{langle v,urangle}{lVert u rVert^2}uright) + frac{(beta+1)}{7}left(w-frac{langle w,urangle}{lVert u rVert^2}uright).
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    From the fact that eigenvalue is zero, this seems trivial. How should I use this fact to compute least square solution?
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:50










  • $begingroup$
    I meant there is no solution of $Ax = u+v+w$. Perhaps I'm missing something. What is your definition of least squre solution?
    $endgroup$
    – Song
    Dec 16 '18 at 12:52










  • $begingroup$
    least square is defined as a vector $x$ such that $|Ax-b|$ is minimized, as I know.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:54












  • $begingroup$
    So basically what I know is that I have to find x which minimizes $||Ax-(u+v+w)||$ from given conditions. Additionally, if we have more than one solution as a least square solution, we should find out which one has minimum length. Formally, $||x||$ should be also minimized. Some of my friends are insisting that solution must be v/3+w/7, but we couldn't prove it. Btw, thanks for your concern :)
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:57










  • $begingroup$
    So the minimization is two-folded. I think I can help.
    $endgroup$
    – Song
    Dec 16 '18 at 12:59



















0












$begingroup$

Consider the basis made up of the eigenvectors, say $B$. Then use the fact that $A$ is diagonisable. Then you can appky least squares formula to $D[x]_B=[Ax]_B=[1 1 1]^T$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for a lesson. However, I still don't really get how am I supposed to apply least squares formula to $PD[x]_B$ after we changed basis to $u,v,w$. Would you mind elaborating a bit more? It would be very helpful.
    $endgroup$
    – Gratus
    Dec 16 '18 at 7:01










  • $begingroup$
    @Gratus See here math.stackexchange.com/questions/72222/…
    $endgroup$
    – AnyAD
    Dec 16 '18 at 8:55










  • $begingroup$
    In this case, it seems $A^T A$ is not invertible. Though I think I got your key idea. Thanks.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:49



















0












$begingroup$

$Ax$ owns to the set $E$ generated by $v$ and $w$.



Let us call $u_0$ the projection of $u$ on $E$.



Then $u =u_0+u_1$ with $u_1$ orthogonal to $E$.



The best you can obtain is $Ax = u_0 + v + w$, projection theorem.

As $u_0$ owns to $E$, it can be represented as $u_0 =alpha v + beta w$, and one solution is
$$ x = frac{1+alpha}{3}v + frac{1+beta}{7}w $$



This simple solution minimizes the error on the output.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can this solution written in forms of $v$ and $w$? Some of my friends are insisting that solution must be $mathbf{v}/3+mathbf{w}/7$, but we couldn't prove it.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:52










  • $begingroup$
    I will try to complete it later
    $endgroup$
    – Damien
    Dec 16 '18 at 13:09











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The problem here is that there is no such vector $x$ that satisfies
$$
Ax = u+v+w.$$ We can see this easily from the fact that
$$operatorname{ran} A =operatorname{span}{v,w}.$$



$textbf{EDIT:}$ Our problem can be written as follows. First, find $y in operatorname{ran}A$ (that is, $y=Ax$ for some $x$)
such that
$$
lVert y-(u+v+w)rVert leq lVert z-(u+v+w)rVert,quadforall zinoperatorname{ran}A.
$$ Then find $xinmathbb{R}^3$ such that $Ax=y$ and
$$
lVert xrVertleq lVert x'rVert, quadforall x':Ax'=y.
$$



Firstly, the solution $y$ is given by the orthogonal projection $P(u+v+w)$ of $u+v+w$ onto $operatorname{ran} A$. What we need to do is to compute $Pu$ since $P(v+w) = v+w$. To do this, let
$$
Pu = alpha v +beta w.$$ It should be that
$$
u-Pu = u-(alpha v +beta w) perp v,w.
$$ This gives a system of equations about $alpha,beta$:
$$
langle u,vrangle=alpha langle v,v rangle +beta langle w,v rangle,
$$ and
$$
langle u,wrangle= alphalangle v,w rangle+ langle w,w rangle.
$$ (If $A$ is normal, then we have $langle u,vrangle=langle v,wrangle=langle w,vrangle=0$. But this is not true in general.)



Solving this equation gives $y =(alpha+1)v + (beta+1)w$. Finally, note that
$$
A^{-1}({y}) = {frac{(alpha+1)}{3}v+frac{(beta+1)}{7}w+gamma u;|;gammainmathbb{R}}.$$
To minimize $lVert x rVert$ over $xin A^{-1}({y})$, it must be that
$$
x =frac{(alpha+1)}{3}v+frac{(beta+1)}{7}w+gamma u perp u.
$$
This gives us
$$
x = frac{(alpha+1)}{3}left(v-frac{langle v,urangle}{lVert u rVert^2}uright) + frac{(beta+1)}{7}left(w-frac{langle w,urangle}{lVert u rVert^2}uright).
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    From the fact that eigenvalue is zero, this seems trivial. How should I use this fact to compute least square solution?
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:50










  • $begingroup$
    I meant there is no solution of $Ax = u+v+w$. Perhaps I'm missing something. What is your definition of least squre solution?
    $endgroup$
    – Song
    Dec 16 '18 at 12:52










  • $begingroup$
    least square is defined as a vector $x$ such that $|Ax-b|$ is minimized, as I know.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:54












  • $begingroup$
    So basically what I know is that I have to find x which minimizes $||Ax-(u+v+w)||$ from given conditions. Additionally, if we have more than one solution as a least square solution, we should find out which one has minimum length. Formally, $||x||$ should be also minimized. Some of my friends are insisting that solution must be v/3+w/7, but we couldn't prove it. Btw, thanks for your concern :)
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:57










  • $begingroup$
    So the minimization is two-folded. I think I can help.
    $endgroup$
    – Song
    Dec 16 '18 at 12:59
















0












$begingroup$

The problem here is that there is no such vector $x$ that satisfies
$$
Ax = u+v+w.$$ We can see this easily from the fact that
$$operatorname{ran} A =operatorname{span}{v,w}.$$



$textbf{EDIT:}$ Our problem can be written as follows. First, find $y in operatorname{ran}A$ (that is, $y=Ax$ for some $x$)
such that
$$
lVert y-(u+v+w)rVert leq lVert z-(u+v+w)rVert,quadforall zinoperatorname{ran}A.
$$ Then find $xinmathbb{R}^3$ such that $Ax=y$ and
$$
lVert xrVertleq lVert x'rVert, quadforall x':Ax'=y.
$$



Firstly, the solution $y$ is given by the orthogonal projection $P(u+v+w)$ of $u+v+w$ onto $operatorname{ran} A$. What we need to do is to compute $Pu$ since $P(v+w) = v+w$. To do this, let
$$
Pu = alpha v +beta w.$$ It should be that
$$
u-Pu = u-(alpha v +beta w) perp v,w.
$$ This gives a system of equations about $alpha,beta$:
$$
langle u,vrangle=alpha langle v,v rangle +beta langle w,v rangle,
$$ and
$$
langle u,wrangle= alphalangle v,w rangle+ langle w,w rangle.
$$ (If $A$ is normal, then we have $langle u,vrangle=langle v,wrangle=langle w,vrangle=0$. But this is not true in general.)



Solving this equation gives $y =(alpha+1)v + (beta+1)w$. Finally, note that
$$
A^{-1}({y}) = {frac{(alpha+1)}{3}v+frac{(beta+1)}{7}w+gamma u;|;gammainmathbb{R}}.$$
To minimize $lVert x rVert$ over $xin A^{-1}({y})$, it must be that
$$
x =frac{(alpha+1)}{3}v+frac{(beta+1)}{7}w+gamma u perp u.
$$
This gives us
$$
x = frac{(alpha+1)}{3}left(v-frac{langle v,urangle}{lVert u rVert^2}uright) + frac{(beta+1)}{7}left(w-frac{langle w,urangle}{lVert u rVert^2}uright).
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    From the fact that eigenvalue is zero, this seems trivial. How should I use this fact to compute least square solution?
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:50










  • $begingroup$
    I meant there is no solution of $Ax = u+v+w$. Perhaps I'm missing something. What is your definition of least squre solution?
    $endgroup$
    – Song
    Dec 16 '18 at 12:52










  • $begingroup$
    least square is defined as a vector $x$ such that $|Ax-b|$ is minimized, as I know.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:54












  • $begingroup$
    So basically what I know is that I have to find x which minimizes $||Ax-(u+v+w)||$ from given conditions. Additionally, if we have more than one solution as a least square solution, we should find out which one has minimum length. Formally, $||x||$ should be also minimized. Some of my friends are insisting that solution must be v/3+w/7, but we couldn't prove it. Btw, thanks for your concern :)
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:57










  • $begingroup$
    So the minimization is two-folded. I think I can help.
    $endgroup$
    – Song
    Dec 16 '18 at 12:59














0












0








0





$begingroup$

The problem here is that there is no such vector $x$ that satisfies
$$
Ax = u+v+w.$$ We can see this easily from the fact that
$$operatorname{ran} A =operatorname{span}{v,w}.$$



$textbf{EDIT:}$ Our problem can be written as follows. First, find $y in operatorname{ran}A$ (that is, $y=Ax$ for some $x$)
such that
$$
lVert y-(u+v+w)rVert leq lVert z-(u+v+w)rVert,quadforall zinoperatorname{ran}A.
$$ Then find $xinmathbb{R}^3$ such that $Ax=y$ and
$$
lVert xrVertleq lVert x'rVert, quadforall x':Ax'=y.
$$



Firstly, the solution $y$ is given by the orthogonal projection $P(u+v+w)$ of $u+v+w$ onto $operatorname{ran} A$. What we need to do is to compute $Pu$ since $P(v+w) = v+w$. To do this, let
$$
Pu = alpha v +beta w.$$ It should be that
$$
u-Pu = u-(alpha v +beta w) perp v,w.
$$ This gives a system of equations about $alpha,beta$:
$$
langle u,vrangle=alpha langle v,v rangle +beta langle w,v rangle,
$$ and
$$
langle u,wrangle= alphalangle v,w rangle+ langle w,w rangle.
$$ (If $A$ is normal, then we have $langle u,vrangle=langle v,wrangle=langle w,vrangle=0$. But this is not true in general.)



Solving this equation gives $y =(alpha+1)v + (beta+1)w$. Finally, note that
$$
A^{-1}({y}) = {frac{(alpha+1)}{3}v+frac{(beta+1)}{7}w+gamma u;|;gammainmathbb{R}}.$$
To minimize $lVert x rVert$ over $xin A^{-1}({y})$, it must be that
$$
x =frac{(alpha+1)}{3}v+frac{(beta+1)}{7}w+gamma u perp u.
$$
This gives us
$$
x = frac{(alpha+1)}{3}left(v-frac{langle v,urangle}{lVert u rVert^2}uright) + frac{(beta+1)}{7}left(w-frac{langle w,urangle}{lVert u rVert^2}uright).
$$






share|cite|improve this answer











$endgroup$



The problem here is that there is no such vector $x$ that satisfies
$$
Ax = u+v+w.$$ We can see this easily from the fact that
$$operatorname{ran} A =operatorname{span}{v,w}.$$



$textbf{EDIT:}$ Our problem can be written as follows. First, find $y in operatorname{ran}A$ (that is, $y=Ax$ for some $x$)
such that
$$
lVert y-(u+v+w)rVert leq lVert z-(u+v+w)rVert,quadforall zinoperatorname{ran}A.
$$ Then find $xinmathbb{R}^3$ such that $Ax=y$ and
$$
lVert xrVertleq lVert x'rVert, quadforall x':Ax'=y.
$$



Firstly, the solution $y$ is given by the orthogonal projection $P(u+v+w)$ of $u+v+w$ onto $operatorname{ran} A$. What we need to do is to compute $Pu$ since $P(v+w) = v+w$. To do this, let
$$
Pu = alpha v +beta w.$$ It should be that
$$
u-Pu = u-(alpha v +beta w) perp v,w.
$$ This gives a system of equations about $alpha,beta$:
$$
langle u,vrangle=alpha langle v,v rangle +beta langle w,v rangle,
$$ and
$$
langle u,wrangle= alphalangle v,w rangle+ langle w,w rangle.
$$ (If $A$ is normal, then we have $langle u,vrangle=langle v,wrangle=langle w,vrangle=0$. But this is not true in general.)



Solving this equation gives $y =(alpha+1)v + (beta+1)w$. Finally, note that
$$
A^{-1}({y}) = {frac{(alpha+1)}{3}v+frac{(beta+1)}{7}w+gamma u;|;gammainmathbb{R}}.$$
To minimize $lVert x rVert$ over $xin A^{-1}({y})$, it must be that
$$
x =frac{(alpha+1)}{3}v+frac{(beta+1)}{7}w+gamma u perp u.
$$
This gives us
$$
x = frac{(alpha+1)}{3}left(v-frac{langle v,urangle}{lVert u rVert^2}uright) + frac{(beta+1)}{7}left(w-frac{langle w,urangle}{lVert u rVert^2}uright).
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 17:41

























answered Dec 16 '18 at 8:58









SongSong

16.8k21145




16.8k21145












  • $begingroup$
    From the fact that eigenvalue is zero, this seems trivial. How should I use this fact to compute least square solution?
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:50










  • $begingroup$
    I meant there is no solution of $Ax = u+v+w$. Perhaps I'm missing something. What is your definition of least squre solution?
    $endgroup$
    – Song
    Dec 16 '18 at 12:52










  • $begingroup$
    least square is defined as a vector $x$ such that $|Ax-b|$ is minimized, as I know.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:54












  • $begingroup$
    So basically what I know is that I have to find x which minimizes $||Ax-(u+v+w)||$ from given conditions. Additionally, if we have more than one solution as a least square solution, we should find out which one has minimum length. Formally, $||x||$ should be also minimized. Some of my friends are insisting that solution must be v/3+w/7, but we couldn't prove it. Btw, thanks for your concern :)
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:57










  • $begingroup$
    So the minimization is two-folded. I think I can help.
    $endgroup$
    – Song
    Dec 16 '18 at 12:59


















  • $begingroup$
    From the fact that eigenvalue is zero, this seems trivial. How should I use this fact to compute least square solution?
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:50










  • $begingroup$
    I meant there is no solution of $Ax = u+v+w$. Perhaps I'm missing something. What is your definition of least squre solution?
    $endgroup$
    – Song
    Dec 16 '18 at 12:52










  • $begingroup$
    least square is defined as a vector $x$ such that $|Ax-b|$ is minimized, as I know.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:54












  • $begingroup$
    So basically what I know is that I have to find x which minimizes $||Ax-(u+v+w)||$ from given conditions. Additionally, if we have more than one solution as a least square solution, we should find out which one has minimum length. Formally, $||x||$ should be also minimized. Some of my friends are insisting that solution must be v/3+w/7, but we couldn't prove it. Btw, thanks for your concern :)
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:57










  • $begingroup$
    So the minimization is two-folded. I think I can help.
    $endgroup$
    – Song
    Dec 16 '18 at 12:59
















$begingroup$
From the fact that eigenvalue is zero, this seems trivial. How should I use this fact to compute least square solution?
$endgroup$
– Gratus
Dec 16 '18 at 12:50




$begingroup$
From the fact that eigenvalue is zero, this seems trivial. How should I use this fact to compute least square solution?
$endgroup$
– Gratus
Dec 16 '18 at 12:50












$begingroup$
I meant there is no solution of $Ax = u+v+w$. Perhaps I'm missing something. What is your definition of least squre solution?
$endgroup$
– Song
Dec 16 '18 at 12:52




$begingroup$
I meant there is no solution of $Ax = u+v+w$. Perhaps I'm missing something. What is your definition of least squre solution?
$endgroup$
– Song
Dec 16 '18 at 12:52












$begingroup$
least square is defined as a vector $x$ such that $|Ax-b|$ is minimized, as I know.
$endgroup$
– Gratus
Dec 16 '18 at 12:54






$begingroup$
least square is defined as a vector $x$ such that $|Ax-b|$ is minimized, as I know.
$endgroup$
– Gratus
Dec 16 '18 at 12:54














$begingroup$
So basically what I know is that I have to find x which minimizes $||Ax-(u+v+w)||$ from given conditions. Additionally, if we have more than one solution as a least square solution, we should find out which one has minimum length. Formally, $||x||$ should be also minimized. Some of my friends are insisting that solution must be v/3+w/7, but we couldn't prove it. Btw, thanks for your concern :)
$endgroup$
– Gratus
Dec 16 '18 at 12:57




$begingroup$
So basically what I know is that I have to find x which minimizes $||Ax-(u+v+w)||$ from given conditions. Additionally, if we have more than one solution as a least square solution, we should find out which one has minimum length. Formally, $||x||$ should be also minimized. Some of my friends are insisting that solution must be v/3+w/7, but we couldn't prove it. Btw, thanks for your concern :)
$endgroup$
– Gratus
Dec 16 '18 at 12:57












$begingroup$
So the minimization is two-folded. I think I can help.
$endgroup$
– Song
Dec 16 '18 at 12:59




$begingroup$
So the minimization is two-folded. I think I can help.
$endgroup$
– Song
Dec 16 '18 at 12:59











0












$begingroup$

Consider the basis made up of the eigenvectors, say $B$. Then use the fact that $A$ is diagonisable. Then you can appky least squares formula to $D[x]_B=[Ax]_B=[1 1 1]^T$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for a lesson. However, I still don't really get how am I supposed to apply least squares formula to $PD[x]_B$ after we changed basis to $u,v,w$. Would you mind elaborating a bit more? It would be very helpful.
    $endgroup$
    – Gratus
    Dec 16 '18 at 7:01










  • $begingroup$
    @Gratus See here math.stackexchange.com/questions/72222/…
    $endgroup$
    – AnyAD
    Dec 16 '18 at 8:55










  • $begingroup$
    In this case, it seems $A^T A$ is not invertible. Though I think I got your key idea. Thanks.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:49
















0












$begingroup$

Consider the basis made up of the eigenvectors, say $B$. Then use the fact that $A$ is diagonisable. Then you can appky least squares formula to $D[x]_B=[Ax]_B=[1 1 1]^T$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for a lesson. However, I still don't really get how am I supposed to apply least squares formula to $PD[x]_B$ after we changed basis to $u,v,w$. Would you mind elaborating a bit more? It would be very helpful.
    $endgroup$
    – Gratus
    Dec 16 '18 at 7:01










  • $begingroup$
    @Gratus See here math.stackexchange.com/questions/72222/…
    $endgroup$
    – AnyAD
    Dec 16 '18 at 8:55










  • $begingroup$
    In this case, it seems $A^T A$ is not invertible. Though I think I got your key idea. Thanks.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:49














0












0








0





$begingroup$

Consider the basis made up of the eigenvectors, say $B$. Then use the fact that $A$ is diagonisable. Then you can appky least squares formula to $D[x]_B=[Ax]_B=[1 1 1]^T$.






share|cite|improve this answer











$endgroup$



Consider the basis made up of the eigenvectors, say $B$. Then use the fact that $A$ is diagonisable. Then you can appky least squares formula to $D[x]_B=[Ax]_B=[1 1 1]^T$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 8:41

























answered Dec 16 '18 at 6:54









AnyADAnyAD

2,098812




2,098812












  • $begingroup$
    Thanks for a lesson. However, I still don't really get how am I supposed to apply least squares formula to $PD[x]_B$ after we changed basis to $u,v,w$. Would you mind elaborating a bit more? It would be very helpful.
    $endgroup$
    – Gratus
    Dec 16 '18 at 7:01










  • $begingroup$
    @Gratus See here math.stackexchange.com/questions/72222/…
    $endgroup$
    – AnyAD
    Dec 16 '18 at 8:55










  • $begingroup$
    In this case, it seems $A^T A$ is not invertible. Though I think I got your key idea. Thanks.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:49


















  • $begingroup$
    Thanks for a lesson. However, I still don't really get how am I supposed to apply least squares formula to $PD[x]_B$ after we changed basis to $u,v,w$. Would you mind elaborating a bit more? It would be very helpful.
    $endgroup$
    – Gratus
    Dec 16 '18 at 7:01










  • $begingroup$
    @Gratus See here math.stackexchange.com/questions/72222/…
    $endgroup$
    – AnyAD
    Dec 16 '18 at 8:55










  • $begingroup$
    In this case, it seems $A^T A$ is not invertible. Though I think I got your key idea. Thanks.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:49
















$begingroup$
Thanks for a lesson. However, I still don't really get how am I supposed to apply least squares formula to $PD[x]_B$ after we changed basis to $u,v,w$. Would you mind elaborating a bit more? It would be very helpful.
$endgroup$
– Gratus
Dec 16 '18 at 7:01




$begingroup$
Thanks for a lesson. However, I still don't really get how am I supposed to apply least squares formula to $PD[x]_B$ after we changed basis to $u,v,w$. Would you mind elaborating a bit more? It would be very helpful.
$endgroup$
– Gratus
Dec 16 '18 at 7:01












$begingroup$
@Gratus See here math.stackexchange.com/questions/72222/…
$endgroup$
– AnyAD
Dec 16 '18 at 8:55




$begingroup$
@Gratus See here math.stackexchange.com/questions/72222/…
$endgroup$
– AnyAD
Dec 16 '18 at 8:55












$begingroup$
In this case, it seems $A^T A$ is not invertible. Though I think I got your key idea. Thanks.
$endgroup$
– Gratus
Dec 16 '18 at 12:49




$begingroup$
In this case, it seems $A^T A$ is not invertible. Though I think I got your key idea. Thanks.
$endgroup$
– Gratus
Dec 16 '18 at 12:49











0












$begingroup$

$Ax$ owns to the set $E$ generated by $v$ and $w$.



Let us call $u_0$ the projection of $u$ on $E$.



Then $u =u_0+u_1$ with $u_1$ orthogonal to $E$.



The best you can obtain is $Ax = u_0 + v + w$, projection theorem.

As $u_0$ owns to $E$, it can be represented as $u_0 =alpha v + beta w$, and one solution is
$$ x = frac{1+alpha}{3}v + frac{1+beta}{7}w $$



This simple solution minimizes the error on the output.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can this solution written in forms of $v$ and $w$? Some of my friends are insisting that solution must be $mathbf{v}/3+mathbf{w}/7$, but we couldn't prove it.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:52










  • $begingroup$
    I will try to complete it later
    $endgroup$
    – Damien
    Dec 16 '18 at 13:09
















0












$begingroup$

$Ax$ owns to the set $E$ generated by $v$ and $w$.



Let us call $u_0$ the projection of $u$ on $E$.



Then $u =u_0+u_1$ with $u_1$ orthogonal to $E$.



The best you can obtain is $Ax = u_0 + v + w$, projection theorem.

As $u_0$ owns to $E$, it can be represented as $u_0 =alpha v + beta w$, and one solution is
$$ x = frac{1+alpha}{3}v + frac{1+beta}{7}w $$



This simple solution minimizes the error on the output.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can this solution written in forms of $v$ and $w$? Some of my friends are insisting that solution must be $mathbf{v}/3+mathbf{w}/7$, but we couldn't prove it.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:52










  • $begingroup$
    I will try to complete it later
    $endgroup$
    – Damien
    Dec 16 '18 at 13:09














0












0








0





$begingroup$

$Ax$ owns to the set $E$ generated by $v$ and $w$.



Let us call $u_0$ the projection of $u$ on $E$.



Then $u =u_0+u_1$ with $u_1$ orthogonal to $E$.



The best you can obtain is $Ax = u_0 + v + w$, projection theorem.

As $u_0$ owns to $E$, it can be represented as $u_0 =alpha v + beta w$, and one solution is
$$ x = frac{1+alpha}{3}v + frac{1+beta}{7}w $$



This simple solution minimizes the error on the output.






share|cite|improve this answer











$endgroup$



$Ax$ owns to the set $E$ generated by $v$ and $w$.



Let us call $u_0$ the projection of $u$ on $E$.



Then $u =u_0+u_1$ with $u_1$ orthogonal to $E$.



The best you can obtain is $Ax = u_0 + v + w$, projection theorem.

As $u_0$ owns to $E$, it can be represented as $u_0 =alpha v + beta w$, and one solution is
$$ x = frac{1+alpha}{3}v + frac{1+beta}{7}w $$



This simple solution minimizes the error on the output.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 17:46

























answered Dec 16 '18 at 7:21









DamienDamien

59714




59714












  • $begingroup$
    Can this solution written in forms of $v$ and $w$? Some of my friends are insisting that solution must be $mathbf{v}/3+mathbf{w}/7$, but we couldn't prove it.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:52










  • $begingroup$
    I will try to complete it later
    $endgroup$
    – Damien
    Dec 16 '18 at 13:09


















  • $begingroup$
    Can this solution written in forms of $v$ and $w$? Some of my friends are insisting that solution must be $mathbf{v}/3+mathbf{w}/7$, but we couldn't prove it.
    $endgroup$
    – Gratus
    Dec 16 '18 at 12:52










  • $begingroup$
    I will try to complete it later
    $endgroup$
    – Damien
    Dec 16 '18 at 13:09
















$begingroup$
Can this solution written in forms of $v$ and $w$? Some of my friends are insisting that solution must be $mathbf{v}/3+mathbf{w}/7$, but we couldn't prove it.
$endgroup$
– Gratus
Dec 16 '18 at 12:52




$begingroup$
Can this solution written in forms of $v$ and $w$? Some of my friends are insisting that solution must be $mathbf{v}/3+mathbf{w}/7$, but we couldn't prove it.
$endgroup$
– Gratus
Dec 16 '18 at 12:52












$begingroup$
I will try to complete it later
$endgroup$
– Damien
Dec 16 '18 at 13:09




$begingroup$
I will try to complete it later
$endgroup$
– Damien
Dec 16 '18 at 13:09


















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