Mixing
Least square solution to the system
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I am trying to solve the following problem:
Let $u_1$ and $u_2$ be two orthogonal vectors in ${rm I!R}^n$ and
set $a_1 = u_1$, $a_2 = u_1 + varepsilon u_2$ for $varepsilon>0$.
Let also $A$ be the matrix with columns $a_1$ and $a_2$ and $b$ a
vector linearly independenet of $a_1$ and $a_2$. Least square solution
is discussed here to the system $Ax = b$ as $varepsilonto0$.
(a) Find the matrix $A^top A$, its inverse, and then $hat{x} =
> (A^top A)^{-1}A^top b$ explicitly. Show that $hat{x}$ explodes as
$varepsilonto0$
(b) Find the projection $Ahat{x}$ of $b$ onto $operatorname{col}(A)$
and check that it does not depend on $varepsilon>0$. Explain the
result.
I have assumed, that $A = begin{pmatrix}u_1 & u_1+varepsilon u_2end{pmatrix}$, therefore:
$A^TA=begin{pmatrix}u_1 \ u_1+varepsilon u_2end{pmatrix}begin{pmatrix}u_1 & u_1+varepsilon u_2end{pmatrix}=begin{pmatrix}u_1^2 & u_1(u_1+varepsilon u_2)\u_1(u_1+varepsilon u_2) & (u_1 + varepsilon u_2)^2end{pmatrix}$
But when I try to compute $(A^TA)^{-1}$, determinant becomes zero, what means that matrix is not invertible. What am I doing wrong? Thanks in advance for any hints!
least-squares
asked yesterday
Michael
1014
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up vote
0
down vote
favorite
I am trying to solve the following problem:
Let $u_1$ and $u_2$ be two orthogonal vectors in ${rm I!R}^n$ and
set $a_1 = u_1$, $a_2 = u_1 + varepsilon u_2$ for $varepsilon>0$.
Let also $A$ be the matrix with columns $a_1$ and $a_2$ and $b$ a
vector linearly independenet of $a_1$ and $a_2$. Least square solution
is discussed here to the system $Ax = b$ as $varepsilonto0$.
(a) Find the matrix $A^top A$, its inverse, and then $hat{x} =
> (A^top A)^{-1}A^top b$ explicitly. Show that $hat{x}$ explodes as
$varepsilonto0$
(b) Find the projection $Ahat{x}$ of $b$ onto $operatorname{col}(A)$
and check that it does not depend on $varepsilon>0$. Explain the
result.
I have assumed, that $A = begin{pmatrix}u_1 & u_1+varepsilon u_2end{pmatrix}$, therefore:
$A^TA=begin{pmatrix}u_1 \ u_1+varepsilon u_2end{pmatrix}begin{pmatrix}u_1 & u_1+varepsilon u_2end{pmatrix}=begin{pmatrix}u_1^2 & u_1(u_1+varepsilon u_2)\u_1(u_1+varepsilon u_2) & (u_1 + varepsilon u_2)^2end{pmatrix}$
But when I try to compute $(A^TA)^{-1}$, determinant becomes zero, what means that matrix is not invertible. What am I doing wrong? Thanks in advance for any hints!
least-squares
asked yesterday
Michael
1014
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I guess that $A^T*A$ always has a zero determinant unless A is invertible.
– NoChance
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to solve the following problem:
Let $u_1$ and $u_2$ be two orthogonal vectors in ${rm I!R}^n$ and
set $a_1 = u_1$, $a_2 = u_1 + varepsilon u_2$ for $varepsilon>0$.
Let also $A$ be the matrix with columns $a_1$ and $a_2$ and $b$ a
vector linearly independenet of $a_1$ and $a_2$. Least square solution
is discussed here to the system $Ax = b$ as $varepsilonto0$.
(a) Find the matrix $A^top A$, its inverse, and then $hat{x} =
> (A^top A)^{-1}A^top b$ explicitly. Show that $hat{x}$ explodes as
$varepsilonto0$
(b) Find the projection $Ahat{x}$ of $b$ onto $operatorname{col}(A)$
and check that it does not depend on $varepsilon>0$. Explain the
result.
I have assumed, that $A = begin{pmatrix}u_1 & u_1+varepsilon u_2end{pmatrix}$, therefore:
$A^TA=begin{pmatrix}u_1 \ u_1+varepsilon u_2end{pmatrix}begin{pmatrix}u_1 & u_1+varepsilon u_2end{pmatrix}=begin{pmatrix}u_1^2 & u_1(u_1+varepsilon u_2)\u_1(u_1+varepsilon u_2) & (u_1 + varepsilon u_2)^2end{pmatrix}$
But when I try to compute $(A^TA)^{-1}$, determinant becomes zero, what means that matrix is not invertible. What am I doing wrong? Thanks in advance for any hints!
least-squares
asked yesterday
Michael
1014
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am trying to solve the following problem:
Let $u_1$ and $u_2$ be two orthogonal vectors in ${rm I!R}^n$ and
set $a_1 = u_1$, $a_2 = u_1 + varepsilon u_2$ for $varepsilon>0$.
Let also $A$ be the matrix with columns $a_1$ and $a_2$ and $b$ a
vector linearly independenet of $a_1$ and $a_2$. Least square solution
is discussed here to the system $Ax = b$ as $varepsilonto0$.
(a) Find the matrix $A^top A$, its inverse, and then $hat{x} =
> (A^top A)^{-1}A^top b$ explicitly. Show that $hat{x}$ explodes as
$varepsilonto0$
(b) Find the projection $Ahat{x}$ of $b$ onto $operatorname{col}(A)$
and check that it does not depend on $varepsilon>0$. Explain the
result.
I have assumed, that $A = begin{pmatrix}u_1 & u_1+varepsilon u_2end{pmatrix}$, therefore:
$A^TA=begin{pmatrix}u_1 \ u_1+varepsilon u_2end{pmatrix}begin{pmatrix}u_1 & u_1+varepsilon u_2end{pmatrix}=begin{pmatrix}u_1^2 & u_1(u_1+varepsilon u_2)\u_1(u_1+varepsilon u_2) & (u_1 + varepsilon u_2)^2end{pmatrix}$
But when I try to compute $(A^TA)^{-1}$, determinant becomes zero, what means that matrix is not invertible. What am I doing wrong? Thanks in advance for any hints!
least-squares
least-squares
asked yesterday
Michael
1014
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Michael
1014
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Michael
1014
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Michael
1014
asked yesterday
Michael
1014
1014
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I guess that $A^T*A$ always has a zero determinant unless A is invertible.
– NoChance
yesterday
add a comment |
I guess that $A^T*A$ always has a zero determinant unless A is invertible.
– NoChance
yesterday
I guess that $A^T*A$ always has a zero determinant unless A is invertible.
– NoChance
yesterday
I guess that $A^T*A$ always has a zero determinant unless A is invertible.
– NoChance
yesterday
add a comment |
1 Answer
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Bear in mind what it means to multiply vectors in the first place. Your determinant is $$(u_1cdot u_1) (u_1cdot u_1 + 2varepsilon u_1cdot u_2+varepsilon^2 u_2cdot u_2)-(u_1cdot u_1 + varepsilon u_1cdot u_2)^2=varepsilon^2 (u_1cdot u_1 u_2cdot u_2-(u_1cdot u_2)^2).$$
answered yesterday
J.G.
18.2k21932
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Bear in mind what it means to multiply vectors in the first place. Your determinant is $$(u_1cdot u_1) (u_1cdot u_1 + 2varepsilon u_1cdot u_2+varepsilon^2 u_2cdot u_2)-(u_1cdot u_1 + varepsilon u_1cdot u_2)^2=varepsilon^2 (u_1cdot u_1 u_2cdot u_2-(u_1cdot u_2)^2).$$
answered yesterday
J.G.
18.2k21932
add a comment |
up vote
0
down vote
Bear in mind what it means to multiply vectors in the first place. Your determinant is $$(u_1cdot u_1) (u_1cdot u_1 + 2varepsilon u_1cdot u_2+varepsilon^2 u_2cdot u_2)-(u_1cdot u_1 + varepsilon u_1cdot u_2)^2=varepsilon^2 (u_1cdot u_1 u_2cdot u_2-(u_1cdot u_2)^2).$$
answered yesterday
J.G.
18.2k21932
add a comment |
up vote
0
down vote
up vote
0
down vote
Bear in mind what it means to multiply vectors in the first place. Your determinant is $$(u_1cdot u_1) (u_1cdot u_1 + 2varepsilon u_1cdot u_2+varepsilon^2 u_2cdot u_2)-(u_1cdot u_1 + varepsilon u_1cdot u_2)^2=varepsilon^2 (u_1cdot u_1 u_2cdot u_2-(u_1cdot u_2)^2).$$
answered yesterday
J.G.
18.2k21932
Bear in mind what it means to multiply vectors in the first place. Your determinant is $$(u_1cdot u_1) (u_1cdot u_1 + 2varepsilon u_1cdot u_2+varepsilon^2 u_2cdot u_2)-(u_1cdot u_1 + varepsilon u_1cdot u_2)^2=varepsilon^2 (u_1cdot u_1 u_2cdot u_2-(u_1cdot u_2)^2).$$
answered yesterday
J.G.
18.2k21932
answered yesterday
J.G.
18.2k21932
answered yesterday
J.G.
18.2k21932
answered yesterday
J.G.
18.2k21932
18.2k21932
add a comment |
add a comment |
Michael is a new contributor. Be nice, and check out our Code of Conduct.
Michael is a new contributor. Be nice, and check out our Code of Conduct.
Michael is a new contributor. Be nice, and check out our Code of Conduct.
Michael is a new contributor. Be nice, and check out our Code of Conduct.
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I guess that $A^T*A$ always has a zero determinant unless A is invertible.
– NoChance
yesterday