Pythagorean theorem in linear regression using matrix notation
$begingroup$
I am reading online scribe notes on linear regression and I am being quite confused (since did not do linear algebra for a while) on how Pythagorean theorem is being applied here:
First of all I am confused of what "cost" is and then how we get from
$$arg min_x ||A(x-x') - bb'||_2^2 $$ to
$$arg min_x ||A(x-x')||_2^2 - ||b'||_2^2 $$ ? E.g I need some intuition and mathematical explanation of what is happening here and how Pythagorean theorem is used in vector space.
Also why $$arg min_x ||Ax - b||_2 equiv arg min_x ||Ax - b||_2^2 $$?
Thank you!
linear-algebra regression
asked Dec 7 '18 at 19:54
YohanRothYohanRoth
6261714
$endgroup$
add a comment |
$begingroup$
I am reading online scribe notes on linear regression and I am being quite confused (since did not do linear algebra for a while) on how Pythagorean theorem is being applied here:
First of all I am confused of what "cost" is and then how we get from
$$arg min_x ||A(x-x') - bb'||_2^2 $$ to
$$arg min_x ||A(x-x')||_2^2 - ||b'||_2^2 $$ ? E.g I need some intuition and mathematical explanation of what is happening here and how Pythagorean theorem is used in vector space.
Also why $$arg min_x ||Ax - b||_2 equiv arg min_x ||Ax - b||_2^2 $$?
Thank you!
linear-algebra regression
asked Dec 7 '18 at 19:54
YohanRothYohanRoth
6261714
$endgroup$
add a comment |
$begingroup$
I am reading online scribe notes on linear regression and I am being quite confused (since did not do linear algebra for a while) on how Pythagorean theorem is being applied here:
First of all I am confused of what "cost" is and then how we get from
$$arg min_x ||A(x-x') - bb'||_2^2 $$ to
$$arg min_x ||A(x-x')||_2^2 - ||b'||_2^2 $$ ? E.g I need some intuition and mathematical explanation of what is happening here and how Pythagorean theorem is used in vector space.
Also why $$arg min_x ||Ax - b||_2 equiv arg min_x ||Ax - b||_2^2 $$?
Thank you!
linear-algebra regression
asked Dec 7 '18 at 19:54
YohanRothYohanRoth
6261714
$endgroup$
I am reading online scribe notes on linear regression and I am being quite confused (since did not do linear algebra for a while) on how Pythagorean theorem is being applied here:
First of all I am confused of what "cost" is and then how we get from
$$arg min_x ||A(x-x') - bb'||_2^2 $$ to
$$arg min_x ||A(x-x')||_2^2 - ||b'||_2^2 $$ ? E.g I need some intuition and mathematical explanation of what is happening here and how Pythagorean theorem is used in vector space.
Also why $$arg min_x ||Ax - b||_2 equiv arg min_x ||Ax - b||_2^2 $$?
Thank you!
linear-algebra regression
linear-algebra regression
asked Dec 7 '18 at 19:54
YohanRothYohanRoth
6261714
asked Dec 7 '18 at 19:54
YohanRothYohanRoth
6261714
asked Dec 7 '18 at 19:54
YohanRothYohanRoth
6261714
asked Dec 7 '18 at 19:54
YohanRothYohanRoth
6261714
asked Dec 7 '18 at 19:54
YohanRothYohanRoth
6261714
6261714
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The cost is $argmin_x ||Ax-b||_2^2$, and one aims at finding a minimizer $x$.
The "Pythagorean theorem" refers to the fact that
$$||x+y||_2^2 = ||x||_2^2 + ||y||_2^2$$ whenever $x perp y$.
In general (Hilbert spaces over $Bbb{C}$), we have
begin{align}
||Ax-b||_2^2 &= langle A(x-x') - b', A(x-x') - b' rangle \
&= langle A(x-x'), A(x-x')rangle - 2mathop{{rm Re}}(langle A(x-x'), b'rangle) + langle b',b'rangle \
&= ||A(x-x')||_2^2 + 0 + ||b'||_2^2 tag{$b' perp A(x-x')$} \
&= ||A(x-x')||_2^2 + ||b'||_2^2
end{align}
To address that last problem, observe that for any strictly increasing function $f$, $x < y iff f(x) < f(y)$ by defintion, so $f(p(x))< f(p(y)) iff p(x) < p(y)$. Take $f(x) = x^2$ on $[0,infty)$ to finish the argument.
answered Dec 7 '18 at 20:21
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
$endgroup$
$begingroup$
I am not very clear on this math notation. What is x' or b'? What is Re? )
$endgroup$
– YohanRoth
Dec 7 '18 at 21:10
$begingroup$
I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 7 '18 at 21:15
$begingroup$
one more question . Why b = Ax' + b' must be orthogonal to col space of A?
$endgroup$
– YohanRoth
Dec 7 '18 at 21:28
$begingroup$
@YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 8 '18 at 9:11
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The cost is $argmin_x ||Ax-b||_2^2$, and one aims at finding a minimizer $x$.
The "Pythagorean theorem" refers to the fact that
$$||x+y||_2^2 = ||x||_2^2 + ||y||_2^2$$ whenever $x perp y$.
In general (Hilbert spaces over $Bbb{C}$), we have
begin{align}
||Ax-b||_2^2 &= langle A(x-x') - b', A(x-x') - b' rangle \
&= langle A(x-x'), A(x-x')rangle - 2mathop{{rm Re}}(langle A(x-x'), b'rangle) + langle b',b'rangle \
&= ||A(x-x')||_2^2 + 0 + ||b'||_2^2 tag{$b' perp A(x-x')$} \
&= ||A(x-x')||_2^2 + ||b'||_2^2
end{align}
To address that last problem, observe that for any strictly increasing function $f$, $x < y iff f(x) < f(y)$ by defintion, so $f(p(x))< f(p(y)) iff p(x) < p(y)$. Take $f(x) = x^2$ on $[0,infty)$ to finish the argument.
answered Dec 7 '18 at 20:21
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
$endgroup$
$begingroup$
I am not very clear on this math notation. What is x' or b'? What is Re? )
$endgroup$
– YohanRoth
Dec 7 '18 at 21:10
$begingroup$
I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 7 '18 at 21:15
$begingroup$
one more question . Why b = Ax' + b' must be orthogonal to col space of A?
$endgroup$
– YohanRoth
Dec 7 '18 at 21:28
$begingroup$
@YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 8 '18 at 9:11
add a comment |
$begingroup$
The cost is $argmin_x ||Ax-b||_2^2$, and one aims at finding a minimizer $x$.
The "Pythagorean theorem" refers to the fact that
$$||x+y||_2^2 = ||x||_2^2 + ||y||_2^2$$ whenever $x perp y$.
In general (Hilbert spaces over $Bbb{C}$), we have
begin{align}
||Ax-b||_2^2 &= langle A(x-x') - b', A(x-x') - b' rangle \
&= langle A(x-x'), A(x-x')rangle - 2mathop{{rm Re}}(langle A(x-x'), b'rangle) + langle b',b'rangle \
&= ||A(x-x')||_2^2 + 0 + ||b'||_2^2 tag{$b' perp A(x-x')$} \
&= ||A(x-x')||_2^2 + ||b'||_2^2
end{align}
To address that last problem, observe that for any strictly increasing function $f$, $x < y iff f(x) < f(y)$ by defintion, so $f(p(x))< f(p(y)) iff p(x) < p(y)$. Take $f(x) = x^2$ on $[0,infty)$ to finish the argument.
answered Dec 7 '18 at 20:21
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
$endgroup$
$begingroup$
I am not very clear on this math notation. What is x' or b'? What is Re? )
$endgroup$
– YohanRoth
Dec 7 '18 at 21:10
$begingroup$
I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 7 '18 at 21:15
$begingroup$
one more question . Why b = Ax' + b' must be orthogonal to col space of A?
$endgroup$
– YohanRoth
Dec 7 '18 at 21:28
$begingroup$
@YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 8 '18 at 9:11
add a comment |
$begingroup$
The cost is $argmin_x ||Ax-b||_2^2$, and one aims at finding a minimizer $x$.
The "Pythagorean theorem" refers to the fact that
$$||x+y||_2^2 = ||x||_2^2 + ||y||_2^2$$ whenever $x perp y$.
In general (Hilbert spaces over $Bbb{C}$), we have
begin{align}
||Ax-b||_2^2 &= langle A(x-x') - b', A(x-x') - b' rangle \
&= langle A(x-x'), A(x-x')rangle - 2mathop{{rm Re}}(langle A(x-x'), b'rangle) + langle b',b'rangle \
&= ||A(x-x')||_2^2 + 0 + ||b'||_2^2 tag{$b' perp A(x-x')$} \
&= ||A(x-x')||_2^2 + ||b'||_2^2
end{align}
To address that last problem, observe that for any strictly increasing function $f$, $x < y iff f(x) < f(y)$ by defintion, so $f(p(x))< f(p(y)) iff p(x) < p(y)$. Take $f(x) = x^2$ on $[0,infty)$ to finish the argument.
answered Dec 7 '18 at 20:21
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
$endgroup$
The cost is $argmin_x ||Ax-b||_2^2$, and one aims at finding a minimizer $x$.
The "Pythagorean theorem" refers to the fact that
$$||x+y||_2^2 = ||x||_2^2 + ||y||_2^2$$ whenever $x perp y$.
In general (Hilbert spaces over $Bbb{C}$), we have
begin{align}
||Ax-b||_2^2 &= langle A(x-x') - b', A(x-x') - b' rangle \
&= langle A(x-x'), A(x-x')rangle - 2mathop{{rm Re}}(langle A(x-x'), b'rangle) + langle b',b'rangle \
&= ||A(x-x')||_2^2 + 0 + ||b'||_2^2 tag{$b' perp A(x-x')$} \
&= ||A(x-x')||_2^2 + ||b'||_2^2
end{align}
To address that last problem, observe that for any strictly increasing function $f$, $x < y iff f(x) < f(y)$ by defintion, so $f(p(x))< f(p(y)) iff p(x) < p(y)$. Take $f(x) = x^2$ on $[0,infty)$ to finish the argument.
answered Dec 7 '18 at 20:21
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Dec 8 '18 at 9:09
answered Dec 7 '18 at 20:21
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Dec 7 '18 at 20:21
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Dec 7 '18 at 20:21
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
$begingroup$
I am not very clear on this math notation. What is x' or b'? What is Re? )
$endgroup$
– YohanRoth
Dec 7 '18 at 21:10
$begingroup$
I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 7 '18 at 21:15
$begingroup$
one more question . Why b = Ax' + b' must be orthogonal to col space of A?
$endgroup$
– YohanRoth
Dec 7 '18 at 21:28
$begingroup$
@YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 8 '18 at 9:11
add a comment |
$begingroup$
I am not very clear on this math notation. What is x' or b'? What is Re? )
$endgroup$
– YohanRoth
Dec 7 '18 at 21:10
$begingroup$
I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 7 '18 at 21:15
$begingroup$
one more question . Why b = Ax' + b' must be orthogonal to col space of A?
$endgroup$
– YohanRoth
Dec 7 '18 at 21:28
$begingroup$
@YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 8 '18 at 9:11
$begingroup$
I am not very clear on this math notation. What is x' or b'? What is Re? )
$endgroup$
– YohanRoth
Dec 7 '18 at 21:10
$begingroup$
I am not very clear on this math notation. What is x' or b'? What is Re? )
$endgroup$
– YohanRoth
Dec 7 '18 at 21:10
$begingroup$
I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 7 '18 at 21:15
$begingroup$
I use $x'$ and $b'$ as in your notes. $Re$ means real part, but that's irrelevant to this question since you're dealing with real-valued data. I put this so that it's also true in Hilbert spaces.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 7 '18 at 21:15
$begingroup$
one more question . Why b = Ax' + b' must be orthogonal to col space of A?
$endgroup$
– YohanRoth
Dec 7 '18 at 21:28
$begingroup$
one more question . Why b = Ax' + b' must be orthogonal to col space of A?
$endgroup$
– YohanRoth
Dec 7 '18 at 21:28
$begingroup$
@YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 8 '18 at 9:11
$begingroup$
@YohanRoth Sorry for late reply. I've made a typo. It should be $b'$ instead, as in the notes.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 8 '18 at 9:11
add a comment |
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