Ellipsoid representation by PSD matrix and by linear mapping
$begingroup$
Consider the following two representations of ellipsoid:
$$E_1 = {x mid x^TSxleq 1, , S succ 0}$$
and $$E_2 = {y mid y = Ax, , |x|leq 1, , det(A) neq 0}$$
If I want $E_1=E_2$, I can do
$$x^TSx = x^TS^{1/2}S^{1/2}x = |S^{1/2}x|^2_2 leq 1$$
and $$y=Ax longrightarrow A^{-1}y = x longrightarrow |A^{-1}y| = |x| le1 longrightarrow |A^{-1}y|^2_2leq 1$$
So comparing both equations,
$$A^{-1} = S^{1/2}$$
In this case, $A$ has to be symmetric.
However, if I do the following
$$|A^{-1}y|^2_2leq 1 longrightarrow y^T(AA^T)^{-1}y leq 1$$
and let $(AA^T)^{-1}=S$ with $S = QLambda Q^T$, $A$ can be chosen as
$$A = QLambda^{-1/2}$$
In this case, $A$ does not have to be symmetric.
What's wrong with the first approach? Or are both correct?
linear-algebra matrices positive-definite positive-semidefinite
asked Nov 29 '18 at 23:38
sleeve chensleeve chen
3,10641852
$endgroup$
add a comment |
$begingroup$
Consider the following two representations of ellipsoid:
$$E_1 = {x mid x^TSxleq 1, , S succ 0}$$
and $$E_2 = {y mid y = Ax, , |x|leq 1, , det(A) neq 0}$$
If I want $E_1=E_2$, I can do
$$x^TSx = x^TS^{1/2}S^{1/2}x = |S^{1/2}x|^2_2 leq 1$$
and $$y=Ax longrightarrow A^{-1}y = x longrightarrow |A^{-1}y| = |x| le1 longrightarrow |A^{-1}y|^2_2leq 1$$
So comparing both equations,
$$A^{-1} = S^{1/2}$$
In this case, $A$ has to be symmetric.
However, if I do the following
$$|A^{-1}y|^2_2leq 1 longrightarrow y^T(AA^T)^{-1}y leq 1$$
and let $(AA^T)^{-1}=S$ with $S = QLambda Q^T$, $A$ can be chosen as
$$A = QLambda^{-1/2}$$
In this case, $A$ does not have to be symmetric.
What's wrong with the first approach? Or are both correct?
linear-algebra matrices positive-definite positive-semidefinite
asked Nov 29 '18 at 23:38
sleeve chensleeve chen
3,10641852
$endgroup$
add a comment |
$begingroup$
Consider the following two representations of ellipsoid:
$$E_1 = {x mid x^TSxleq 1, , S succ 0}$$
and $$E_2 = {y mid y = Ax, , |x|leq 1, , det(A) neq 0}$$
If I want $E_1=E_2$, I can do
$$x^TSx = x^TS^{1/2}S^{1/2}x = |S^{1/2}x|^2_2 leq 1$$
and $$y=Ax longrightarrow A^{-1}y = x longrightarrow |A^{-1}y| = |x| le1 longrightarrow |A^{-1}y|^2_2leq 1$$
So comparing both equations,
$$A^{-1} = S^{1/2}$$
In this case, $A$ has to be symmetric.
However, if I do the following
$$|A^{-1}y|^2_2leq 1 longrightarrow y^T(AA^T)^{-1}y leq 1$$
and let $(AA^T)^{-1}=S$ with $S = QLambda Q^T$, $A$ can be chosen as
$$A = QLambda^{-1/2}$$
In this case, $A$ does not have to be symmetric.
What's wrong with the first approach? Or are both correct?
linear-algebra matrices positive-definite positive-semidefinite
asked Nov 29 '18 at 23:38
sleeve chensleeve chen
3,10641852
$endgroup$
Consider the following two representations of ellipsoid:
$$E_1 = {x mid x^TSxleq 1, , S succ 0}$$
and $$E_2 = {y mid y = Ax, , |x|leq 1, , det(A) neq 0}$$
If I want $E_1=E_2$, I can do
$$x^TSx = x^TS^{1/2}S^{1/2}x = |S^{1/2}x|^2_2 leq 1$$
and $$y=Ax longrightarrow A^{-1}y = x longrightarrow |A^{-1}y| = |x| le1 longrightarrow |A^{-1}y|^2_2leq 1$$
So comparing both equations,
$$A^{-1} = S^{1/2}$$
In this case, $A$ has to be symmetric.
However, if I do the following
$$|A^{-1}y|^2_2leq 1 longrightarrow y^T(AA^T)^{-1}y leq 1$$
and let $(AA^T)^{-1}=S$ with $S = QLambda Q^T$, $A$ can be chosen as
$$A = QLambda^{-1/2}$$
In this case, $A$ does not have to be symmetric.
What's wrong with the first approach? Or are both correct?
linear-algebra matrices positive-definite positive-semidefinite
linear-algebra matrices positive-definite positive-semidefinite
asked Nov 29 '18 at 23:38
sleeve chensleeve chen
3,10641852
asked Nov 29 '18 at 23:38
sleeve chensleeve chen
3,10641852
edited Nov 30 '18 at 0:05
sleeve chen
asked Nov 29 '18 at 23:38
sleeve chensleeve chen
3,10641852
asked Nov 29 '18 at 23:38
sleeve chensleeve chen
3,10641852
asked Nov 29 '18 at 23:38
sleeve chensleeve chen
3,10641852
3,10641852
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
$S$ does not have to be symmetric, and neither does $A$.
Consider for instance:
$$x^2+2y^2+2xy=
begin{pmatrix}x¥d{pmatrix}begin{pmatrix}1&2\0&2end{pmatrix}begin{pmatrix}x\yend{pmatrix}
=begin{pmatrix}x¥d{pmatrix}begin{pmatrix}1&1\1&2end{pmatrix}begin{pmatrix}x\yend{pmatrix}
$$
It's just that we usually choose $S$ to be symmetric, since then we can do a useful eigenvalue decomposition.
answered Nov 30 '18 at 0:08
I like SerenaI like Serena
4,1771722
$endgroup$
$begingroup$
That said, only the symmetric part of $S$ actually contributes to the quadratic form, which is why one usually assumes that it’s symmetric, anyway.
$endgroup$
– amd
Nov 30 '18 at 0:41
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$S$ does not have to be symmetric, and neither does $A$.
Consider for instance:
$$x^2+2y^2+2xy=
begin{pmatrix}x¥d{pmatrix}begin{pmatrix}1&2\0&2end{pmatrix}begin{pmatrix}x\yend{pmatrix}
=begin{pmatrix}x¥d{pmatrix}begin{pmatrix}1&1\1&2end{pmatrix}begin{pmatrix}x\yend{pmatrix}
$$
It's just that we usually choose $S$ to be symmetric, since then we can do a useful eigenvalue decomposition.
answered Nov 30 '18 at 0:08
I like SerenaI like Serena
4,1771722
$endgroup$
$begingroup$
That said, only the symmetric part of $S$ actually contributes to the quadratic form, which is why one usually assumes that it’s symmetric, anyway.
$endgroup$
– amd
Nov 30 '18 at 0:41
add a comment |
$begingroup$
$S$ does not have to be symmetric, and neither does $A$.
Consider for instance:
$$x^2+2y^2+2xy=
begin{pmatrix}x¥d{pmatrix}begin{pmatrix}1&2\0&2end{pmatrix}begin{pmatrix}x\yend{pmatrix}
=begin{pmatrix}x¥d{pmatrix}begin{pmatrix}1&1\1&2end{pmatrix}begin{pmatrix}x\yend{pmatrix}
$$
It's just that we usually choose $S$ to be symmetric, since then we can do a useful eigenvalue decomposition.
answered Nov 30 '18 at 0:08
I like SerenaI like Serena
4,1771722
$endgroup$
$begingroup$
That said, only the symmetric part of $S$ actually contributes to the quadratic form, which is why one usually assumes that it’s symmetric, anyway.
$endgroup$
– amd
Nov 30 '18 at 0:41
add a comment |
$begingroup$
$S$ does not have to be symmetric, and neither does $A$.
Consider for instance:
$$x^2+2y^2+2xy=
begin{pmatrix}x¥d{pmatrix}begin{pmatrix}1&2\0&2end{pmatrix}begin{pmatrix}x\yend{pmatrix}
=begin{pmatrix}x¥d{pmatrix}begin{pmatrix}1&1\1&2end{pmatrix}begin{pmatrix}x\yend{pmatrix}
$$
It's just that we usually choose $S$ to be symmetric, since then we can do a useful eigenvalue decomposition.
answered Nov 30 '18 at 0:08
I like SerenaI like Serena
4,1771722
$endgroup$
$S$ does not have to be symmetric, and neither does $A$.
Consider for instance:
$$x^2+2y^2+2xy=
begin{pmatrix}x¥d{pmatrix}begin{pmatrix}1&2\0&2end{pmatrix}begin{pmatrix}x\yend{pmatrix}
=begin{pmatrix}x¥d{pmatrix}begin{pmatrix}1&1\1&2end{pmatrix}begin{pmatrix}x\yend{pmatrix}
$$
It's just that we usually choose $S$ to be symmetric, since then we can do a useful eigenvalue decomposition.
answered Nov 30 '18 at 0:08
I like SerenaI like Serena
4,1771722
answered Nov 30 '18 at 0:08
I like SerenaI like Serena
4,1771722
answered Nov 30 '18 at 0:08
I like SerenaI like Serena
4,1771722
answered Nov 30 '18 at 0:08
I like SerenaI like Serena
4,1771722
4,1771722
$begingroup$
That said, only the symmetric part of $S$ actually contributes to the quadratic form, which is why one usually assumes that it’s symmetric, anyway.
$endgroup$
– amd
Nov 30 '18 at 0:41
add a comment |
$begingroup$
That said, only the symmetric part of $S$ actually contributes to the quadratic form, which is why one usually assumes that it’s symmetric, anyway.
$endgroup$
– amd
Nov 30 '18 at 0:41
$begingroup$
That said, only the symmetric part of $S$ actually contributes to the quadratic form, which is why one usually assumes that it’s symmetric, anyway.
$endgroup$
– amd
Nov 30 '18 at 0:41
$begingroup$
That said, only the symmetric part of $S$ actually contributes to the quadratic form, which is why one usually assumes that it’s symmetric, anyway.
$endgroup$
– amd
Nov 30 '18 at 0:41
add a comment |
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