${xmid x^TAxleq 1} = {xmid x^TBxleq 1} Rightarrow A = B$, where $Asucc 0, Bsucc 0$
$begingroup$
I want show that $${xmid x^TAxleq 1}_{epsilon_A} = {xmid x^TBxleq 1}_{epsilon_B} Rightarrow A = B,$$ where $Asucc 0, Bsucc 0$ (positive definite).
Can I prove this by arguing the following?
- We know $epsilon_A$, $epsilon_B$ are ellipsoids.
- Let eigen-decomposition of $A$ be $A = QLambda Q^T$.
- We know for the set $epsilon_A$, the semi-axis is $a_i = lambda_i^{-1/2}q_i$
- Due to $epsilon_A = epsilon_B$, so both have the same semi-axis, i.e., $b_i = lambda_i^{-1/2}q_i$.
$A$ and $B$ have the same eigenvectors and eigenvalues
$A = B$.
Could anyone please give me any suggestions?
matrices quadratic-forms positive-definite
asked Dec 5 '18 at 3:10
sleeve chensleeve chen
3,07641852
$endgroup$
add a comment |
$begingroup$
I want show that $${xmid x^TAxleq 1}_{epsilon_A} = {xmid x^TBxleq 1}_{epsilon_B} Rightarrow A = B,$$ where $Asucc 0, Bsucc 0$ (positive definite).
Can I prove this by arguing the following?
- We know $epsilon_A$, $epsilon_B$ are ellipsoids.
- Let eigen-decomposition of $A$ be $A = QLambda Q^T$.
- We know for the set $epsilon_A$, the semi-axis is $a_i = lambda_i^{-1/2}q_i$
- Due to $epsilon_A = epsilon_B$, so both have the same semi-axis, i.e., $b_i = lambda_i^{-1/2}q_i$.
$A$ and $B$ have the same eigenvectors and eigenvalues
$A = B$.
Could anyone please give me any suggestions?
matrices quadratic-forms positive-definite
asked Dec 5 '18 at 3:10
sleeve chensleeve chen
3,07641852
$endgroup$
add a comment |
$begingroup$
I want show that $${xmid x^TAxleq 1}_{epsilon_A} = {xmid x^TBxleq 1}_{epsilon_B} Rightarrow A = B,$$ where $Asucc 0, Bsucc 0$ (positive definite).
Can I prove this by arguing the following?
- We know $epsilon_A$, $epsilon_B$ are ellipsoids.
- Let eigen-decomposition of $A$ be $A = QLambda Q^T$.
- We know for the set $epsilon_A$, the semi-axis is $a_i = lambda_i^{-1/2}q_i$
- Due to $epsilon_A = epsilon_B$, so both have the same semi-axis, i.e., $b_i = lambda_i^{-1/2}q_i$.
$A$ and $B$ have the same eigenvectors and eigenvalues
$A = B$.
Could anyone please give me any suggestions?
matrices quadratic-forms positive-definite
asked Dec 5 '18 at 3:10
sleeve chensleeve chen
3,07641852
$endgroup$
I want show that $${xmid x^TAxleq 1}_{epsilon_A} = {xmid x^TBxleq 1}_{epsilon_B} Rightarrow A = B,$$ where $Asucc 0, Bsucc 0$ (positive definite).
Can I prove this by arguing the following?
- We know $epsilon_A$, $epsilon_B$ are ellipsoids.
- Let eigen-decomposition of $A$ be $A = QLambda Q^T$.
- We know for the set $epsilon_A$, the semi-axis is $a_i = lambda_i^{-1/2}q_i$
- Due to $epsilon_A = epsilon_B$, so both have the same semi-axis, i.e., $b_i = lambda_i^{-1/2}q_i$.
$A$ and $B$ have the same eigenvectors and eigenvalues
$A = B$.
Could anyone please give me any suggestions?
matrices quadratic-forms positive-definite
matrices quadratic-forms positive-definite
asked Dec 5 '18 at 3:10
sleeve chensleeve chen
3,07641852
asked Dec 5 '18 at 3:10
sleeve chensleeve chen
3,07641852
asked Dec 5 '18 at 3:10
sleeve chensleeve chen
3,07641852
asked Dec 5 '18 at 3:10
sleeve chensleeve chen
3,07641852
asked Dec 5 '18 at 3:10
sleeve chensleeve chen
3,07641852
3,07641852
add a comment |
add a comment |
1 Answer
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$begingroup$
For $xne 0$, the quadratic form $Q_A(x)=x^TAx$ can be evaluated by observing that $Q_A(tx)=t^2Q_A(x)$ and hence $$Q_A(x)=inf{t^{-2}: Q_A(tx)le 1} = inf { t^{-2}: txin epsilon_A}.$$ Since $epsilon_A=epsilon_B$, the quadratic forms $Q_A(x)$ and $Q_B(x)$ are equal. Hence the corresponding bilinear forms $x^TAy$ and $x^TBy$ are equal, and finally the matrices $A$ and $B$.
answered Dec 5 '18 at 3:54
kimchi loverkimchi lover
9,74631128
$endgroup$
add a comment |
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$begingroup$
For $xne 0$, the quadratic form $Q_A(x)=x^TAx$ can be evaluated by observing that $Q_A(tx)=t^2Q_A(x)$ and hence $$Q_A(x)=inf{t^{-2}: Q_A(tx)le 1} = inf { t^{-2}: txin epsilon_A}.$$ Since $epsilon_A=epsilon_B$, the quadratic forms $Q_A(x)$ and $Q_B(x)$ are equal. Hence the corresponding bilinear forms $x^TAy$ and $x^TBy$ are equal, and finally the matrices $A$ and $B$.
answered Dec 5 '18 at 3:54
kimchi loverkimchi lover
9,74631128
$endgroup$
add a comment |
$begingroup$
For $xne 0$, the quadratic form $Q_A(x)=x^TAx$ can be evaluated by observing that $Q_A(tx)=t^2Q_A(x)$ and hence $$Q_A(x)=inf{t^{-2}: Q_A(tx)le 1} = inf { t^{-2}: txin epsilon_A}.$$ Since $epsilon_A=epsilon_B$, the quadratic forms $Q_A(x)$ and $Q_B(x)$ are equal. Hence the corresponding bilinear forms $x^TAy$ and $x^TBy$ are equal, and finally the matrices $A$ and $B$.
answered Dec 5 '18 at 3:54
kimchi loverkimchi lover
9,74631128
$endgroup$
add a comment |
$begingroup$
For $xne 0$, the quadratic form $Q_A(x)=x^TAx$ can be evaluated by observing that $Q_A(tx)=t^2Q_A(x)$ and hence $$Q_A(x)=inf{t^{-2}: Q_A(tx)le 1} = inf { t^{-2}: txin epsilon_A}.$$ Since $epsilon_A=epsilon_B$, the quadratic forms $Q_A(x)$ and $Q_B(x)$ are equal. Hence the corresponding bilinear forms $x^TAy$ and $x^TBy$ are equal, and finally the matrices $A$ and $B$.
answered Dec 5 '18 at 3:54
kimchi loverkimchi lover
9,74631128
$endgroup$
For $xne 0$, the quadratic form $Q_A(x)=x^TAx$ can be evaluated by observing that $Q_A(tx)=t^2Q_A(x)$ and hence $$Q_A(x)=inf{t^{-2}: Q_A(tx)le 1} = inf { t^{-2}: txin epsilon_A}.$$ Since $epsilon_A=epsilon_B$, the quadratic forms $Q_A(x)$ and $Q_B(x)$ are equal. Hence the corresponding bilinear forms $x^TAy$ and $x^TBy$ are equal, and finally the matrices $A$ and $B$.
answered Dec 5 '18 at 3:54
kimchi loverkimchi lover
9,74631128
answered Dec 5 '18 at 3:54
kimchi loverkimchi lover
9,74631128
answered Dec 5 '18 at 3:54
kimchi loverkimchi lover
9,74631128
answered Dec 5 '18 at 3:54
kimchi loverkimchi lover
9,74631128
9,74631128
add a comment |
add a comment |
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