Why do these conditions ensure symmetric positive-definiteness?
$begingroup$
Forgive me, if I have made a blunder or missed something obvious, I'm not a mathematician!
I'm trying to understand 2 seemingly simple lines of maths - and understand how the conclusions are drawn from these. So here are the three lines (or rather two lines of maths, one statement at the end):
Let:
$L(n) := { A in M(n) : A^T = A }$
$P(n) := { A in L(n) : A > 0 } $
"The boundary of $P(n)$ is the set of singular positive semidefinite matrices"
So from what I understand here, the set L contains symmetric ($A^T = A$) n dimensional square matrices with real entries ($A epsilon M(n)$). Then I have presumed that the part, $A > 0$, refers to the determinant of A, and not A itself.
I am also inferring (possibly incorrectly) that P(n) is therefore the space of SPD matrices..?
So why does ensuring we have square, real, symmetric matrices whose determinant is above 0 (this means they're invertible..?) ensure that the eigenvalues of said matrices are above 0 (my understanding of what makes a matrix positive definite)?
positive-definite symmetric-matrices
edited Dec 6 '18 at 17:52
zxmkn
330212
asked Dec 6 '18 at 15:26
bidbybidby
876
$endgroup$
add a comment |
$begingroup$
Forgive me, if I have made a blunder or missed something obvious, I'm not a mathematician!
I'm trying to understand 2 seemingly simple lines of maths - and understand how the conclusions are drawn from these. So here are the three lines (or rather two lines of maths, one statement at the end):
Let:
$L(n) := { A in M(n) : A^T = A }$
$P(n) := { A in L(n) : A > 0 } $
"The boundary of $P(n)$ is the set of singular positive semidefinite matrices"
So from what I understand here, the set L contains symmetric ($A^T = A$) n dimensional square matrices with real entries ($A epsilon M(n)$). Then I have presumed that the part, $A > 0$, refers to the determinant of A, and not A itself.
I am also inferring (possibly incorrectly) that P(n) is therefore the space of SPD matrices..?
So why does ensuring we have square, real, symmetric matrices whose determinant is above 0 (this means they're invertible..?) ensure that the eigenvalues of said matrices are above 0 (my understanding of what makes a matrix positive definite)?
positive-definite symmetric-matrices
edited Dec 6 '18 at 17:52
zxmkn
330212
asked Dec 6 '18 at 15:26
bidbybidby
876
$endgroup$
1
$begingroup$
I think you are misreading $A>0$ as "$A$ has a positive determinant" when you should read it as "$A$ is strictly positive definite". The author might have written $Asucc 0$ to make this somewhat clearer.
$endgroup$
– kimchi lover
Dec 6 '18 at 15:30
$begingroup$
Unfortunately the author used >, but you may be right! If so then my question is rather moot. Would it not make any sense to draw the SPD conclusions from my original interpretation?
$endgroup$
– bidby
Dec 6 '18 at 15:34
$begingroup$
It's all nonsense under the determinant interpretation (the 2 by 2 diagonal matrix with -1, -1 on the diagonal has positive determinant and is not PSD or PD; the diagonal matrix with -1 and 0 on the diagonal is not in the closure of the PSD matrices). Under the other interpretation it all makes sense, in a trite way.
$endgroup$
– kimchi lover
Dec 6 '18 at 15:42
add a comment |
$begingroup$
Forgive me, if I have made a blunder or missed something obvious, I'm not a mathematician!
I'm trying to understand 2 seemingly simple lines of maths - and understand how the conclusions are drawn from these. So here are the three lines (or rather two lines of maths, one statement at the end):
Let:
$L(n) := { A in M(n) : A^T = A }$
$P(n) := { A in L(n) : A > 0 } $
"The boundary of $P(n)$ is the set of singular positive semidefinite matrices"
So from what I understand here, the set L contains symmetric ($A^T = A$) n dimensional square matrices with real entries ($A epsilon M(n)$). Then I have presumed that the part, $A > 0$, refers to the determinant of A, and not A itself.
I am also inferring (possibly incorrectly) that P(n) is therefore the space of SPD matrices..?
So why does ensuring we have square, real, symmetric matrices whose determinant is above 0 (this means they're invertible..?) ensure that the eigenvalues of said matrices are above 0 (my understanding of what makes a matrix positive definite)?
positive-definite symmetric-matrices
edited Dec 6 '18 at 17:52
zxmkn
330212
asked Dec 6 '18 at 15:26
bidbybidby
876
$endgroup$
Forgive me, if I have made a blunder or missed something obvious, I'm not a mathematician!
I'm trying to understand 2 seemingly simple lines of maths - and understand how the conclusions are drawn from these. So here are the three lines (or rather two lines of maths, one statement at the end):
Let:
$L(n) := { A in M(n) : A^T = A }$
$P(n) := { A in L(n) : A > 0 } $
"The boundary of $P(n)$ is the set of singular positive semidefinite matrices"
So from what I understand here, the set L contains symmetric ($A^T = A$) n dimensional square matrices with real entries ($A epsilon M(n)$). Then I have presumed that the part, $A > 0$, refers to the determinant of A, and not A itself.
I am also inferring (possibly incorrectly) that P(n) is therefore the space of SPD matrices..?
So why does ensuring we have square, real, symmetric matrices whose determinant is above 0 (this means they're invertible..?) ensure that the eigenvalues of said matrices are above 0 (my understanding of what makes a matrix positive definite)?
positive-definite symmetric-matrices
positive-definite symmetric-matrices
edited Dec 6 '18 at 17:52
zxmkn
330212
asked Dec 6 '18 at 15:26
bidbybidby
876
edited Dec 6 '18 at 17:52
zxmkn
330212
asked Dec 6 '18 at 15:26
bidbybidby
876
edited Dec 6 '18 at 17:52
zxmkn
330212
edited Dec 6 '18 at 17:52
zxmkn
330212
edited Dec 6 '18 at 17:52
zxmkn
330212
330212
asked Dec 6 '18 at 15:26
bidbybidby
876
asked Dec 6 '18 at 15:26
bidbybidby
876
asked Dec 6 '18 at 15:26
bidbybidby
876
876
1
$begingroup$
I think you are misreading $A>0$ as "$A$ has a positive determinant" when you should read it as "$A$ is strictly positive definite". The author might have written $Asucc 0$ to make this somewhat clearer.
$endgroup$
– kimchi lover
Dec 6 '18 at 15:30
$begingroup$
Unfortunately the author used >, but you may be right! If so then my question is rather moot. Would it not make any sense to draw the SPD conclusions from my original interpretation?
$endgroup$
– bidby
Dec 6 '18 at 15:34
$begingroup$
It's all nonsense under the determinant interpretation (the 2 by 2 diagonal matrix with -1, -1 on the diagonal has positive determinant and is not PSD or PD; the diagonal matrix with -1 and 0 on the diagonal is not in the closure of the PSD matrices). Under the other interpretation it all makes sense, in a trite way.
$endgroup$
– kimchi lover
Dec 6 '18 at 15:42
add a comment |
1
$begingroup$
I think you are misreading $A>0$ as "$A$ has a positive determinant" when you should read it as "$A$ is strictly positive definite". The author might have written $Asucc 0$ to make this somewhat clearer.
$endgroup$
– kimchi lover
Dec 6 '18 at 15:30
$begingroup$
Unfortunately the author used >, but you may be right! If so then my question is rather moot. Would it not make any sense to draw the SPD conclusions from my original interpretation?
$endgroup$
– bidby
Dec 6 '18 at 15:34
$begingroup$
It's all nonsense under the determinant interpretation (the 2 by 2 diagonal matrix with -1, -1 on the diagonal has positive determinant and is not PSD or PD; the diagonal matrix with -1 and 0 on the diagonal is not in the closure of the PSD matrices). Under the other interpretation it all makes sense, in a trite way.
$endgroup$
– kimchi lover
Dec 6 '18 at 15:42
1
1
$begingroup$
I think you are misreading $A>0$ as "$A$ has a positive determinant" when you should read it as "$A$ is strictly positive definite". The author might have written $Asucc 0$ to make this somewhat clearer.
$endgroup$
– kimchi lover
Dec 6 '18 at 15:30
$begingroup$
I think you are misreading $A>0$ as "$A$ has a positive determinant" when you should read it as "$A$ is strictly positive definite". The author might have written $Asucc 0$ to make this somewhat clearer.
$endgroup$
– kimchi lover
Dec 6 '18 at 15:30
$begingroup$
Unfortunately the author used >, but you may be right! If so then my question is rather moot. Would it not make any sense to draw the SPD conclusions from my original interpretation?
$endgroup$
– bidby
Dec 6 '18 at 15:34
$begingroup$
Unfortunately the author used >, but you may be right! If so then my question is rather moot. Would it not make any sense to draw the SPD conclusions from my original interpretation?
$endgroup$
– bidby
Dec 6 '18 at 15:34
$begingroup$
It's all nonsense under the determinant interpretation (the 2 by 2 diagonal matrix with -1, -1 on the diagonal has positive determinant and is not PSD or PD; the diagonal matrix with -1 and 0 on the diagonal is not in the closure of the PSD matrices). Under the other interpretation it all makes sense, in a trite way.
$endgroup$
– kimchi lover
Dec 6 '18 at 15:42
$begingroup$
It's all nonsense under the determinant interpretation (the 2 by 2 diagonal matrix with -1, -1 on the diagonal has positive determinant and is not PSD or PD; the diagonal matrix with -1 and 0 on the diagonal is not in the closure of the PSD matrices). Under the other interpretation it all makes sense, in a trite way.
$endgroup$
– kimchi lover
Dec 6 '18 at 15:42
add a comment |
1 Answer
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$begingroup$
As others have stated, $P(n)$ is intended to represent the set of all positive definite $n times n$ matrices (not the set of all matrices with positive determinant, as bidby initially thought). We let $A succ 0$ denote a positive definite matrix and $A succeq 0$ denote a positive semidefinite matrix.
Claim: $partial P(n) overset{textrm{Def}}{=} overline{P(n)} setminus mathring{P(n)} = {A in mathbb{R}^{n times n} | A succeq 0, nexists A^{-1}}$.
Proof: Let $A in partial P(n)$, $(A_k)_{k in mathbb{N}} subseteq P(n)$, and $z in mathbb{R}^n$. Then
$$
z^T A z = z^T left( lim_{k to infty} A_k right) z = lim_{ktoinfty} underbrace{left( z^T A_k z right)}_{geq 0} geq 0,
$$
since the functions $f : mathbb{R}^n to mathbb{R}, ; x mapsto z^T x$ and $g : mathbb{R}^{n times n} to mathbb{R}^n, ; B mapsto B z^T$ are linear (and linear transformations between finite dimensional vector spaces are continuous). Therefore, $A$ is positive semidefinite.
We now show that $A$ is singular. Since $A$ is positive semidefinite, all its eigenvalues are non-negative. We know that $A$ is not positive definite, since $mathring{P(n)} = P(n)$ by this post. But if the eigenvalues were all positive, then $A$ would be positive definite. Hence, $A$ has $0$ as an eigenvalue -- i.e. $exists$ eigenvector $v in mathbb{R}^n setminus {0}$ with $Av = 0 cdot v = 0$. Therefore, $textrm{Ker}(A) neq {0}$, so $A$ is singular (see 3. here).
(A trivial example of such an $A$ and such a sequence $(A_n)$ is $A := 0 in mathbb{R}^{1 times 1}$ and $A_n := 1/n$.)
answered Dec 6 '18 at 17:27
zxmknzxmkn
330212
$endgroup$
add a comment |
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$begingroup$
As others have stated, $P(n)$ is intended to represent the set of all positive definite $n times n$ matrices (not the set of all matrices with positive determinant, as bidby initially thought). We let $A succ 0$ denote a positive definite matrix and $A succeq 0$ denote a positive semidefinite matrix.
Claim: $partial P(n) overset{textrm{Def}}{=} overline{P(n)} setminus mathring{P(n)} = {A in mathbb{R}^{n times n} | A succeq 0, nexists A^{-1}}$.
Proof: Let $A in partial P(n)$, $(A_k)_{k in mathbb{N}} subseteq P(n)$, and $z in mathbb{R}^n$. Then
$$
z^T A z = z^T left( lim_{k to infty} A_k right) z = lim_{ktoinfty} underbrace{left( z^T A_k z right)}_{geq 0} geq 0,
$$
since the functions $f : mathbb{R}^n to mathbb{R}, ; x mapsto z^T x$ and $g : mathbb{R}^{n times n} to mathbb{R}^n, ; B mapsto B z^T$ are linear (and linear transformations between finite dimensional vector spaces are continuous). Therefore, $A$ is positive semidefinite.
We now show that $A$ is singular. Since $A$ is positive semidefinite, all its eigenvalues are non-negative. We know that $A$ is not positive definite, since $mathring{P(n)} = P(n)$ by this post. But if the eigenvalues were all positive, then $A$ would be positive definite. Hence, $A$ has $0$ as an eigenvalue -- i.e. $exists$ eigenvector $v in mathbb{R}^n setminus {0}$ with $Av = 0 cdot v = 0$. Therefore, $textrm{Ker}(A) neq {0}$, so $A$ is singular (see 3. here).
(A trivial example of such an $A$ and such a sequence $(A_n)$ is $A := 0 in mathbb{R}^{1 times 1}$ and $A_n := 1/n$.)
answered Dec 6 '18 at 17:27
zxmknzxmkn
330212
$endgroup$
add a comment |
$begingroup$
As others have stated, $P(n)$ is intended to represent the set of all positive definite $n times n$ matrices (not the set of all matrices with positive determinant, as bidby initially thought). We let $A succ 0$ denote a positive definite matrix and $A succeq 0$ denote a positive semidefinite matrix.
Claim: $partial P(n) overset{textrm{Def}}{=} overline{P(n)} setminus mathring{P(n)} = {A in mathbb{R}^{n times n} | A succeq 0, nexists A^{-1}}$.
Proof: Let $A in partial P(n)$, $(A_k)_{k in mathbb{N}} subseteq P(n)$, and $z in mathbb{R}^n$. Then
$$
z^T A z = z^T left( lim_{k to infty} A_k right) z = lim_{ktoinfty} underbrace{left( z^T A_k z right)}_{geq 0} geq 0,
$$
since the functions $f : mathbb{R}^n to mathbb{R}, ; x mapsto z^T x$ and $g : mathbb{R}^{n times n} to mathbb{R}^n, ; B mapsto B z^T$ are linear (and linear transformations between finite dimensional vector spaces are continuous). Therefore, $A$ is positive semidefinite.
We now show that $A$ is singular. Since $A$ is positive semidefinite, all its eigenvalues are non-negative. We know that $A$ is not positive definite, since $mathring{P(n)} = P(n)$ by this post. But if the eigenvalues were all positive, then $A$ would be positive definite. Hence, $A$ has $0$ as an eigenvalue -- i.e. $exists$ eigenvector $v in mathbb{R}^n setminus {0}$ with $Av = 0 cdot v = 0$. Therefore, $textrm{Ker}(A) neq {0}$, so $A$ is singular (see 3. here).
(A trivial example of such an $A$ and such a sequence $(A_n)$ is $A := 0 in mathbb{R}^{1 times 1}$ and $A_n := 1/n$.)
answered Dec 6 '18 at 17:27
zxmknzxmkn
330212
$endgroup$
add a comment |
$begingroup$
As others have stated, $P(n)$ is intended to represent the set of all positive definite $n times n$ matrices (not the set of all matrices with positive determinant, as bidby initially thought). We let $A succ 0$ denote a positive definite matrix and $A succeq 0$ denote a positive semidefinite matrix.
Claim: $partial P(n) overset{textrm{Def}}{=} overline{P(n)} setminus mathring{P(n)} = {A in mathbb{R}^{n times n} | A succeq 0, nexists A^{-1}}$.
Proof: Let $A in partial P(n)$, $(A_k)_{k in mathbb{N}} subseteq P(n)$, and $z in mathbb{R}^n$. Then
$$
z^T A z = z^T left( lim_{k to infty} A_k right) z = lim_{ktoinfty} underbrace{left( z^T A_k z right)}_{geq 0} geq 0,
$$
since the functions $f : mathbb{R}^n to mathbb{R}, ; x mapsto z^T x$ and $g : mathbb{R}^{n times n} to mathbb{R}^n, ; B mapsto B z^T$ are linear (and linear transformations between finite dimensional vector spaces are continuous). Therefore, $A$ is positive semidefinite.
We now show that $A$ is singular. Since $A$ is positive semidefinite, all its eigenvalues are non-negative. We know that $A$ is not positive definite, since $mathring{P(n)} = P(n)$ by this post. But if the eigenvalues were all positive, then $A$ would be positive definite. Hence, $A$ has $0$ as an eigenvalue -- i.e. $exists$ eigenvector $v in mathbb{R}^n setminus {0}$ with $Av = 0 cdot v = 0$. Therefore, $textrm{Ker}(A) neq {0}$, so $A$ is singular (see 3. here).
(A trivial example of such an $A$ and such a sequence $(A_n)$ is $A := 0 in mathbb{R}^{1 times 1}$ and $A_n := 1/n$.)
answered Dec 6 '18 at 17:27
zxmknzxmkn
330212
$endgroup$
As others have stated, $P(n)$ is intended to represent the set of all positive definite $n times n$ matrices (not the set of all matrices with positive determinant, as bidby initially thought). We let $A succ 0$ denote a positive definite matrix and $A succeq 0$ denote a positive semidefinite matrix.
Claim: $partial P(n) overset{textrm{Def}}{=} overline{P(n)} setminus mathring{P(n)} = {A in mathbb{R}^{n times n} | A succeq 0, nexists A^{-1}}$.
Proof: Let $A in partial P(n)$, $(A_k)_{k in mathbb{N}} subseteq P(n)$, and $z in mathbb{R}^n$. Then
$$
z^T A z = z^T left( lim_{k to infty} A_k right) z = lim_{ktoinfty} underbrace{left( z^T A_k z right)}_{geq 0} geq 0,
$$
since the functions $f : mathbb{R}^n to mathbb{R}, ; x mapsto z^T x$ and $g : mathbb{R}^{n times n} to mathbb{R}^n, ; B mapsto B z^T$ are linear (and linear transformations between finite dimensional vector spaces are continuous). Therefore, $A$ is positive semidefinite.
We now show that $A$ is singular. Since $A$ is positive semidefinite, all its eigenvalues are non-negative. We know that $A$ is not positive definite, since $mathring{P(n)} = P(n)$ by this post. But if the eigenvalues were all positive, then $A$ would be positive definite. Hence, $A$ has $0$ as an eigenvalue -- i.e. $exists$ eigenvector $v in mathbb{R}^n setminus {0}$ with $Av = 0 cdot v = 0$. Therefore, $textrm{Ker}(A) neq {0}$, so $A$ is singular (see 3. here).
(A trivial example of such an $A$ and such a sequence $(A_n)$ is $A := 0 in mathbb{R}^{1 times 1}$ and $A_n := 1/n$.)
answered Dec 6 '18 at 17:27
zxmknzxmkn
330212
answered Dec 6 '18 at 17:27
zxmknzxmkn
330212
answered Dec 6 '18 at 17:27
zxmknzxmkn
330212
answered Dec 6 '18 at 17:27
zxmknzxmkn
330212
330212
add a comment |
add a comment |
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$begingroup$
I think you are misreading $A>0$ as "$A$ has a positive determinant" when you should read it as "$A$ is strictly positive definite". The author might have written $Asucc 0$ to make this somewhat clearer.
$endgroup$
– kimchi lover
Dec 6 '18 at 15:30
$begingroup$
Unfortunately the author used >, but you may be right! If so then my question is rather moot. Would it not make any sense to draw the SPD conclusions from my original interpretation?
$endgroup$
– bidby
Dec 6 '18 at 15:34
$begingroup$
It's all nonsense under the determinant interpretation (the 2 by 2 diagonal matrix with -1, -1 on the diagonal has positive determinant and is not PSD or PD; the diagonal matrix with -1 and 0 on the diagonal is not in the closure of the PSD matrices). Under the other interpretation it all makes sense, in a trite way.
$endgroup$
– kimchi lover
Dec 6 '18 at 15:42