Mixing
Matrix with nonnegative symmetric part and semisimplicty of the eigenvalue 0
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Let B be a real square matrix with non-negative symmetric part, i.e. for all vectors $X$, $X^top B Xgeq 0$. We also assume that $B$ is singular. I am wondering if the eigenvalue $0$ of $B$ is necessary semi-simple, i.e. is the dimension of the kernel of $B$ equal to algebraic multiplicity of the eigenvalue $0$.
I am unable to prove it, nor I am able to find a counterexample. A counterexample would necessary be non-symmetric, and of dimension at least $3$, and at this point, the differences between diagonalising the matrix $B$ and the quadratic form $X^top BX$ are wrecking my brain.
linear-algebra matrices
asked Apr 19 '18 at 11:38
Armand KoenigArmand Koenig
31616
$endgroup$
add a comment |
$begingroup$
Let B be a real square matrix with non-negative symmetric part, i.e. for all vectors $X$, $X^top B Xgeq 0$. We also assume that $B$ is singular. I am wondering if the eigenvalue $0$ of $B$ is necessary semi-simple, i.e. is the dimension of the kernel of $B$ equal to algebraic multiplicity of the eigenvalue $0$.
I am unable to prove it, nor I am able to find a counterexample. A counterexample would necessary be non-symmetric, and of dimension at least $3$, and at this point, the differences between diagonalising the matrix $B$ and the quadratic form $X^top BX$ are wrecking my brain.
linear-algebra matrices
asked Apr 19 '18 at 11:38
Armand KoenigArmand Koenig
31616
$endgroup$
add a comment |
$begingroup$
Let B be a real square matrix with non-negative symmetric part, i.e. for all vectors $X$, $X^top B Xgeq 0$. We also assume that $B$ is singular. I am wondering if the eigenvalue $0$ of $B$ is necessary semi-simple, i.e. is the dimension of the kernel of $B$ equal to algebraic multiplicity of the eigenvalue $0$.
I am unable to prove it, nor I am able to find a counterexample. A counterexample would necessary be non-symmetric, and of dimension at least $3$, and at this point, the differences between diagonalising the matrix $B$ and the quadratic form $X^top BX$ are wrecking my brain.
linear-algebra matrices
asked Apr 19 '18 at 11:38
Armand KoenigArmand Koenig
31616
$endgroup$
Let B be a real square matrix with non-negative symmetric part, i.e. for all vectors $X$, $X^top B Xgeq 0$. We also assume that $B$ is singular. I am wondering if the eigenvalue $0$ of $B$ is necessary semi-simple, i.e. is the dimension of the kernel of $B$ equal to algebraic multiplicity of the eigenvalue $0$.
I am unable to prove it, nor I am able to find a counterexample. A counterexample would necessary be non-symmetric, and of dimension at least $3$, and at this point, the differences between diagonalising the matrix $B$ and the quadratic form $X^top BX$ are wrecking my brain.
linear-algebra matrices
linear-algebra matrices
asked Apr 19 '18 at 11:38
Armand KoenigArmand Koenig
31616
asked Apr 19 '18 at 11:38
Armand KoenigArmand Koenig
31616
asked Apr 19 '18 at 11:38
Armand KoenigArmand Koenig
31616
asked Apr 19 '18 at 11:38
Armand KoenigArmand Koenig
31616
asked Apr 19 '18 at 11:38
Armand KoenigArmand Koenig
31616
31616
add a comment |
add a comment |
1 Answer
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I came back to this and found the answer: yes, if $B$ is singular and has nonnegative symmetric part, then the eigenvalue $0$ of $B$ is semisimple. This relies on the following observations:
Claim 1: If $0$ is not a semisimple eigenvalue of $B$, then the resolvent $R(z) = (B-z)^{-1}$ has a pole of order at least two at zero.
Claim 2: If $B$ has nonnegative symmetric part, then there exists $C>0$ such that for $z$ small enough, $|R(z)|leq C|z|^{-1}$.
We use the following standard theorem:
Proposition: if there exists $c>0$ such that for all vectors $X$, $|AX|geq c|X|$, then $A$ is nonsingular, and if $c$ is the greatest real that satisfies the previous inequality, then $|A^{-1}| = c^{-1}$.
Proof of claim 1: By assumption, $ker(B)neq ker(B^n)$, so let us choose $X_0neq 0$ such that $ BX_0neq 0$ and $B^2 X_0 = 0$. Then, for $zinmathbb C^ast$,
$$(B-z)(BX_0-zX_0) = -z^2X_0.$$
Since for $z$ small $|BX_0-zX_0|geq c>0$, this implies that for $z$ small $|R(z)|geq frac{c}{|X_0|}|z|^{-2}$.
Proof of claim 2: We have for all $zin(-infty,0)$ and vector $X$:
begin{align*}
|(B-z)X|^2 &= ((B-z)^ast(B-z)X|X)\
&=(B^ast BX|x)-bar z(BX|X) - z(B^top X|X) + |z|^2(X|X)\
&= |BX|^2 -z((B^top+B)X|X) + |z|^2|X|^2qquadtext{since }bar z =z\
&geq |z|^2|X|^2qquadtext{since }z<0
end{align*}
So, for $z<0$, $|R(z)|leq |z|^{-1}$. Now if we write the resolvent as a Laurent series $R(z) = sum_{k=-n}^{+infty}R_k z^k$, the previous inequality implies that $R_{-n} = R_{-n+1}=dots=R_{-2} =0$, which concludes the proof. (Note: we can find the fact that for $k<-n$, $R_k = 0$ in Kato's "Perturbation theory for linear operators" p. 39 eqs. (5.18) and (5.20).)
answered Jan 28 at 23:07
Armand KoenigArmand Koenig
31616
$endgroup$
add a comment |
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$begingroup$
I came back to this and found the answer: yes, if $B$ is singular and has nonnegative symmetric part, then the eigenvalue $0$ of $B$ is semisimple. This relies on the following observations:
Claim 1: If $0$ is not a semisimple eigenvalue of $B$, then the resolvent $R(z) = (B-z)^{-1}$ has a pole of order at least two at zero.
Claim 2: If $B$ has nonnegative symmetric part, then there exists $C>0$ such that for $z$ small enough, $|R(z)|leq C|z|^{-1}$.
We use the following standard theorem:
Proposition: if there exists $c>0$ such that for all vectors $X$, $|AX|geq c|X|$, then $A$ is nonsingular, and if $c$ is the greatest real that satisfies the previous inequality, then $|A^{-1}| = c^{-1}$.
Proof of claim 1: By assumption, $ker(B)neq ker(B^n)$, so let us choose $X_0neq 0$ such that $ BX_0neq 0$ and $B^2 X_0 = 0$. Then, for $zinmathbb C^ast$,
$$(B-z)(BX_0-zX_0) = -z^2X_0.$$
Since for $z$ small $|BX_0-zX_0|geq c>0$, this implies that for $z$ small $|R(z)|geq frac{c}{|X_0|}|z|^{-2}$.
Proof of claim 2: We have for all $zin(-infty,0)$ and vector $X$:
begin{align*}
|(B-z)X|^2 &= ((B-z)^ast(B-z)X|X)\
&=(B^ast BX|x)-bar z(BX|X) - z(B^top X|X) + |z|^2(X|X)\
&= |BX|^2 -z((B^top+B)X|X) + |z|^2|X|^2qquadtext{since }bar z =z\
&geq |z|^2|X|^2qquadtext{since }z<0
end{align*}
So, for $z<0$, $|R(z)|leq |z|^{-1}$. Now if we write the resolvent as a Laurent series $R(z) = sum_{k=-n}^{+infty}R_k z^k$, the previous inequality implies that $R_{-n} = R_{-n+1}=dots=R_{-2} =0$, which concludes the proof. (Note: we can find the fact that for $k<-n$, $R_k = 0$ in Kato's "Perturbation theory for linear operators" p. 39 eqs. (5.18) and (5.20).)
answered Jan 28 at 23:07
Armand KoenigArmand Koenig
31616
$endgroup$
add a comment |
$begingroup$
I came back to this and found the answer: yes, if $B$ is singular and has nonnegative symmetric part, then the eigenvalue $0$ of $B$ is semisimple. This relies on the following observations:
Claim 1: If $0$ is not a semisimple eigenvalue of $B$, then the resolvent $R(z) = (B-z)^{-1}$ has a pole of order at least two at zero.
Claim 2: If $B$ has nonnegative symmetric part, then there exists $C>0$ such that for $z$ small enough, $|R(z)|leq C|z|^{-1}$.
We use the following standard theorem:
Proposition: if there exists $c>0$ such that for all vectors $X$, $|AX|geq c|X|$, then $A$ is nonsingular, and if $c$ is the greatest real that satisfies the previous inequality, then $|A^{-1}| = c^{-1}$.
Proof of claim 1: By assumption, $ker(B)neq ker(B^n)$, so let us choose $X_0neq 0$ such that $ BX_0neq 0$ and $B^2 X_0 = 0$. Then, for $zinmathbb C^ast$,
$$(B-z)(BX_0-zX_0) = -z^2X_0.$$
Since for $z$ small $|BX_0-zX_0|geq c>0$, this implies that for $z$ small $|R(z)|geq frac{c}{|X_0|}|z|^{-2}$.
Proof of claim 2: We have for all $zin(-infty,0)$ and vector $X$:
begin{align*}
|(B-z)X|^2 &= ((B-z)^ast(B-z)X|X)\
&=(B^ast BX|x)-bar z(BX|X) - z(B^top X|X) + |z|^2(X|X)\
&= |BX|^2 -z((B^top+B)X|X) + |z|^2|X|^2qquadtext{since }bar z =z\
&geq |z|^2|X|^2qquadtext{since }z<0
end{align*}
So, for $z<0$, $|R(z)|leq |z|^{-1}$. Now if we write the resolvent as a Laurent series $R(z) = sum_{k=-n}^{+infty}R_k z^k$, the previous inequality implies that $R_{-n} = R_{-n+1}=dots=R_{-2} =0$, which concludes the proof. (Note: we can find the fact that for $k<-n$, $R_k = 0$ in Kato's "Perturbation theory for linear operators" p. 39 eqs. (5.18) and (5.20).)
answered Jan 28 at 23:07
Armand KoenigArmand Koenig
31616
$endgroup$
add a comment |
$begingroup$
I came back to this and found the answer: yes, if $B$ is singular and has nonnegative symmetric part, then the eigenvalue $0$ of $B$ is semisimple. This relies on the following observations:
Claim 1: If $0$ is not a semisimple eigenvalue of $B$, then the resolvent $R(z) = (B-z)^{-1}$ has a pole of order at least two at zero.
Claim 2: If $B$ has nonnegative symmetric part, then there exists $C>0$ such that for $z$ small enough, $|R(z)|leq C|z|^{-1}$.
We use the following standard theorem:
Proposition: if there exists $c>0$ such that for all vectors $X$, $|AX|geq c|X|$, then $A$ is nonsingular, and if $c$ is the greatest real that satisfies the previous inequality, then $|A^{-1}| = c^{-1}$.
Proof of claim 1: By assumption, $ker(B)neq ker(B^n)$, so let us choose $X_0neq 0$ such that $ BX_0neq 0$ and $B^2 X_0 = 0$. Then, for $zinmathbb C^ast$,
$$(B-z)(BX_0-zX_0) = -z^2X_0.$$
Since for $z$ small $|BX_0-zX_0|geq c>0$, this implies that for $z$ small $|R(z)|geq frac{c}{|X_0|}|z|^{-2}$.
Proof of claim 2: We have for all $zin(-infty,0)$ and vector $X$:
begin{align*}
|(B-z)X|^2 &= ((B-z)^ast(B-z)X|X)\
&=(B^ast BX|x)-bar z(BX|X) - z(B^top X|X) + |z|^2(X|X)\
&= |BX|^2 -z((B^top+B)X|X) + |z|^2|X|^2qquadtext{since }bar z =z\
&geq |z|^2|X|^2qquadtext{since }z<0
end{align*}
So, for $z<0$, $|R(z)|leq |z|^{-1}$. Now if we write the resolvent as a Laurent series $R(z) = sum_{k=-n}^{+infty}R_k z^k$, the previous inequality implies that $R_{-n} = R_{-n+1}=dots=R_{-2} =0$, which concludes the proof. (Note: we can find the fact that for $k<-n$, $R_k = 0$ in Kato's "Perturbation theory for linear operators" p. 39 eqs. (5.18) and (5.20).)
answered Jan 28 at 23:07
Armand KoenigArmand Koenig
31616
$endgroup$
I came back to this and found the answer: yes, if $B$ is singular and has nonnegative symmetric part, then the eigenvalue $0$ of $B$ is semisimple. This relies on the following observations:
Claim 1: If $0$ is not a semisimple eigenvalue of $B$, then the resolvent $R(z) = (B-z)^{-1}$ has a pole of order at least two at zero.
Claim 2: If $B$ has nonnegative symmetric part, then there exists $C>0$ such that for $z$ small enough, $|R(z)|leq C|z|^{-1}$.
We use the following standard theorem:
Proposition: if there exists $c>0$ such that for all vectors $X$, $|AX|geq c|X|$, then $A$ is nonsingular, and if $c$ is the greatest real that satisfies the previous inequality, then $|A^{-1}| = c^{-1}$.
Proof of claim 1: By assumption, $ker(B)neq ker(B^n)$, so let us choose $X_0neq 0$ such that $ BX_0neq 0$ and $B^2 X_0 = 0$. Then, for $zinmathbb C^ast$,
$$(B-z)(BX_0-zX_0) = -z^2X_0.$$
Since for $z$ small $|BX_0-zX_0|geq c>0$, this implies that for $z$ small $|R(z)|geq frac{c}{|X_0|}|z|^{-2}$.
Proof of claim 2: We have for all $zin(-infty,0)$ and vector $X$:
begin{align*}
|(B-z)X|^2 &= ((B-z)^ast(B-z)X|X)\
&=(B^ast BX|x)-bar z(BX|X) - z(B^top X|X) + |z|^2(X|X)\
&= |BX|^2 -z((B^top+B)X|X) + |z|^2|X|^2qquadtext{since }bar z =z\
&geq |z|^2|X|^2qquadtext{since }z<0
end{align*}
So, for $z<0$, $|R(z)|leq |z|^{-1}$. Now if we write the resolvent as a Laurent series $R(z) = sum_{k=-n}^{+infty}R_k z^k$, the previous inequality implies that $R_{-n} = R_{-n+1}=dots=R_{-2} =0$, which concludes the proof. (Note: we can find the fact that for $k<-n$, $R_k = 0$ in Kato's "Perturbation theory for linear operators" p. 39 eqs. (5.18) and (5.20).)
answered Jan 28 at 23:07
Armand KoenigArmand Koenig
31616
answered Jan 28 at 23:07
Armand KoenigArmand Koenig
31616
answered Jan 28 at 23:07
Armand KoenigArmand Koenig
31616
answered Jan 28 at 23:07
Armand KoenigArmand Koenig
31616
31616
add a comment |
add a comment |
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