Fix $v in mathbb R^n$, the probability matrix $A$ with normally distributed entries has an eigenbasis that $v$ projects nontrivially is $1$trace class norms of random matricesCan't show that these matrices are diagonalizable.Show $A$ is diagonalizableGiven that $A$ is normal, show that there exists a matrix $B$ such that $B^k = A$, where $k in mathbb N$Hermitian Operators and the Spectral Theorem$A^topA = (A^topA)^2$ mistakeSquare root of operators over $mathbbC^n$Spectrum of a matrix with normally distributed entriesAre the sets $X: max_i textRelambda_i (B+AX) < 0$ and $X: rho(B+AX) < 1$ homeomorphic?Is the function $A mapsto sumlimits_j=0^infty langle A^j v, A^j v rangle$ differentiable everywhere?
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Fix $v in mathbb R^n$, the probability matrix $A$ with normally distributed entries has an eigenbasis that $v$ projects nontrivially is $1$
trace class norms of random matricesCan't show that these matrices are diagonalizable.Show $A$ is diagonalizableGiven that $A$ is normal, show that there exists a matrix $B$ such that $B^k = A$, where $k in mathbb N$Hermitian Operators and the Spectral Theorem$A^topA = (A^topA)^2$ mistakeSquare root of operators over $mathbbC^n$Spectrum of a matrix with normally distributed entriesAre the sets $X: max_i textRelambda_i (B+AX) < 0$ and $X: rho(B+AX) < 1$ homeomorphic?Is the function $A mapsto sumlimits_j=0^infty langle A^j v, A^j v rangle$ differentiable everywhere?
$begingroup$
Suppose $v in mathbb R^n$ is a fixed nonzero vector and $A in M_n(mathbb R)$ is a random matrix where each entry is taken from a standard normal distribution over $mathbb R$. We know that the probability $A$ is diagonalizable is $1$. My question is: if we concern those matrices that are diagonalizable (thus with an eigenbasis), what is the probability that those with an eigenbasis and $v$ projects nontrivially on each eigenvector in the eigenbasis, i.e., the probability measure of the set
beginalign*
A in M_n(mathbb R): & texteach entry is generated from a standard normal,\
& textif $A=SLambda S^-1$ (spectral decomposition) then $Sv$ has no zero component.
endalign*
linear-algebra probability-theory random-matrices
$endgroup$
add a comment |
$begingroup$
Suppose $v in mathbb R^n$ is a fixed nonzero vector and $A in M_n(mathbb R)$ is a random matrix where each entry is taken from a standard normal distribution over $mathbb R$. We know that the probability $A$ is diagonalizable is $1$. My question is: if we concern those matrices that are diagonalizable (thus with an eigenbasis), what is the probability that those with an eigenbasis and $v$ projects nontrivially on each eigenvector in the eigenbasis, i.e., the probability measure of the set
beginalign*
A in M_n(mathbb R): & texteach entry is generated from a standard normal,\
& textif $A=SLambda S^-1$ (spectral decomposition) then $Sv$ has no zero component.
endalign*
linear-algebra probability-theory random-matrices
$endgroup$
$begingroup$
Unless $v=0$, the probability that $v$ is is in any proper subspace defined by $A$, is zero. Use this for the finitely many orthogonal complements of the eigenspaces of $A$.
$endgroup$
– Hagen von Eitzen
Mar 14 at 5:30
$begingroup$
@HagenvonEitzen: Thanks for your comment. But I could not understand what you mean by "proper subspace defined by $A$"? Do you mean the eigenspaces of $A$? I cannot see how this can be argued. Could you point me a reference?
$endgroup$
– MyCindy2012
Mar 14 at 5:36
$begingroup$
The probability that $A$ can be diagonalized over $mathbbR$ tends to $0$ when $n$ tends to $infty$.
$endgroup$
– loup blanc
Mar 15 at 9:40
add a comment |
$begingroup$
Suppose $v in mathbb R^n$ is a fixed nonzero vector and $A in M_n(mathbb R)$ is a random matrix where each entry is taken from a standard normal distribution over $mathbb R$. We know that the probability $A$ is diagonalizable is $1$. My question is: if we concern those matrices that are diagonalizable (thus with an eigenbasis), what is the probability that those with an eigenbasis and $v$ projects nontrivially on each eigenvector in the eigenbasis, i.e., the probability measure of the set
beginalign*
A in M_n(mathbb R): & texteach entry is generated from a standard normal,\
& textif $A=SLambda S^-1$ (spectral decomposition) then $Sv$ has no zero component.
endalign*
linear-algebra probability-theory random-matrices
$endgroup$
Suppose $v in mathbb R^n$ is a fixed nonzero vector and $A in M_n(mathbb R)$ is a random matrix where each entry is taken from a standard normal distribution over $mathbb R$. We know that the probability $A$ is diagonalizable is $1$. My question is: if we concern those matrices that are diagonalizable (thus with an eigenbasis), what is the probability that those with an eigenbasis and $v$ projects nontrivially on each eigenvector in the eigenbasis, i.e., the probability measure of the set
beginalign*
A in M_n(mathbb R): & texteach entry is generated from a standard normal,\
& textif $A=SLambda S^-1$ (spectral decomposition) then $Sv$ has no zero component.
endalign*
linear-algebra probability-theory random-matrices
linear-algebra probability-theory random-matrices
edited Mar 14 at 5:34
MyCindy2012
asked Mar 14 at 5:06
MyCindy2012MyCindy2012
9211
9211
$begingroup$
Unless $v=0$, the probability that $v$ is is in any proper subspace defined by $A$, is zero. Use this for the finitely many orthogonal complements of the eigenspaces of $A$.
$endgroup$
– Hagen von Eitzen
Mar 14 at 5:30
$begingroup$
@HagenvonEitzen: Thanks for your comment. But I could not understand what you mean by "proper subspace defined by $A$"? Do you mean the eigenspaces of $A$? I cannot see how this can be argued. Could you point me a reference?
$endgroup$
– MyCindy2012
Mar 14 at 5:36
$begingroup$
The probability that $A$ can be diagonalized over $mathbbR$ tends to $0$ when $n$ tends to $infty$.
$endgroup$
– loup blanc
Mar 15 at 9:40
add a comment |
$begingroup$
Unless $v=0$, the probability that $v$ is is in any proper subspace defined by $A$, is zero. Use this for the finitely many orthogonal complements of the eigenspaces of $A$.
$endgroup$
– Hagen von Eitzen
Mar 14 at 5:30
$begingroup$
@HagenvonEitzen: Thanks for your comment. But I could not understand what you mean by "proper subspace defined by $A$"? Do you mean the eigenspaces of $A$? I cannot see how this can be argued. Could you point me a reference?
$endgroup$
– MyCindy2012
Mar 14 at 5:36
$begingroup$
The probability that $A$ can be diagonalized over $mathbbR$ tends to $0$ when $n$ tends to $infty$.
$endgroup$
– loup blanc
Mar 15 at 9:40
$begingroup$
Unless $v=0$, the probability that $v$ is is in any proper subspace defined by $A$, is zero. Use this for the finitely many orthogonal complements of the eigenspaces of $A$.
$endgroup$
– Hagen von Eitzen
Mar 14 at 5:30
$begingroup$
Unless $v=0$, the probability that $v$ is is in any proper subspace defined by $A$, is zero. Use this for the finitely many orthogonal complements of the eigenspaces of $A$.
$endgroup$
– Hagen von Eitzen
Mar 14 at 5:30
$begingroup$
@HagenvonEitzen: Thanks for your comment. But I could not understand what you mean by "proper subspace defined by $A$"? Do you mean the eigenspaces of $A$? I cannot see how this can be argued. Could you point me a reference?
$endgroup$
– MyCindy2012
Mar 14 at 5:36
$begingroup$
@HagenvonEitzen: Thanks for your comment. But I could not understand what you mean by "proper subspace defined by $A$"? Do you mean the eigenspaces of $A$? I cannot see how this can be argued. Could you point me a reference?
$endgroup$
– MyCindy2012
Mar 14 at 5:36
$begingroup$
The probability that $A$ can be diagonalized over $mathbbR$ tends to $0$ when $n$ tends to $infty$.
$endgroup$
– loup blanc
Mar 15 at 9:40
$begingroup$
The probability that $A$ can be diagonalized over $mathbbR$ tends to $0$ when $n$ tends to $infty$.
$endgroup$
– loup blanc
Mar 15 at 9:40
add a comment |
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$begingroup$
Unless $v=0$, the probability that $v$ is is in any proper subspace defined by $A$, is zero. Use this for the finitely many orthogonal complements of the eigenspaces of $A$.
$endgroup$
– Hagen von Eitzen
Mar 14 at 5:30
$begingroup$
@HagenvonEitzen: Thanks for your comment. But I could not understand what you mean by "proper subspace defined by $A$"? Do you mean the eigenspaces of $A$? I cannot see how this can be argued. Could you point me a reference?
$endgroup$
– MyCindy2012
Mar 14 at 5:36
$begingroup$
The probability that $A$ can be diagonalized over $mathbbR$ tends to $0$ when $n$ tends to $infty$.
$endgroup$
– loup blanc
Mar 15 at 9:40