The diagonalizable matrices are not dense in the square real matrices
$begingroup$
Suppose that $n ge 2$. How to prove that the set $mathcal D subset M_n(mathbb R)$ of the diagonalizable real matrices is not dense in $M_n(mathbb R)$?
general-topology matrices diagonalization
asked Mar 13 '17 at 8:41
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
|
show 3 more comments
$begingroup$
Suppose that $n ge 2$. How to prove that the set $mathcal D subset M_n(mathbb R)$ of the diagonalizable real matrices is not dense in $M_n(mathbb R)$?
general-topology matrices diagonalization
asked Mar 13 '17 at 8:41
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
Can you find a non-zero linear functional which vanishes on diagonal matrices? Then, we can use Hahn-Banach theorem to conclude non-density.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 13 '17 at 8:46
$begingroup$
@Nitrogen $mathcal D$ is not a closet subset. The sequence of diagonalizable matrices $A_nbegin{pmatrix} 1 - 1/n & 1\ 0 & 1+1/nend{pmatrix}$ converges to a non diagonalizable matrix.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:00
$begingroup$
@астонвіллаолофмэллбэрг Such a linear functional can't exist.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:03
$begingroup$
But it must, right? If you take the closure of the diagonalizable matrices, then this won't be the entire space, so you can apply Hahn-Banach, I think. (I could still be wrong, but I certainly think it applies in this context).
$endgroup$
– астон вілла олоф мэллбэрг
Mar 13 '17 at 9:13
1
$begingroup$
@астонвіллаолофмэллбэрг $mathcal D$ is not convex.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:41
|
show 3 more comments
$begingroup$
Suppose that $n ge 2$. How to prove that the set $mathcal D subset M_n(mathbb R)$ of the diagonalizable real matrices is not dense in $M_n(mathbb R)$?
general-topology matrices diagonalization
asked Mar 13 '17 at 8:41
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
Suppose that $n ge 2$. How to prove that the set $mathcal D subset M_n(mathbb R)$ of the diagonalizable real matrices is not dense in $M_n(mathbb R)$?
general-topology matrices diagonalization
general-topology matrices diagonalization
asked Mar 13 '17 at 8:41
mathcounterexamples.netmathcounterexamples.net
26.9k22158
asked Mar 13 '17 at 8:41
mathcounterexamples.netmathcounterexamples.net
26.9k22158
asked Mar 13 '17 at 8:41
mathcounterexamples.netmathcounterexamples.net
26.9k22158
asked Mar 13 '17 at 8:41
mathcounterexamples.netmathcounterexamples.net
26.9k22158
asked Mar 13 '17 at 8:41
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
$begingroup$
Can you find a non-zero linear functional which vanishes on diagonal matrices? Then, we can use Hahn-Banach theorem to conclude non-density.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 13 '17 at 8:46
$begingroup$
@Nitrogen $mathcal D$ is not a closet subset. The sequence of diagonalizable matrices $A_nbegin{pmatrix} 1 - 1/n & 1\ 0 & 1+1/nend{pmatrix}$ converges to a non diagonalizable matrix.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:00
$begingroup$
@астонвіллаолофмэллбэрг Such a linear functional can't exist.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:03
$begingroup$
But it must, right? If you take the closure of the diagonalizable matrices, then this won't be the entire space, so you can apply Hahn-Banach, I think. (I could still be wrong, but I certainly think it applies in this context).
$endgroup$
– астон вілла олоф мэллбэрг
Mar 13 '17 at 9:13
1
$begingroup$
@астонвіллаолофмэллбэрг $mathcal D$ is not convex.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:41
|
show 3 more comments
$begingroup$
Can you find a non-zero linear functional which vanishes on diagonal matrices? Then, we can use Hahn-Banach theorem to conclude non-density.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 13 '17 at 8:46
$begingroup$
@Nitrogen $mathcal D$ is not a closet subset. The sequence of diagonalizable matrices $A_nbegin{pmatrix} 1 - 1/n & 1\ 0 & 1+1/nend{pmatrix}$ converges to a non diagonalizable matrix.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:00
$begingroup$
@астонвіллаолофмэллбэрг Such a linear functional can't exist.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:03
$begingroup$
But it must, right? If you take the closure of the diagonalizable matrices, then this won't be the entire space, so you can apply Hahn-Banach, I think. (I could still be wrong, but I certainly think it applies in this context).
$endgroup$
– астон вілла олоф мэллбэрг
Mar 13 '17 at 9:13
1
$begingroup$
@астонвіллаолофмэллбэрг $mathcal D$ is not convex.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:41
$begingroup$
Can you find a non-zero linear functional which vanishes on diagonal matrices? Then, we can use Hahn-Banach theorem to conclude non-density.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 13 '17 at 8:46
$begingroup$
Can you find a non-zero linear functional which vanishes on diagonal matrices? Then, we can use Hahn-Banach theorem to conclude non-density.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 13 '17 at 8:46
$begingroup$
@Nitrogen $mathcal D$ is not a closet subset. The sequence of diagonalizable matrices $A_nbegin{pmatrix} 1 - 1/n & 1\ 0 & 1+1/nend{pmatrix}$ converges to a non diagonalizable matrix.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:00
$begingroup$
@Nitrogen $mathcal D$ is not a closet subset. The sequence of diagonalizable matrices $A_nbegin{pmatrix} 1 - 1/n & 1\ 0 & 1+1/nend{pmatrix}$ converges to a non diagonalizable matrix.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:00
$begingroup$
@астонвіллаолофмэллбэрг Such a linear functional can't exist.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:03
$begingroup$
@астонвіллаолофмэллбэрг Such a linear functional can't exist.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:03
$begingroup$
But it must, right? If you take the closure of the diagonalizable matrices, then this won't be the entire space, so you can apply Hahn-Banach, I think. (I could still be wrong, but I certainly think it applies in this context).
$endgroup$
– астон вілла олоф мэллбэрг
Mar 13 '17 at 9:13
$begingroup$
But it must, right? If you take the closure of the diagonalizable matrices, then this won't be the entire space, so you can apply Hahn-Banach, I think. (I could still be wrong, but I certainly think it applies in this context).
$endgroup$
– астон вілла олоф мэллбэрг
Mar 13 '17 at 9:13
1
1
$begingroup$
@астонвіллаолофмэллбэрг $mathcal D$ is not convex.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:41
$begingroup$
@астонвіллаолофмэллбэрг $mathcal D$ is not convex.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:41
|
show 3 more comments
1 Answer
1
active
oldest
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$begingroup$
Consider the matrix :
$$R=pmatrix{0 & -1cr 1 & 0}in M_2(mathbb{R})$$
which is not diagonalizable since its characteristic polynomial $X^2+1$ does not split in $mathbb{R}[X]$.
Suppose there exists a sequence $(D_n)$ of diagonalizable matrices in $M_2(mathbb{R})$, which converges to $R$.
For every $n$, the characteristic polynomial of $D_n$ has nonnegative discriminant and by continuity of the determinant, it should be the same for $R$, but this is not the case.
This proves that the set $mathcal{D}_2$ of all diagonalizable matrices in $M_2(mathbb{R})$ is not dense in $M_2(mathbb{R})$.
answered Mar 13 '17 at 9:03
AdrenAdren
5,413519
$endgroup$
$begingroup$
Merci! Qu'est-ce qui t'a fait penser à utiliser le discriminant?
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:12
$begingroup$
@mathcounterexamples.net: Le discriminant donne un moyen naturel de savoir si un trinôme est scindé. Si une suite de matrices réelles dont les polynômes caractéristiques sont tous scindés dans $mathbb{R}[X]$ converge, sa limite limite possède ainsi la même propriété (en taille $nge2$ quelconque, on peut utiliser le même outil : le discriminant d'un polynôme $P$ est le résultant de $P$ et de $P'$). D'où l'idée d'aller chercher une matrice réelle dont le polynôme caractéristique n'est pas scindé dans $mathbb{R}[X]$.
$endgroup$
– Adren
Mar 13 '17 at 9:19
add a comment |
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1 Answer
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$begingroup$
Consider the matrix :
$$R=pmatrix{0 & -1cr 1 & 0}in M_2(mathbb{R})$$
which is not diagonalizable since its characteristic polynomial $X^2+1$ does not split in $mathbb{R}[X]$.
Suppose there exists a sequence $(D_n)$ of diagonalizable matrices in $M_2(mathbb{R})$, which converges to $R$.
For every $n$, the characteristic polynomial of $D_n$ has nonnegative discriminant and by continuity of the determinant, it should be the same for $R$, but this is not the case.
This proves that the set $mathcal{D}_2$ of all diagonalizable matrices in $M_2(mathbb{R})$ is not dense in $M_2(mathbb{R})$.
answered Mar 13 '17 at 9:03
AdrenAdren
5,413519
$endgroup$
$begingroup$
Merci! Qu'est-ce qui t'a fait penser à utiliser le discriminant?
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:12
$begingroup$
@mathcounterexamples.net: Le discriminant donne un moyen naturel de savoir si un trinôme est scindé. Si une suite de matrices réelles dont les polynômes caractéristiques sont tous scindés dans $mathbb{R}[X]$ converge, sa limite limite possède ainsi la même propriété (en taille $nge2$ quelconque, on peut utiliser le même outil : le discriminant d'un polynôme $P$ est le résultant de $P$ et de $P'$). D'où l'idée d'aller chercher une matrice réelle dont le polynôme caractéristique n'est pas scindé dans $mathbb{R}[X]$.
$endgroup$
– Adren
Mar 13 '17 at 9:19
add a comment |
$begingroup$
Consider the matrix :
$$R=pmatrix{0 & -1cr 1 & 0}in M_2(mathbb{R})$$
which is not diagonalizable since its characteristic polynomial $X^2+1$ does not split in $mathbb{R}[X]$.
Suppose there exists a sequence $(D_n)$ of diagonalizable matrices in $M_2(mathbb{R})$, which converges to $R$.
For every $n$, the characteristic polynomial of $D_n$ has nonnegative discriminant and by continuity of the determinant, it should be the same for $R$, but this is not the case.
This proves that the set $mathcal{D}_2$ of all diagonalizable matrices in $M_2(mathbb{R})$ is not dense in $M_2(mathbb{R})$.
answered Mar 13 '17 at 9:03
AdrenAdren
5,413519
$endgroup$
$begingroup$
Merci! Qu'est-ce qui t'a fait penser à utiliser le discriminant?
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:12
$begingroup$
@mathcounterexamples.net: Le discriminant donne un moyen naturel de savoir si un trinôme est scindé. Si une suite de matrices réelles dont les polynômes caractéristiques sont tous scindés dans $mathbb{R}[X]$ converge, sa limite limite possède ainsi la même propriété (en taille $nge2$ quelconque, on peut utiliser le même outil : le discriminant d'un polynôme $P$ est le résultant de $P$ et de $P'$). D'où l'idée d'aller chercher une matrice réelle dont le polynôme caractéristique n'est pas scindé dans $mathbb{R}[X]$.
$endgroup$
– Adren
Mar 13 '17 at 9:19
add a comment |
$begingroup$
Consider the matrix :
$$R=pmatrix{0 & -1cr 1 & 0}in M_2(mathbb{R})$$
which is not diagonalizable since its characteristic polynomial $X^2+1$ does not split in $mathbb{R}[X]$.
Suppose there exists a sequence $(D_n)$ of diagonalizable matrices in $M_2(mathbb{R})$, which converges to $R$.
For every $n$, the characteristic polynomial of $D_n$ has nonnegative discriminant and by continuity of the determinant, it should be the same for $R$, but this is not the case.
This proves that the set $mathcal{D}_2$ of all diagonalizable matrices in $M_2(mathbb{R})$ is not dense in $M_2(mathbb{R})$.
answered Mar 13 '17 at 9:03
AdrenAdren
5,413519
$endgroup$
Consider the matrix :
$$R=pmatrix{0 & -1cr 1 & 0}in M_2(mathbb{R})$$
which is not diagonalizable since its characteristic polynomial $X^2+1$ does not split in $mathbb{R}[X]$.
Suppose there exists a sequence $(D_n)$ of diagonalizable matrices in $M_2(mathbb{R})$, which converges to $R$.
For every $n$, the characteristic polynomial of $D_n$ has nonnegative discriminant and by continuity of the determinant, it should be the same for $R$, but this is not the case.
This proves that the set $mathcal{D}_2$ of all diagonalizable matrices in $M_2(mathbb{R})$ is not dense in $M_2(mathbb{R})$.
answered Mar 13 '17 at 9:03
AdrenAdren
5,413519
answered Mar 13 '17 at 9:03
AdrenAdren
5,413519
answered Mar 13 '17 at 9:03
AdrenAdren
5,413519
answered Mar 13 '17 at 9:03
AdrenAdren
5,413519
5,413519
$begingroup$
Merci! Qu'est-ce qui t'a fait penser à utiliser le discriminant?
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:12
$begingroup$
@mathcounterexamples.net: Le discriminant donne un moyen naturel de savoir si un trinôme est scindé. Si une suite de matrices réelles dont les polynômes caractéristiques sont tous scindés dans $mathbb{R}[X]$ converge, sa limite limite possède ainsi la même propriété (en taille $nge2$ quelconque, on peut utiliser le même outil : le discriminant d'un polynôme $P$ est le résultant de $P$ et de $P'$). D'où l'idée d'aller chercher une matrice réelle dont le polynôme caractéristique n'est pas scindé dans $mathbb{R}[X]$.
$endgroup$
– Adren
Mar 13 '17 at 9:19
add a comment |
$begingroup$
Merci! Qu'est-ce qui t'a fait penser à utiliser le discriminant?
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:12
$begingroup$
@mathcounterexamples.net: Le discriminant donne un moyen naturel de savoir si un trinôme est scindé. Si une suite de matrices réelles dont les polynômes caractéristiques sont tous scindés dans $mathbb{R}[X]$ converge, sa limite limite possède ainsi la même propriété (en taille $nge2$ quelconque, on peut utiliser le même outil : le discriminant d'un polynôme $P$ est le résultant de $P$ et de $P'$). D'où l'idée d'aller chercher une matrice réelle dont le polynôme caractéristique n'est pas scindé dans $mathbb{R}[X]$.
$endgroup$
– Adren
Mar 13 '17 at 9:19
$begingroup$
Merci! Qu'est-ce qui t'a fait penser à utiliser le discriminant?
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:12
$begingroup$
Merci! Qu'est-ce qui t'a fait penser à utiliser le discriminant?
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:12
$begingroup$
@mathcounterexamples.net: Le discriminant donne un moyen naturel de savoir si un trinôme est scindé. Si une suite de matrices réelles dont les polynômes caractéristiques sont tous scindés dans $mathbb{R}[X]$ converge, sa limite limite possède ainsi la même propriété (en taille $nge2$ quelconque, on peut utiliser le même outil : le discriminant d'un polynôme $P$ est le résultant de $P$ et de $P'$). D'où l'idée d'aller chercher une matrice réelle dont le polynôme caractéristique n'est pas scindé dans $mathbb{R}[X]$.
$endgroup$
– Adren
Mar 13 '17 at 9:19
$begingroup$
@mathcounterexamples.net: Le discriminant donne un moyen naturel de savoir si un trinôme est scindé. Si une suite de matrices réelles dont les polynômes caractéristiques sont tous scindés dans $mathbb{R}[X]$ converge, sa limite limite possède ainsi la même propriété (en taille $nge2$ quelconque, on peut utiliser le même outil : le discriminant d'un polynôme $P$ est le résultant de $P$ et de $P'$). D'où l'idée d'aller chercher une matrice réelle dont le polynôme caractéristique n'est pas scindé dans $mathbb{R}[X]$.
$endgroup$
– Adren
Mar 13 '17 at 9:19
add a comment |
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$begingroup$
Can you find a non-zero linear functional which vanishes on diagonal matrices? Then, we can use Hahn-Banach theorem to conclude non-density.
$endgroup$
– астон вілла олоф мэллбэрг
Mar 13 '17 at 8:46
$begingroup$
@Nitrogen $mathcal D$ is not a closet subset. The sequence of diagonalizable matrices $A_nbegin{pmatrix} 1 - 1/n & 1\ 0 & 1+1/nend{pmatrix}$ converges to a non diagonalizable matrix.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:00
$begingroup$
@астонвіллаолофмэллбэрг Such a linear functional can't exist.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:03
$begingroup$
But it must, right? If you take the closure of the diagonalizable matrices, then this won't be the entire space, so you can apply Hahn-Banach, I think. (I could still be wrong, but I certainly think it applies in this context).
$endgroup$
– астон вілла олоф мэллбэрг
Mar 13 '17 at 9:13
1
$begingroup$
@астонвіллаолофмэллбэрг $mathcal D$ is not convex.
$endgroup$
– mathcounterexamples.net
Mar 13 '17 at 9:41