The diagonalizable matrices are not dense in the square real matrices












1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














1












1








1


2



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















5












$begingroup$

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 $M_2(mathbb{R})$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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 $M_2(mathbb{R})$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















5












$begingroup$

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 $M_2(mathbb{R})$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















5












5








5





$begingroup$

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 $M_2(mathbb{R})$.






share|cite|improve this answer









$endgroup$



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 $M_2(mathbb{R})$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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