Mixing
Are all matrices almost diagonalizable?
$begingroup$
Every real $2$ by $2$ matrix that is not diagonalizable is similar to the $2$ by $2$ jordan canonical form,
$$
J_2=begin{bmatrix}s&1\0&send{bmatrix},
$$
where $s$ is the eigenvalue (with multiplicity $2$). My question: Is $J_2$ almost diagonalizable? I mean is it similar to
$$
begin{bmatrix}s&epsilon\0&send{bmatrix}
$$
for every $epsilon>0$ no matter how small? And what happens in higher dimensions, is every matrix similar to an arbitrarily close to diagonal matrix?
My guess is yes but I just can't find the similarity transformation. It could be something very simple.
Thanks in advance, all ideas welcome.
linear-algebra matrices geometry eigenvalues-eigenvectors diagonalization
edited Jan 18 at 10:47
Rodrigo de Azevedo
13.1k41959
asked Jan 18 at 10:17
plus1plus1
3911
$endgroup$
add a comment |
$begingroup$
Every real $2$ by $2$ matrix that is not diagonalizable is similar to the $2$ by $2$ jordan canonical form,
$$
J_2=begin{bmatrix}s&1\0&send{bmatrix},
$$
where $s$ is the eigenvalue (with multiplicity $2$). My question: Is $J_2$ almost diagonalizable? I mean is it similar to
$$
begin{bmatrix}s&epsilon\0&send{bmatrix}
$$
for every $epsilon>0$ no matter how small? And what happens in higher dimensions, is every matrix similar to an arbitrarily close to diagonal matrix?
My guess is yes but I just can't find the similarity transformation. It could be something very simple.
Thanks in advance, all ideas welcome.
linear-algebra matrices geometry eigenvalues-eigenvectors diagonalization
edited Jan 18 at 10:47
Rodrigo de Azevedo
13.1k41959
asked Jan 18 at 10:17
plus1plus1
3911
$endgroup$
add a comment |
$begingroup$
Every real $2$ by $2$ matrix that is not diagonalizable is similar to the $2$ by $2$ jordan canonical form,
$$
J_2=begin{bmatrix}s&1\0&send{bmatrix},
$$
where $s$ is the eigenvalue (with multiplicity $2$). My question: Is $J_2$ almost diagonalizable? I mean is it similar to
$$
begin{bmatrix}s&epsilon\0&send{bmatrix}
$$
for every $epsilon>0$ no matter how small? And what happens in higher dimensions, is every matrix similar to an arbitrarily close to diagonal matrix?
My guess is yes but I just can't find the similarity transformation. It could be something very simple.
Thanks in advance, all ideas welcome.
linear-algebra matrices geometry eigenvalues-eigenvectors diagonalization
edited Jan 18 at 10:47
Rodrigo de Azevedo
13.1k41959
asked Jan 18 at 10:17
plus1plus1
3911
$endgroup$
Every real $2$ by $2$ matrix that is not diagonalizable is similar to the $2$ by $2$ jordan canonical form,
$$
J_2=begin{bmatrix}s&1\0&send{bmatrix},
$$
where $s$ is the eigenvalue (with multiplicity $2$). My question: Is $J_2$ almost diagonalizable? I mean is it similar to
$$
begin{bmatrix}s&epsilon\0&send{bmatrix}
$$
for every $epsilon>0$ no matter how small? And what happens in higher dimensions, is every matrix similar to an arbitrarily close to diagonal matrix?
My guess is yes but I just can't find the similarity transformation. It could be something very simple.
Thanks in advance, all ideas welcome.
linear-algebra matrices geometry eigenvalues-eigenvectors diagonalization
linear-algebra matrices geometry eigenvalues-eigenvectors diagonalization
edited Jan 18 at 10:47
Rodrigo de Azevedo
13.1k41959
asked Jan 18 at 10:17
plus1plus1
3911
edited Jan 18 at 10:47
Rodrigo de Azevedo
13.1k41959
asked Jan 18 at 10:17
plus1plus1
3911
edited Jan 18 at 10:47
Rodrigo de Azevedo
13.1k41959
edited Jan 18 at 10:47
Rodrigo de Azevedo
13.1k41959
edited Jan 18 at 10:47
Rodrigo de Azevedo
13.1k41959
13.1k41959
asked Jan 18 at 10:17
plus1plus1
3911
asked Jan 18 at 10:17
plus1plus1
3911
asked Jan 18 at 10:17
plus1plus1
3911
3911
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Yes, this is true and you gave the answer almost by yourself. The key is the Jordan normal form. To get $epsilon$ instead of $1$, you just need to take the Jordan normal form of $frac{1}{epsilon}A$ and then multiply by $epsilon$.
answered Jan 18 at 10:20
KlausKlaus
2,0099
$endgroup$
$begingroup$
Nice one! thanks!
$endgroup$
– plus1
Jan 18 at 10:34
add a comment |
$begingroup$
Not all real matrices are close to matrices diagonalizable over $mathbb{R}$; there are open sets on which every matrix has irreducible quadratic factors.
For example, $begin{bmatrix}0&1\-1&0end{bmatrix}$ is not diagonalizable over $mathbb{R}$, because its characteristic polynomial $x^2+1$ doesn't factor. Moreover, adding anything with coefficients less than $frac13$ won't change that; the discriminant will still be negative.
Over $mathbb{C}$, where every polynomial factors, the answer is yes; we can perturb the matrix slightly by adding small random elements to the diagonal so that the eigenvalues all become different.
answered Jan 18 at 10:52
jmerryjmerry
10.8k1225
$endgroup$
add a comment |
$begingroup$
Yes, it is true. For every $ntimes n$ matrix $A$ and for every $varepsilon>0$, there is some diagonal matrix $D$ such that $A$ is similar to a matrix $D^star$ such that $lVert D-D^starrVert<varepsilon$. Here$$lVert MrVert=sqrt{sum_{i,j=1}^nlvert m_{ij}rvert^2},$$if $M=(m_{ij})_{1leqslant i,jleqslant n}$.
answered Jan 18 at 10:26
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
I guess it's also true for any other norm. I buy that but how do you prove it?
$endgroup$
– plus1
Jan 18 at 10:35
$begingroup$
On a finite dimensional real vector space, all norms are equivalente. So, yes, if it holds for one of them, it holds for all of them. Concerning a proof, your idea is good: consider Jordan matrices with the $1$'s above the main diagonal replaced by a tiny real number.
$endgroup$
– José Carlos Santos
Jan 18 at 10:45
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, this is true and you gave the answer almost by yourself. The key is the Jordan normal form. To get $epsilon$ instead of $1$, you just need to take the Jordan normal form of $frac{1}{epsilon}A$ and then multiply by $epsilon$.
answered Jan 18 at 10:20
KlausKlaus
2,0099
$endgroup$
$begingroup$
Nice one! thanks!
$endgroup$
– plus1
Jan 18 at 10:34
add a comment |
$begingroup$
Yes, this is true and you gave the answer almost by yourself. The key is the Jordan normal form. To get $epsilon$ instead of $1$, you just need to take the Jordan normal form of $frac{1}{epsilon}A$ and then multiply by $epsilon$.
answered Jan 18 at 10:20
KlausKlaus
2,0099
$endgroup$
$begingroup$
Nice one! thanks!
$endgroup$
– plus1
Jan 18 at 10:34
add a comment |
$begingroup$
Yes, this is true and you gave the answer almost by yourself. The key is the Jordan normal form. To get $epsilon$ instead of $1$, you just need to take the Jordan normal form of $frac{1}{epsilon}A$ and then multiply by $epsilon$.
answered Jan 18 at 10:20
KlausKlaus
2,0099
$endgroup$
Yes, this is true and you gave the answer almost by yourself. The key is the Jordan normal form. To get $epsilon$ instead of $1$, you just need to take the Jordan normal form of $frac{1}{epsilon}A$ and then multiply by $epsilon$.
answered Jan 18 at 10:20
KlausKlaus
2,0099
answered Jan 18 at 10:20
KlausKlaus
2,0099
answered Jan 18 at 10:20
KlausKlaus
2,0099
answered Jan 18 at 10:20
KlausKlaus
2,0099
2,0099
$begingroup$
Nice one! thanks!
$endgroup$
– plus1
Jan 18 at 10:34
add a comment |
$begingroup$
Nice one! thanks!
$endgroup$
– plus1
Jan 18 at 10:34
$begingroup$
Nice one! thanks!
$endgroup$
– plus1
Jan 18 at 10:34
$begingroup$
Nice one! thanks!
$endgroup$
– plus1
Jan 18 at 10:34
add a comment |
$begingroup$
Not all real matrices are close to matrices diagonalizable over $mathbb{R}$; there are open sets on which every matrix has irreducible quadratic factors.
For example, $begin{bmatrix}0&1\-1&0end{bmatrix}$ is not diagonalizable over $mathbb{R}$, because its characteristic polynomial $x^2+1$ doesn't factor. Moreover, adding anything with coefficients less than $frac13$ won't change that; the discriminant will still be negative.
Over $mathbb{C}$, where every polynomial factors, the answer is yes; we can perturb the matrix slightly by adding small random elements to the diagonal so that the eigenvalues all become different.
answered Jan 18 at 10:52
jmerryjmerry
10.8k1225
$endgroup$
add a comment |
$begingroup$
Not all real matrices are close to matrices diagonalizable over $mathbb{R}$; there are open sets on which every matrix has irreducible quadratic factors.
For example, $begin{bmatrix}0&1\-1&0end{bmatrix}$ is not diagonalizable over $mathbb{R}$, because its characteristic polynomial $x^2+1$ doesn't factor. Moreover, adding anything with coefficients less than $frac13$ won't change that; the discriminant will still be negative.
Over $mathbb{C}$, where every polynomial factors, the answer is yes; we can perturb the matrix slightly by adding small random elements to the diagonal so that the eigenvalues all become different.
answered Jan 18 at 10:52
jmerryjmerry
10.8k1225
$endgroup$
add a comment |
$begingroup$
Not all real matrices are close to matrices diagonalizable over $mathbb{R}$; there are open sets on which every matrix has irreducible quadratic factors.
For example, $begin{bmatrix}0&1\-1&0end{bmatrix}$ is not diagonalizable over $mathbb{R}$, because its characteristic polynomial $x^2+1$ doesn't factor. Moreover, adding anything with coefficients less than $frac13$ won't change that; the discriminant will still be negative.
Over $mathbb{C}$, where every polynomial factors, the answer is yes; we can perturb the matrix slightly by adding small random elements to the diagonal so that the eigenvalues all become different.
answered Jan 18 at 10:52
jmerryjmerry
10.8k1225
$endgroup$
Not all real matrices are close to matrices diagonalizable over $mathbb{R}$; there are open sets on which every matrix has irreducible quadratic factors.
For example, $begin{bmatrix}0&1\-1&0end{bmatrix}$ is not diagonalizable over $mathbb{R}$, because its characteristic polynomial $x^2+1$ doesn't factor. Moreover, adding anything with coefficients less than $frac13$ won't change that; the discriminant will still be negative.
Over $mathbb{C}$, where every polynomial factors, the answer is yes; we can perturb the matrix slightly by adding small random elements to the diagonal so that the eigenvalues all become different.
answered Jan 18 at 10:52
jmerryjmerry
10.8k1225
answered Jan 18 at 10:52
jmerryjmerry
10.8k1225
answered Jan 18 at 10:52
jmerryjmerry
10.8k1225
answered Jan 18 at 10:52
jmerryjmerry
10.8k1225
10.8k1225
add a comment |
add a comment |
$begingroup$
Yes, it is true. For every $ntimes n$ matrix $A$ and for every $varepsilon>0$, there is some diagonal matrix $D$ such that $A$ is similar to a matrix $D^star$ such that $lVert D-D^starrVert<varepsilon$. Here$$lVert MrVert=sqrt{sum_{i,j=1}^nlvert m_{ij}rvert^2},$$if $M=(m_{ij})_{1leqslant i,jleqslant n}$.
answered Jan 18 at 10:26
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
I guess it's also true for any other norm. I buy that but how do you prove it?
$endgroup$
– plus1
Jan 18 at 10:35
$begingroup$
On a finite dimensional real vector space, all norms are equivalente. So, yes, if it holds for one of them, it holds for all of them. Concerning a proof, your idea is good: consider Jordan matrices with the $1$'s above the main diagonal replaced by a tiny real number.
$endgroup$
– José Carlos Santos
Jan 18 at 10:45
add a comment |
$begingroup$
Yes, it is true. For every $ntimes n$ matrix $A$ and for every $varepsilon>0$, there is some diagonal matrix $D$ such that $A$ is similar to a matrix $D^star$ such that $lVert D-D^starrVert<varepsilon$. Here$$lVert MrVert=sqrt{sum_{i,j=1}^nlvert m_{ij}rvert^2},$$if $M=(m_{ij})_{1leqslant i,jleqslant n}$.
answered Jan 18 at 10:26
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
I guess it's also true for any other norm. I buy that but how do you prove it?
$endgroup$
– plus1
Jan 18 at 10:35
$begingroup$
On a finite dimensional real vector space, all norms are equivalente. So, yes, if it holds for one of them, it holds for all of them. Concerning a proof, your idea is good: consider Jordan matrices with the $1$'s above the main diagonal replaced by a tiny real number.
$endgroup$
– José Carlos Santos
Jan 18 at 10:45
add a comment |
$begingroup$
Yes, it is true. For every $ntimes n$ matrix $A$ and for every $varepsilon>0$, there is some diagonal matrix $D$ such that $A$ is similar to a matrix $D^star$ such that $lVert D-D^starrVert<varepsilon$. Here$$lVert MrVert=sqrt{sum_{i,j=1}^nlvert m_{ij}rvert^2},$$if $M=(m_{ij})_{1leqslant i,jleqslant n}$.
answered Jan 18 at 10:26
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
Yes, it is true. For every $ntimes n$ matrix $A$ and for every $varepsilon>0$, there is some diagonal matrix $D$ such that $A$ is similar to a matrix $D^star$ such that $lVert D-D^starrVert<varepsilon$. Here$$lVert MrVert=sqrt{sum_{i,j=1}^nlvert m_{ij}rvert^2},$$if $M=(m_{ij})_{1leqslant i,jleqslant n}$.
answered Jan 18 at 10:26
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 18 at 10:26
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 18 at 10:26
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 18 at 10:26
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
I guess it's also true for any other norm. I buy that but how do you prove it?
$endgroup$
– plus1
Jan 18 at 10:35
$begingroup$
On a finite dimensional real vector space, all norms are equivalente. So, yes, if it holds for one of them, it holds for all of them. Concerning a proof, your idea is good: consider Jordan matrices with the $1$'s above the main diagonal replaced by a tiny real number.
$endgroup$
– José Carlos Santos
Jan 18 at 10:45
add a comment |
$begingroup$
I guess it's also true for any other norm. I buy that but how do you prove it?
$endgroup$
– plus1
Jan 18 at 10:35
$begingroup$
On a finite dimensional real vector space, all norms are equivalente. So, yes, if it holds for one of them, it holds for all of them. Concerning a proof, your idea is good: consider Jordan matrices with the $1$'s above the main diagonal replaced by a tiny real number.
$endgroup$
– José Carlos Santos
Jan 18 at 10:45
$begingroup$
I guess it's also true for any other norm. I buy that but how do you prove it?
$endgroup$
– plus1
Jan 18 at 10:35
$begingroup$
I guess it's also true for any other norm. I buy that but how do you prove it?
$endgroup$
– plus1
Jan 18 at 10:35
$begingroup$
On a finite dimensional real vector space, all norms are equivalente. So, yes, if it holds for one of them, it holds for all of them. Concerning a proof, your idea is good: consider Jordan matrices with the $1$'s above the main diagonal replaced by a tiny real number.
$endgroup$
– José Carlos Santos
Jan 18 at 10:45
$begingroup$
On a finite dimensional real vector space, all norms are equivalente. So, yes, if it holds for one of them, it holds for all of them. Concerning a proof, your idea is good: consider Jordan matrices with the $1$'s above the main diagonal replaced by a tiny real number.
$endgroup$
– José Carlos Santos
Jan 18 at 10:45
add a comment |
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