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Are these square matrices always diagonalisable?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30 pm US/Eastern)Conditions for diagonalizability of $ntimes n$ anti-diagonal matricesFinding eigenvalues/vectors of a matrix and proving it is not diagonalisable.Finding if a linear transformation is diagonalisableThe diagonalisation of the two matricesIf $A in K^{n times n}$ is diagonalisable, the dimension of the subspace of its commuting matrices is $geq n$ nCalculating the eigenvalues of a diagonalisable linear operator $L$.How to find eigenvalues for mod 2 field?Find for which real parameter a matrix is diagonalisableSquare Roots of a Matrix: Diagonalisable Solutions.eigenvalues and eigenvectors of Diagonalisable matrices
$begingroup$
When trying to solve a physics problem on decoupling a system of ODEs, I found myself needing to address the following problem:
Let $A_nin M_n(mathbb R)$ be the matrix with all $1$s above its main diagonal, all $-1$s below its diagonal, and $0$s everywhere else. Is $A_n$ always diagonalisable? If so, what is its diagonalisation (equivalently: what are its eigenvalues and corresponding eigenvectors)?
For example,
$$A_3=begin{bmatrix}0&1&0\-1&0&1\0&-1&0end{bmatrix},quad A_5begin{bmatrix}0&1&0&0&0\-1&0&1&0&0\0&-1&0&1&0\0&0&-1&0&1\0&0&0&-1&0end{bmatrix}.$$
Assuming my code is correct, Mathematica has been able to verify that $A_n$ is always diagonalisable up to $n=1000$. If we use $chi_n(t)inmathbb Z[t]$ to denote the characteristic polynomial of $A_n$, a straightforward evaluation also shows that
$$chi_n(t)=-tchi_{n-1}(t)+chi_{n-2}(t)tag{1}$$
for all $ngeq4$. Furthermore, note that $A_n=-A_n^t$ so that, in the case where the dimension is even,
$$det(A_{2n}-lambda I)=det(A_{2n}^t-lambda I)=det(-A_{2n}-lambda I)=det(A_{2n}+lambda I).$$
This implies that whenever $lambda$ is an eigenvalue of $A_{2n}$, so is $-lambda$. In other words, $chi_{2n}(t)$ is always of the form $(t^2-lambda _1^2)(t^2-lambda_2^2)dotsm(t^2-lambda_n^2)$ for some $lambda_i$.
And this is where I am stuck. In order for $A_n$ to be diagonalisable, we must have that all the eigenvalues are distinct, but trying to use the recurrence $(1)$ and strong induction, or trying to use the formula for the even case have not helped at all. It seems like the most probable line of attack would be to somehow show that
$$chi_{2n}'(t)=2tsum_{k=1}^nfrac{chi_{2n}(t)}{t^2-lambda_k^2}$$
never shares a common zero with $chi_{2n}$ (which would resolve the even case), though I don't see how to make this work.
Note: I do not have any clue how to actually find the eigenvalues/eigenvectors even in the case where the $A_n$ are diagonalisable. As such even if someone cannot answer the second part of the question, but can prove that the $A_n$ are diagonalisable, I would appreciate that as an answer as well. Above I tried to look at the special case where the dimension is even, though of course the proof for all odd and even $n$ is more valuable. Even if this is not possible, for my purposes I just need an unbounded subset $Ssubseteqmathbb Z$ for which the conclusion is proven for $nin S$, so any such approach is welcome too.
Thank you in advance!
linear-algebra eigenvalues-eigenvectors determinant diagonalization
$endgroup$
add a comment |
$begingroup$
When trying to solve a physics problem on decoupling a system of ODEs, I found myself needing to address the following problem:
Let $A_nin M_n(mathbb R)$ be the matrix with all $1$s above its main diagonal, all $-1$s below its diagonal, and $0$s everywhere else. Is $A_n$ always diagonalisable? If so, what is its diagonalisation (equivalently: what are its eigenvalues and corresponding eigenvectors)?
For example,
$$A_3=begin{bmatrix}0&1&0\-1&0&1\0&-1&0end{bmatrix},quad A_5begin{bmatrix}0&1&0&0&0\-1&0&1&0&0\0&-1&0&1&0\0&0&-1&0&1\0&0&0&-1&0end{bmatrix}.$$
Assuming my code is correct, Mathematica has been able to verify that $A_n$ is always diagonalisable up to $n=1000$. If we use $chi_n(t)inmathbb Z[t]$ to denote the characteristic polynomial of $A_n$, a straightforward evaluation also shows that
$$chi_n(t)=-tchi_{n-1}(t)+chi_{n-2}(t)tag{1}$$
for all $ngeq4$. Furthermore, note that $A_n=-A_n^t$ so that, in the case where the dimension is even,
$$det(A_{2n}-lambda I)=det(A_{2n}^t-lambda I)=det(-A_{2n}-lambda I)=det(A_{2n}+lambda I).$$
This implies that whenever $lambda$ is an eigenvalue of $A_{2n}$, so is $-lambda$. In other words, $chi_{2n}(t)$ is always of the form $(t^2-lambda _1^2)(t^2-lambda_2^2)dotsm(t^2-lambda_n^2)$ for some $lambda_i$.
And this is where I am stuck. In order for $A_n$ to be diagonalisable, we must have that all the eigenvalues are distinct, but trying to use the recurrence $(1)$ and strong induction, or trying to use the formula for the even case have not helped at all. It seems like the most probable line of attack would be to somehow show that
$$chi_{2n}'(t)=2tsum_{k=1}^nfrac{chi_{2n}(t)}{t^2-lambda_k^2}$$
never shares a common zero with $chi_{2n}$ (which would resolve the even case), though I don't see how to make this work.
Note: I do not have any clue how to actually find the eigenvalues/eigenvectors even in the case where the $A_n$ are diagonalisable. As such even if someone cannot answer the second part of the question, but can prove that the $A_n$ are diagonalisable, I would appreciate that as an answer as well. Above I tried to look at the special case where the dimension is even, though of course the proof for all odd and even $n$ is more valuable. Even if this is not possible, for my purposes I just need an unbounded subset $Ssubseteqmathbb Z$ for which the conclusion is proven for $nin S$, so any such approach is welcome too.
Thank you in advance!
linear-algebra eigenvalues-eigenvectors determinant diagonalization
$endgroup$
add a comment |
$begingroup$
When trying to solve a physics problem on decoupling a system of ODEs, I found myself needing to address the following problem:
Let $A_nin M_n(mathbb R)$ be the matrix with all $1$s above its main diagonal, all $-1$s below its diagonal, and $0$s everywhere else. Is $A_n$ always diagonalisable? If so, what is its diagonalisation (equivalently: what are its eigenvalues and corresponding eigenvectors)?
For example,
$$A_3=begin{bmatrix}0&1&0\-1&0&1\0&-1&0end{bmatrix},quad A_5begin{bmatrix}0&1&0&0&0\-1&0&1&0&0\0&-1&0&1&0\0&0&-1&0&1\0&0&0&-1&0end{bmatrix}.$$
Assuming my code is correct, Mathematica has been able to verify that $A_n$ is always diagonalisable up to $n=1000$. If we use $chi_n(t)inmathbb Z[t]$ to denote the characteristic polynomial of $A_n$, a straightforward evaluation also shows that
$$chi_n(t)=-tchi_{n-1}(t)+chi_{n-2}(t)tag{1}$$
for all $ngeq4$. Furthermore, note that $A_n=-A_n^t$ so that, in the case where the dimension is even,
$$det(A_{2n}-lambda I)=det(A_{2n}^t-lambda I)=det(-A_{2n}-lambda I)=det(A_{2n}+lambda I).$$
This implies that whenever $lambda$ is an eigenvalue of $A_{2n}$, so is $-lambda$. In other words, $chi_{2n}(t)$ is always of the form $(t^2-lambda _1^2)(t^2-lambda_2^2)dotsm(t^2-lambda_n^2)$ for some $lambda_i$.
And this is where I am stuck. In order for $A_n$ to be diagonalisable, we must have that all the eigenvalues are distinct, but trying to use the recurrence $(1)$ and strong induction, or trying to use the formula for the even case have not helped at all. It seems like the most probable line of attack would be to somehow show that
$$chi_{2n}'(t)=2tsum_{k=1}^nfrac{chi_{2n}(t)}{t^2-lambda_k^2}$$
never shares a common zero with $chi_{2n}$ (which would resolve the even case), though I don't see how to make this work.
Note: I do not have any clue how to actually find the eigenvalues/eigenvectors even in the case where the $A_n$ are diagonalisable. As such even if someone cannot answer the second part of the question, but can prove that the $A_n$ are diagonalisable, I would appreciate that as an answer as well. Above I tried to look at the special case where the dimension is even, though of course the proof for all odd and even $n$ is more valuable. Even if this is not possible, for my purposes I just need an unbounded subset $Ssubseteqmathbb Z$ for which the conclusion is proven for $nin S$, so any such approach is welcome too.
Thank you in advance!
linear-algebra eigenvalues-eigenvectors determinant diagonalization
$endgroup$
When trying to solve a physics problem on decoupling a system of ODEs, I found myself needing to address the following problem:
Let $A_nin M_n(mathbb R)$ be the matrix with all $1$s above its main diagonal, all $-1$s below its diagonal, and $0$s everywhere else. Is $A_n$ always diagonalisable? If so, what is its diagonalisation (equivalently: what are its eigenvalues and corresponding eigenvectors)?
For example,
$$A_3=begin{bmatrix}0&1&0\-1&0&1\0&-1&0end{bmatrix},quad A_5begin{bmatrix}0&1&0&0&0\-1&0&1&0&0\0&-1&0&1&0\0&0&-1&0&1\0&0&0&-1&0end{bmatrix}.$$
Assuming my code is correct, Mathematica has been able to verify that $A_n$ is always diagonalisable up to $n=1000$. If we use $chi_n(t)inmathbb Z[t]$ to denote the characteristic polynomial of $A_n$, a straightforward evaluation also shows that
$$chi_n(t)=-tchi_{n-1}(t)+chi_{n-2}(t)tag{1}$$
for all $ngeq4$. Furthermore, note that $A_n=-A_n^t$ so that, in the case where the dimension is even,
$$det(A_{2n}-lambda I)=det(A_{2n}^t-lambda I)=det(-A_{2n}-lambda I)=det(A_{2n}+lambda I).$$
This implies that whenever $lambda$ is an eigenvalue of $A_{2n}$, so is $-lambda$. In other words, $chi_{2n}(t)$ is always of the form $(t^2-lambda _1^2)(t^2-lambda_2^2)dotsm(t^2-lambda_n^2)$ for some $lambda_i$.
And this is where I am stuck. In order for $A_n$ to be diagonalisable, we must have that all the eigenvalues are distinct, but trying to use the recurrence $(1)$ and strong induction, or trying to use the formula for the even case have not helped at all. It seems like the most probable line of attack would be to somehow show that
$$chi_{2n}'(t)=2tsum_{k=1}^nfrac{chi_{2n}(t)}{t^2-lambda_k^2}$$
never shares a common zero with $chi_{2n}$ (which would resolve the even case), though I don't see how to make this work.
Note: I do not have any clue how to actually find the eigenvalues/eigenvectors even in the case where the $A_n$ are diagonalisable. As such even if someone cannot answer the second part of the question, but can prove that the $A_n$ are diagonalisable, I would appreciate that as an answer as well. Above I tried to look at the special case where the dimension is even, though of course the proof for all odd and even $n$ is more valuable. Even if this is not possible, for my purposes I just need an unbounded subset $Ssubseteqmathbb Z$ for which the conclusion is proven for $nin S$, so any such approach is welcome too.
Thank you in advance!
linear-algebra eigenvalues-eigenvectors determinant diagonalization
linear-algebra eigenvalues-eigenvectors determinant diagonalization
asked 55 mins ago
YiFanYiFan
5,6152829
5,6152829
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3 Answers
3
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$begingroup$
The matrix $A_n$ is a tridiagonal Toeplitz matrix with diagonal entries $delta = 0$ and off-diagonal entries $tau = 1$ and $sigma = -1$. Hence, we can use the formula in this paper to show that the eigenvalues are $$lambda_k = 2icosleft(dfrac{kpi}{n+1}right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfrac{mkpi}{n+1}right).$$
$endgroup$
add a comment |
$begingroup$
All those matrices are anti-symmetric and therefore they are normal matrices. And every normal matrix is diagonalizable over $mathbb C$, by the spectral theorem.
$endgroup$
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
1
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
30 mins ago
add a comment |
$begingroup$
Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article.
New contributor
$endgroup$
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
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
$begingroup$
The matrix $A_n$ is a tridiagonal Toeplitz matrix with diagonal entries $delta = 0$ and off-diagonal entries $tau = 1$ and $sigma = -1$. Hence, we can use the formula in this paper to show that the eigenvalues are $$lambda_k = 2icosleft(dfrac{kpi}{n+1}right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfrac{mkpi}{n+1}right).$$
$endgroup$
add a comment |
$begingroup$
The matrix $A_n$ is a tridiagonal Toeplitz matrix with diagonal entries $delta = 0$ and off-diagonal entries $tau = 1$ and $sigma = -1$. Hence, we can use the formula in this paper to show that the eigenvalues are $$lambda_k = 2icosleft(dfrac{kpi}{n+1}right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfrac{mkpi}{n+1}right).$$
$endgroup$
add a comment |
$begingroup$
The matrix $A_n$ is a tridiagonal Toeplitz matrix with diagonal entries $delta = 0$ and off-diagonal entries $tau = 1$ and $sigma = -1$. Hence, we can use the formula in this paper to show that the eigenvalues are $$lambda_k = 2icosleft(dfrac{kpi}{n+1}right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfrac{mkpi}{n+1}right).$$
$endgroup$
The matrix $A_n$ is a tridiagonal Toeplitz matrix with diagonal entries $delta = 0$ and off-diagonal entries $tau = 1$ and $sigma = -1$. Hence, we can use the formula in this paper to show that the eigenvalues are $$lambda_k = 2icosleft(dfrac{kpi}{n+1}right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfrac{mkpi}{n+1}right).$$
edited 40 mins ago
answered 45 mins ago
JimmyK4542JimmyK4542
41.5k246108
41.5k246108
add a comment |
add a comment |
$begingroup$
All those matrices are anti-symmetric and therefore they are normal matrices. And every normal matrix is diagonalizable over $mathbb C$, by the spectral theorem.
$endgroup$
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
1
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
30 mins ago
add a comment |
$begingroup$
All those matrices are anti-symmetric and therefore they are normal matrices. And every normal matrix is diagonalizable over $mathbb C$, by the spectral theorem.
$endgroup$
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
1
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
30 mins ago
add a comment |
$begingroup$
All those matrices are anti-symmetric and therefore they are normal matrices. And every normal matrix is diagonalizable over $mathbb C$, by the spectral theorem.
$endgroup$
All those matrices are anti-symmetric and therefore they are normal matrices. And every normal matrix is diagonalizable over $mathbb C$, by the spectral theorem.
answered 42 mins ago
José Carlos SantosJosé Carlos Santos
177k24138250
177k24138250
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
1
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
30 mins ago
add a comment |
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
1
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
30 mins ago
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
1
1
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
30 mins ago
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
30 mins ago
add a comment |
$begingroup$
Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article.
New contributor
$endgroup$
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
add a comment |
$begingroup$
Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article.
New contributor
$endgroup$
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
add a comment |
$begingroup$
Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article.
New contributor
$endgroup$
Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article.
New contributor
New contributor
answered 40 mins ago
gcousingcousin
1012
1012
New contributor
New contributor
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
add a comment |
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
$begingroup$
Thank you very much for the quick response! I hope you don't mind that I accepted JimmyK4542's answer, since it also gives explicitly the eigenvectors and eigenvalues.
$endgroup$
– YiFan
31 mins ago
add a comment |
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