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Are these square matrices always diagonalisable?
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,
$$A_3=beginbmatrix0&1&0\-1&0&1\0&-1&0endbmatrix,quad A_5beginbmatrix0&1&0&0&0\-1&0&1&0&0\0&-1&0&1&0\0&0&-1&0&1\0&0&0&-1&0endbmatrix.$$
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
$$chi_n(t)=-tchi_n-1(t)+chi_n-2(t)tag1$$
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
$$chi_2n'(t)=2tsum_k=1^nfracchi_2n(t)t^2-lambda_k^2$$
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
asked Apr 23 at 22:43
YiFanYiFan
5,6842831
$endgroup$
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,
$$A_3=beginbmatrix0&1&0\-1&0&1\0&-1&0endbmatrix,quad A_5beginbmatrix0&1&0&0&0\-1&0&1&0&0\0&-1&0&1&0\0&0&-1&0&1\0&0&0&-1&0endbmatrix.$$
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
$$chi_n(t)=-tchi_n-1(t)+chi_n-2(t)tag1$$
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
$$chi_2n'(t)=2tsum_k=1^nfracchi_2n(t)t^2-lambda_k^2$$
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
asked Apr 23 at 22:43
YiFanYiFan
5,6842831
$endgroup$
3
$begingroup$
All eigenvalues distinct is a sufficient but not necessary condition for a matrix to be diagonalizable.
$endgroup$
– Henning Makholm
Apr 24 at 0:23
$begingroup$
@HenningMakholm that's a very good point. But before the responses to the question, it's the only method I knew (hence all my approaches were based on that).
$endgroup$
– YiFan
Apr 24 at 4:37
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,
$$A_3=beginbmatrix0&1&0\-1&0&1\0&-1&0endbmatrix,quad A_5beginbmatrix0&1&0&0&0\-1&0&1&0&0\0&-1&0&1&0\0&0&-1&0&1\0&0&0&-1&0endbmatrix.$$
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
$$chi_n(t)=-tchi_n-1(t)+chi_n-2(t)tag1$$
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
$$chi_2n'(t)=2tsum_k=1^nfracchi_2n(t)t^2-lambda_k^2$$
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
asked Apr 23 at 22:43
YiFanYiFan
5,6842831
$endgroup$
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,
$$A_3=beginbmatrix0&1&0\-1&0&1\0&-1&0endbmatrix,quad A_5beginbmatrix0&1&0&0&0\-1&0&1&0&0\0&-1&0&1&0\0&0&-1&0&1\0&0&0&-1&0endbmatrix.$$
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
$$chi_n(t)=-tchi_n-1(t)+chi_n-2(t)tag1$$
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
$$chi_2n'(t)=2tsum_k=1^nfracchi_2n(t)t^2-lambda_k^2$$
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 Apr 23 at 22:43
YiFanYiFan
5,6842831
asked Apr 23 at 22:43
YiFanYiFan
5,6842831
asked Apr 23 at 22:43
YiFanYiFan
5,6842831
asked Apr 23 at 22:43
YiFanYiFan
5,6842831
asked Apr 23 at 22:43
YiFanYiFan
5,6842831
5,6842831
3
$begingroup$
All eigenvalues distinct is a sufficient but not necessary condition for a matrix to be diagonalizable.
$endgroup$
– Henning Makholm
Apr 24 at 0:23
$begingroup$
@HenningMakholm that's a very good point. But before the responses to the question, it's the only method I knew (hence all my approaches were based on that).
$endgroup$
– YiFan
Apr 24 at 4:37
add a comment |
3
$begingroup$
All eigenvalues distinct is a sufficient but not necessary condition for a matrix to be diagonalizable.
$endgroup$
– Henning Makholm
Apr 24 at 0:23
$begingroup$
@HenningMakholm that's a very good point. But before the responses to the question, it's the only method I knew (hence all my approaches were based on that).
$endgroup$
– YiFan
Apr 24 at 4:37
3
3
$begingroup$
All eigenvalues distinct is a sufficient but not necessary condition for a matrix to be diagonalizable.
$endgroup$
– Henning Makholm
Apr 24 at 0:23
$begingroup$
All eigenvalues distinct is a sufficient but not necessary condition for a matrix to be diagonalizable.
$endgroup$
– Henning Makholm
Apr 24 at 0:23
$begingroup$
@HenningMakholm that's a very good point. But before the responses to the question, it's the only method I knew (hence all my approaches were based on that).
$endgroup$
– YiFan
Apr 24 at 4:37
$begingroup$
@HenningMakholm that's a very good point. But before the responses to the question, it's the only method I knew (hence all my approaches were based on that).
$endgroup$
– YiFan
Apr 24 at 4:37
add a comment |
3 Answers
3
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 $$lambda_k = 2icosleft(dfrackpin+1right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfracmkpin+1right).$$
answered Apr 23 at 22:52
JimmyK4542JimmyK4542
41.8k248110
$endgroup$
1
$begingroup$
There's probably a connection to be made between these matrices and the Chebyshev polynomials, but I can't quite nail it down. Note that roots of the Chebyshev polynomial $T_n(lambda)$ are given by $lambda = cos( k pi/n)$ for $k in mathcalZ$. It's also possible to show that the characteristic polynomials $P_n(lambda)$ of the given matrices obey the recursion relation $P_n(lambda) = lambda P_n-1(lambda) + P_n-2(lambda)$; this is quite similar to the Chebyshev recursion relation $T_n(lambda) = 2 lambda T_n-1(lambda) - T_n-2(lambda)$.
$endgroup$
– Michael Seifert
Apr 24 at 13:58
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.
answered Apr 23 at 22:56
José Carlos SantosJosé Carlos Santos
178k24139253
$endgroup$
$begingroup$
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
Apr 23 at 23:06
4
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
Apr 23 at 23:08
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.
answered Apr 23 at 22:57
gcousingcousin
1412
New contributor
gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
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
Apr 23 at 23:06
$begingroup$
I think, compared to the above detailed answer, the only interest of mine is to show that you could have found all these informations on your own using google, as soon as you would have observed that your matrices were skew-symmetric,
$endgroup$
– gcousin
Apr 24 at 17:30
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 $$lambda_k = 2icosleft(dfrackpin+1right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfracmkpin+1right).$$
answered Apr 23 at 22:52
JimmyK4542JimmyK4542
41.8k248110
$endgroup$
1
$begingroup$
There's probably a connection to be made between these matrices and the Chebyshev polynomials, but I can't quite nail it down. Note that roots of the Chebyshev polynomial $T_n(lambda)$ are given by $lambda = cos( k pi/n)$ for $k in mathcalZ$. It's also possible to show that the characteristic polynomials $P_n(lambda)$ of the given matrices obey the recursion relation $P_n(lambda) = lambda P_n-1(lambda) + P_n-2(lambda)$; this is quite similar to the Chebyshev recursion relation $T_n(lambda) = 2 lambda T_n-1(lambda) - T_n-2(lambda)$.
$endgroup$
– Michael Seifert
Apr 24 at 13:58
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 $$lambda_k = 2icosleft(dfrackpin+1right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfracmkpin+1right).$$
answered Apr 23 at 22:52
JimmyK4542JimmyK4542
41.8k248110
$endgroup$
1
$begingroup$
There's probably a connection to be made between these matrices and the Chebyshev polynomials, but I can't quite nail it down. Note that roots of the Chebyshev polynomial $T_n(lambda)$ are given by $lambda = cos( k pi/n)$ for $k in mathcalZ$. It's also possible to show that the characteristic polynomials $P_n(lambda)$ of the given matrices obey the recursion relation $P_n(lambda) = lambda P_n-1(lambda) + P_n-2(lambda)$; this is quite similar to the Chebyshev recursion relation $T_n(lambda) = 2 lambda T_n-1(lambda) - T_n-2(lambda)$.
$endgroup$
– Michael Seifert
Apr 24 at 13:58
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 $$lambda_k = 2icosleft(dfrackpin+1right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfracmkpin+1right).$$
answered Apr 23 at 22:52
JimmyK4542JimmyK4542
41.8k248110
$endgroup$
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 $$lambda_k = 2icosleft(dfrackpin+1right),$$ for $k = 1,ldots,n$, and the corresponding eigenvectors $v_1,ldots,v_n$ have entries $$v_k[m] = i^msinleft(dfracmkpin+1right).$$
answered Apr 23 at 22:52
JimmyK4542JimmyK4542
41.8k248110
edited Apr 23 at 22:58
answered Apr 23 at 22:52
JimmyK4542JimmyK4542
41.8k248110
answered Apr 23 at 22:52
JimmyK4542JimmyK4542
41.8k248110
answered Apr 23 at 22:52
JimmyK4542JimmyK4542
41.8k248110
41.8k248110
1
$begingroup$
There's probably a connection to be made between these matrices and the Chebyshev polynomials, but I can't quite nail it down. Note that roots of the Chebyshev polynomial $T_n(lambda)$ are given by $lambda = cos( k pi/n)$ for $k in mathcalZ$. It's also possible to show that the characteristic polynomials $P_n(lambda)$ of the given matrices obey the recursion relation $P_n(lambda) = lambda P_n-1(lambda) + P_n-2(lambda)$; this is quite similar to the Chebyshev recursion relation $T_n(lambda) = 2 lambda T_n-1(lambda) - T_n-2(lambda)$.
$endgroup$
– Michael Seifert
Apr 24 at 13:58
add a comment |
1
$begingroup$
There's probably a connection to be made between these matrices and the Chebyshev polynomials, but I can't quite nail it down. Note that roots of the Chebyshev polynomial $T_n(lambda)$ are given by $lambda = cos( k pi/n)$ for $k in mathcalZ$. It's also possible to show that the characteristic polynomials $P_n(lambda)$ of the given matrices obey the recursion relation $P_n(lambda) = lambda P_n-1(lambda) + P_n-2(lambda)$; this is quite similar to the Chebyshev recursion relation $T_n(lambda) = 2 lambda T_n-1(lambda) - T_n-2(lambda)$.
$endgroup$
– Michael Seifert
Apr 24 at 13:58
1
1
$begingroup$
There's probably a connection to be made between these matrices and the Chebyshev polynomials, but I can't quite nail it down. Note that roots of the Chebyshev polynomial $T_n(lambda)$ are given by $lambda = cos( k pi/n)$ for $k in mathcalZ$. It's also possible to show that the characteristic polynomials $P_n(lambda)$ of the given matrices obey the recursion relation $P_n(lambda) = lambda P_n-1(lambda) + P_n-2(lambda)$; this is quite similar to the Chebyshev recursion relation $T_n(lambda) = 2 lambda T_n-1(lambda) - T_n-2(lambda)$.
$endgroup$
– Michael Seifert
Apr 24 at 13:58
$begingroup$
There's probably a connection to be made between these matrices and the Chebyshev polynomials, but I can't quite nail it down. Note that roots of the Chebyshev polynomial $T_n(lambda)$ are given by $lambda = cos( k pi/n)$ for $k in mathcalZ$. It's also possible to show that the characteristic polynomials $P_n(lambda)$ of the given matrices obey the recursion relation $P_n(lambda) = lambda P_n-1(lambda) + P_n-2(lambda)$; this is quite similar to the Chebyshev recursion relation $T_n(lambda) = 2 lambda T_n-1(lambda) - T_n-2(lambda)$.
$endgroup$
– Michael Seifert
Apr 24 at 13:58
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.
answered Apr 23 at 22:56
José Carlos SantosJosé Carlos Santos
178k24139253
$endgroup$
$begingroup$
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
Apr 23 at 23:06
4
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
Apr 23 at 23:08
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.
answered Apr 23 at 22:56
José Carlos SantosJosé Carlos Santos
178k24139253
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$begingroup$
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.
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– YiFan
Apr 23 at 23:06
4
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
Apr 23 at 23:08
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.
answered Apr 23 at 22:56
José Carlos SantosJosé Carlos Santos
178k24139253
$endgroup$
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 Apr 23 at 22:56
José Carlos SantosJosé Carlos Santos
178k24139253
answered Apr 23 at 22:56
José Carlos SantosJosé Carlos Santos
178k24139253
answered Apr 23 at 22:56
José Carlos SantosJosé Carlos Santos
178k24139253
answered Apr 23 at 22:56
José Carlos SantosJosé Carlos Santos
178k24139253
178k24139253
$begingroup$
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
Apr 23 at 23:06
4
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
Apr 23 at 23:08
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
Apr 23 at 23:06
4
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
Apr 23 at 23:08
$begingroup$
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
Apr 23 at 23:06
$begingroup$
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
Apr 23 at 23:06
4
4
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
Apr 23 at 23:08
$begingroup$
I would have been surprised if you had not accepted that answer, since it provides more information than mine.
$endgroup$
– José Carlos Santos
Apr 23 at 23:08
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.
answered Apr 23 at 22:57
gcousingcousin
1412
New contributor
gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
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
Apr 23 at 23:06
$begingroup$
I think, compared to the above detailed answer, the only interest of mine is to show that you could have found all these informations on your own using google, as soon as you would have observed that your matrices were skew-symmetric,
$endgroup$
– gcousin
Apr 24 at 17:30
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.
answered Apr 23 at 22:57
gcousingcousin
1412
New contributor
gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
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
Apr 23 at 23:06
$begingroup$
I think, compared to the above detailed answer, the only interest of mine is to show that you could have found all these informations on your own using google, as soon as you would have observed that your matrices were skew-symmetric,
$endgroup$
– gcousin
Apr 24 at 17:30
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.
answered Apr 23 at 22:57
gcousingcousin
1412
New contributor
gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article.
answered Apr 23 at 22:57
gcousingcousin
1412
New contributor
gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 23 at 22:57
gcousingcousin
1412
New contributor
gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 23 at 22:57
gcousingcousin
1412
answered Apr 23 at 22:57
gcousingcousin
1412
1412
New contributor
gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
gcousin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
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
Apr 23 at 23:06
$begingroup$
I think, compared to the above detailed answer, the only interest of mine is to show that you could have found all these informations on your own using google, as soon as you would have observed that your matrices were skew-symmetric,
$endgroup$
– gcousin
Apr 24 at 17:30
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
Apr 23 at 23:06
$begingroup$
I think, compared to the above detailed answer, the only interest of mine is to show that you could have found all these informations on your own using google, as soon as you would have observed that your matrices were skew-symmetric,
$endgroup$
– gcousin
Apr 24 at 17:30
$begingroup$
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
Apr 23 at 23:06
$begingroup$
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
Apr 23 at 23:06
$begingroup$
I think, compared to the above detailed answer, the only interest of mine is to show that you could have found all these informations on your own using google, as soon as you would have observed that your matrices were skew-symmetric,
$endgroup$
– gcousin
Apr 24 at 17:30
$begingroup$
I think, compared to the above detailed answer, the only interest of mine is to show that you could have found all these informations on your own using google, as soon as you would have observed that your matrices were skew-symmetric,
$endgroup$
– gcousin
Apr 24 at 17:30
add a comment |
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All eigenvalues distinct is a sufficient but not necessary condition for a matrix to be diagonalizable.
$endgroup$
– Henning Makholm
Apr 24 at 0:23
$begingroup$
@HenningMakholm that's a very good point. But before the responses to the question, it's the only method I knew (hence all my approaches were based on that).
$endgroup$
– YiFan
Apr 24 at 4:37