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I ran the following code:

A = matrix(QQbar, [[2,-1,0],[-1,2,1],[0,-1,2]]) D, P = A.diagonalization()

This generated a value error saying that A is not diagonalisable. To see if there are workarounds, I replaced the diagonalization method by eigenmatrix_right(), and the results were weird.

D was twice the identity matrix, and P had zero vectors in its creation. I think writing QQbar to make the thing work in algebraic numbers is not proving fruitful at all, because replacing it with QQ yields exactly the same result.

I ran the following code:

A = matrix(QQbar, [[2,-1,0],[-1,2,1],[0,-1,2]]) D, P = A.diagonalization()

This generated a value error saying that A is not diagonalisable. To see if there are workarounds, I replaced the diagonalization method by eigenmatrix_right(), and the results were weird.

D was twice the identity matrix, and P had zero vectors in its creation. I think writing QQbar to make the thing work in algebraic numbers is not proving fruitful at all, because replacing it with QQ yields exactly the same result.

I ran the following code:

A = matrix(QQbar, [[2,-1,0],[-1,2,1],[0,-1,2]]) D, P = A.diagonalization()

This generated a value error saying that A is not diagonalisable. To see if there are workarounds, I replaced the diagonalization method by eigenmatrix_right(), and the results were weird.

D was twice the identity matrix, and P had zero vectors in its creation. I think writing QQbar to make the thing work in algebraic numbers is not proving fruitful at all, because replacing it with QQ yields exactly the same result.

UPD1: I thought that the issue might lie with the implementation of QQbar, so I tried it Q[sqrt(2)] instead, which does indeed contain the eigenvalues of my matrix: 2, 2 + \sqrt(2), 2 - \sqrt(2), and the entries of the corresponding eigenvectors as well, but to no avail. I think it's the matrix construction mechanism itself which is faulty here.here. Unless addressed soon, I suppose I would remove the QQbar tag from the question then.

I ran the following code:

A = matrix(QQbar, [[2,-1,0],[-1,2,1],[0,-1,2]]) [[2,-1,0],[-1,2,1],[0,-1,2]]); D, P = A.diagonalization()

This generated a value error saying that A is not diagonalisable. To see if there are workarounds, I replaced the diagonalization method by eigenmatrix_right(), and the results were weird.

D was twice the identity matrix, and P had zero vectors in its creation. I think writing QQbar to make the thing work in algebraic numbers is not proving fruitful at all, because replacing it with QQ yields exactly the same result.

UPD1: I thought that the issue might lie with the implementation of QQbar, so I tried it Q[sqrt(2)] instead, which does indeed contain the eigenvalues of my matrix: 2, 2 + \sqrt(2), 2 - \sqrt(2), and the entries of the corresponding eigenvectors as well, but to no avail. I think it's the matrix construction mechanism itself which is faulty here. Unless addressed soon, I suppose I would remove the QQbar tag from the question then.

I ran the following code:

A = matrix(QQbar, [[2,-1,0],[-1,2,1],[0,-1,2]]); D, P = A.diagonalization()

This generated a value error saying that A is not diagonalisable. To see if there are workarounds, I replaced the diagonalization method by eigenmatrix_right(), and the results were weird.

D was twice the identity matrix, and P had zero vectors in its creation. I think writing QQbar to make the thing work in algebraic numbers is not proving fruitful at all, because replacing it with QQ yields exactly the same result.

UPD1: I thought that the issue might lie with the implementation of QQbar, so I tried it Q[sqrt(2)] instead, which does indeed contain the eigenvalues of my matrix: 2, 2 + \sqrt(2), 2 - \sqrt(2), and the entries of the corresponding eigenvectors as well, but to no avail. I think it's the matrix construction mechanism itself which is faulty here. Unless addressed soon, I suppose I would remove the QQbar tag from the question then.