triangularizable造句
例句與造句
- Any complex square matrix is triangularizable.
- *PM : commuting matrices are simultaneously triangularizable, id = 7294-- WP guess : commuting matrices are simultaneously triangularizable-- Status:
- *PM : commuting matrices are simultaneously triangularizable, id = 7294-- WP guess : commuting matrices are simultaneously triangularizable-- Status:
- In other words, ? is triangularizable if there exists a basis such that every element of ? has an upper-triangular matrix representation in that basis.
- *PM : proof that commuting matrices are simultaneously triangularizable, id = 7301-- WP guess : proof that commuting matrices are simultaneously triangularizable-- Status:
- It's difficult to find triangularizable in a sentence. 用triangularizable造句挺難的
- *PM : proof that commuting matrices are simultaneously triangularizable, id = 7301-- WP guess : proof that commuting matrices are simultaneously triangularizable-- Status:
- However, while a smooth manifold is not a PL manifold, it carries a canonical PL structure it is uniquely triangularizable; conversely, not every PL manifold is smoothable.
- The basic result is that ( over an algebraically closed field ), the commuting matrices A, B or more generally A _ 1, \ ldots, A _ k are simultaneously triangularizable.
- This follows by combining the theorems that, over an algebraically closed field, commuting matrices are simultaneously triangularizable and a normal matrix is diagonalizable the added result is that these can both be done simultaneously.
- This is generalized by Lie's theorem, which shows that any representation of a solvable Lie algebra is simultaneously upper triangularizable, the case of commuting matrices being the abelian Lie algebra case, abelian being a fortiori solvable.
- In fact, by Engel's theorem, any finite-dimensional nilpotent Lie algebra is conjugate to a subalgebra of the strictly upper triangular matrices, that is to say, a finite-dimensional nilpotent Lie algebra is simultaneously strictly upper triangularizable.
- This can be proven by using induction on the fact that " A " has an eigenvector, by taking the quotient space by the eigenvector and inducting to show that " A " stabilises a flag, and is thus triangularizable with respect to a basis for that flag.
- One direction is clear : if the matrices are simultaneously triangularisable, then [ A _ i, A _ j ] is " strictly " upper triangularizable ( hence nilpotent ), which is preserved by multiplication by any A _ k or combination thereof it will still have 0s on the diagonal in the triangularizing basis.
- Over an algebraically closed field such as the complex numbers, all semisimple Lie algebras are splittable ( indeed, not only does the Cartan subalgebra act by triangularizable matrices, but even stronger, it acts by diagonalizable ones ) and all splittings are conjugate; thus split Lie algebras are of most interest for non-algebraically closed fields.