diagonalizable matrices造句

This also allows one to exponentiate diagonalizable matrices.
For diagonalizable matrices, an even better method is to use the eigenvalue decomposition of.
Diagonalizable matrices are dense in the set of all matrices, and this consideration proves the whole theorem.
The diagonal form for diagonalizable matrices, for instance normal matrices, is a special case of the Jordan normal form.
The following theorem characterises simultaneously diagonalisable matrices : A set of diagonalizable matrices commutes if and only if the set is simultaneously diagonalisable.
It's difficult to find diagonalizable matrices in a sentence. 用diagonalizable matrices造句挺难的

For diagonalizable matrices it's easy to demonstrate this, but I'm curious if it's true more generally.
One can also say that the diagonalizable matrices form a dense subset with respect to the Zariski topology : the complement lies inside the set where the discriminant of the characteristic polynomial vanishes, which is a hypersurface.
Diagonalizable matrices and maps are of interest because diagonal matrices are especially easy to handle : scales the space, as does a " homogeneous dilation ", but by a different factor in each direction, determined by the scale factors on each axis ( diagonal entries ).
Now if any of the expressions that the theorem equates to would not reduce to a null expression, in other words if it would be a nonzero polynomial in the coefficients of the matrix, then the set of complex matrices for which this expression happens to give would not be dense in the set of all matrices, which would contradict the fact that the theorem holds for all diagonalizable matrices.