Linear mapping/Trigonalizable/Characterizations/1/Fact/Proof

Proof

Form (1) to (2). Let be a basis with the property that the describing matrix of with respect to this matrix is an upper triangular form. The action of this matrix shows directly that the linear subspaces

are -invariant and that they form an invariant flag.

From (2) to (1). Let

be a -invariant flag. Due to fact, there exists a basis of such that

Since this flag is invariant, we have

Therefore, the describing matrix of with respect to this basis is upper triangular.

From (1) to (3). The characteristic polynomial of is equal to the characteristic polynomial , where denotes a describing matrix with respect to an arbitrary basis. We may assume that is an upper triangular matrix. Then, according to fact, the characteristic polynomial is the product of the linear factors arising from the diagonal entries.

From (3) to (4). Due to fact, the minimal polynomial divides the characteristic polynomial.

From (4) to (2). We prove the statement by induction over , the cases

being clear. Due to the condition and fact and fact, the mapping has an eigenvalue. Because of fact, there exists an -dimensional linear subspace

which is -invariant. Due to fact, the minimal polynomial of the restriction divides the minimal polynomial of , therefore, it also splits into linear factors. By the induction hypothesis, there exists an -invariant flag

and so this is also a -invariant flag.

The supplement follows as follows. Suppose that the trigonalizable mapping is described by the matrix with respect to the basis , and by the upper triangular matrix with respect to the basis Then, because of fact the relation holds, where is the transformation matrix of the base change.