Reduction of matrix

Recalls

Endomorphism and matric representation

Being oneself the endomorphisms of vector space of finished size, they are current and very practical to represent them matriciellement . The matrix of such a endomorphism depends then on the base chosen to represent it. The theory of reduction of the endomorphisms in finished dimension consists in seeking a base in which this matrix is simplest possible: in the best of the cases, a diagonal matrix (of which only the diagonal elements can not be null, it acts then of a diagonalisation ), if not a higher triangular matrix (of which only the diagonal and on-diagonal elements can not be null, it acts then of trigonalisation ).

Matrices and endomorphisms

Being given a vector Space $E$ of dimension $n$, brought back to a bases $\backslash \; beta$, with all Endomorphisme of $E$, one can associate a square matrix of order $n$ representative $u$ in this base. It is called the matrix of $u$ in $\backslash \; beta$. Its vectors column are the images in $\backslash \; beta$ of each ordered vector of this same base.

Reciprocally, being given a square matrix $A$ order $n$, and a base $\backslash \; beta$ of $E$, there exists a single endomorphism whose matrix in this base is $A$.

In addition, being given a square matrix $A$ order $n$ on a body $\backslash \; mathbb\; K$, one calls endomorphism canonically associated with $A$ the endomorphism with $\backslash \; mathbb\; \{K\}\; ^n$ whose matrix in the canonical base of $\backslash \; mathbb\; \{K\}\; ^n$ is $A$. It is the linear application $\backslash \; mathbb\; \{K\}\; ^n\; \backslash \; longrightarrow\; \backslash \; mathbb\; \{K\}\; ^n\; \backslash ,\; \backslash ,\; X\; \backslash \; longmapsto\; has\; \backslash ,\; X$.

Diagonalisation

A square matrix $A$ is known as diagonalisable if and only if the endomorphism which is canonically associated for him is diagonalisable: it is then similar to a diagonal matrix, i.e. if there exists an invertible matrix $P$ such as the matrix $P^\; \{-\; 1\}\; AP$ is diagonal.

Trigonalisation

When a endomorphism is not diagonalisable, one can wonder whether there exists a base compared to which its matrix is triangular.

Definition:

A endomorphism $u$ of $E$ is known as trigonalisable if there exists a base of $E$ compared to which the matrix of $u$ is triangular higher. A square matrix $A$ is known as trigonalisable if it is similar to a higher triangular matrix, i.e. if there exists an invertible matrix $P$ such as $P^\; \{-\; 1\}\; AP$ is triangular.

Let us suppose the matrix $A$, with coefficients in a body $K$, trigonalisable and similar to the triangular matrix $T$. The eigenvalues of $T$ are the elements of its diagonal, they are thus elements of the body $K$. As $A$ and $T$ are similar, they have the same eigenvalues and of this fact the eigenvalues of $A$ belong all to the body $K$. Consequently, so that the matrix $A$ is trigonalisable, it is necessary that the roots of its characteristic polynomial are in the body $K$. This condition is always checked when $K$ is the body of the complex numbers.

It is shown that this condition is also sufficient to ensure the trigonalisability.

Proposal 3:

That is to say $u$ a endomorphism of a vectorial $K$-espace of finished size. The following conditions are equivalent:

1. $u$ is trigonalisable
2. the polynomial characteristic of $u$ is divided in $K$, i.e. he is written in the form of the product of polynomials of degree 1 with coefficients in the body $K$.

Applications of the reduction of endomorphism

• Calculation of the powers of a square matrix.

• Exponential of a matrix.

Random links:

The Calumny of Calls (Botticelli) | Villeveyrac | Pinel-Hauterive | David Loosli | Roger Aaron Brown | Une_pitié_grave

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