Main room

The main space is a term from linear algebra and a generalization of the eigenspace . Main rooms play a major role in setting up the Jordanian normal form and calculating an associated basis.

Definition of the main room

If a linear mapping from a finite-dimensional vector space in itself is an eigenvalue of and denotes the algebraic multiplicity of the eigenvalue , then one calls the kernel of the -fold execution of main space to the eigenvalue , i.e. H. ${\displaystyle F\colon V\to V}$ ${\displaystyle V}$ ${\displaystyle \lambda }$ ${\displaystyle F}$ ${\displaystyle r}$ ${\displaystyle \lambda }$ ${\displaystyle r}$ ${\displaystyle (F-\lambda \,\mathrm {id} )}$ ${\displaystyle \lambda }$

${\displaystyle \operatorname {Hau} (F,\lambda ):=\{v\in V\mid (F-\lambda \,\mathrm {id} )^{r}(v)=0{\text{ für ein }}r\in \mathbb {N} \}}$.

Here is the identity mapping on . The main space is thus spanned by precisely those vectors for which applies. In particular, the eigenspace for an eigenvalue is a subspace of the main space for this eigenvalue. ${\displaystyle \mathrm {id} }$ ${\displaystyle V}$ ${\displaystyle v}$ ${\displaystyle (F-\lambda \,\mathrm {id} )^{r}(v)=0}$

Main vector

The elements of the main space are also sometimes called main vectors . A level or a level can be assigned to these main vectors. Let be an endomorphism and an eigenvalue of the endomorphism. A vector is called the main vector of the stage if ${\displaystyle F}$ ${\displaystyle \lambda }$ ${\displaystyle v}$ ${\displaystyle p}$

${\displaystyle (F-\lambda \,\mathrm {id} )^{p}(v)=0}$

but

${\displaystyle (F-\lambda \,\mathrm {id} )^{p-1}(v)\neq 0}$

applies. All eigenvectors are thus main vectors of level 1.

Theorem about main space decomposition

Let it be an endomorphism and its characteristic polynomial ${\displaystyle F}$

${\displaystyle \chi _{F}(t)=\pm \prod _{j=1}^{k}(t-\lambda _{j})^{r_{j}}}$

decay completely into linear factors with different pairs . Then: ${\displaystyle \lambda _{1}\ldots \lambda _{k}\in K}$

1. The main room is -invariant, that is . ${\displaystyle F}$ ${\displaystyle F\left(\operatorname {Hau} (F,\lambda _{i})\right)\subset \operatorname {Hau} (F,\lambda _{i})}$
2. The dimensions of the main rooms are consistent with the multiplicities of the zeros of the characteristic polynomial match, so . ${\displaystyle \dim \left(\operatorname {Hau} (F,\lambda _{i})\right)=r_{i}}$
3. The main spaces form a direct decomposition ( inner direct sum ) of . So it applies . ${\displaystyle V}$ ${\displaystyle V=\operatorname {Hau} (F,\lambda _{1})\oplus \cdots \oplus \operatorname {Hau} (F,\lambda _{k})}$
4. The endomorphism has a decomposition . In it is diagonalizable , is nilpotent , and it is true . ${\displaystyle F}$ ${\displaystyle F=F_{D}+F_{N}}$ ${\displaystyle F_{D}}$ ${\displaystyle F_{N}}$ ${\displaystyle F_{D}\circ F_{N}=F_{N}\circ F_{D}}$

example

Given a matrix whose characteristic polynomial breaks down into linear factors: ${\displaystyle A\in \mathbb {R} ^{6\times 6}}$

${\displaystyle \det \left(A-\lambda I\right)=(\lambda -2)^{3}(\lambda -4)^{3}}$.

In addition, the following should apply:

${\displaystyle {\begin{aligned}\dim \operatorname {Ker} \left(A-2I\right)&=2\,,\quad \dim \operatorname {Ker} \left(A-2I\right)^{2}=3\,,\quad \dim \operatorname {Ker} \left(A-2I\right)^{3}=3\\\dim \operatorname {Ker} \left(A-4I\right)&=1\,,\quad \dim \operatorname {Ker} \left(A-4I\right)^{2}=2\,,\quad \dim \operatorname {Ker} \left(A-4I\right)^{3}=3\\\end{aligned}}}$

The algebraic multiplicity of the eigenvalue 2 is 3 and that of the eigenvalue 4 is 3. The eigenspaces have the dimension 2 or 1, ie smaller than the respective algebraic multiplicity, which is why the matrix cannot be diagonalized. But the Jordan normal form can be constructed ${\displaystyle J}$

${\displaystyle J={\begin{bmatrix}2&0&0&0&0&0\\0&2&1&0&0&0\\0&0&2&0&0&0\\0&0&0&4&1&0\\0&0&0&0&4&1\\0&0&0&0&0&4\end{bmatrix}}}$

via a similarity transformation with the transformation matrix ${\displaystyle P}$

${\displaystyle P^{-1}AP=J\quad \Longleftrightarrow \quad AP=PJ}$,

where the column vectors of the main vectors correspond to: ${\displaystyle P}$ ${\displaystyle p_{i}}$

${\displaystyle P={\begin{bmatrix}p_{1}&p_{2}&p_{3}&p_{4}&p_{5}&p_{6}\end{bmatrix}}}$

The transformation reads with the help of the main vectors: ${\displaystyle AP=PJ}$

${\displaystyle A{\begin{bmatrix}p_{1}&p_{2}&p_{3}&p_{4}&p_{5}&p_{6}\end{bmatrix}}={\begin{bmatrix}p_{1}&p_{2}&p_{3}&p_{4}&p_{5}&p_{6}\end{bmatrix}}J={\begin{bmatrix}2p_{1}&2p_{2}&2p_{3}+p_{2}&4p_{4}&4p_{5}+p_{4}&4p_{6}+p_{5}\end{bmatrix}}}$

So it follows:

${\displaystyle {\begin{aligned}\left(A-2I\right)p_{1}&=0\\\left(A-2I\right)p_{2}&=0\\\left(A-2I\right)p_{3}&=p_{2}\quad \Rightarrow \quad \left(A-2I\right)^{2}p_{3}=\left(A-2I\right)p_{2}=0\\\left(A-4I\right)p_{4}&=0\\\left(A-4I\right)p_{5}&=p_{4}\quad \Rightarrow \quad \left(A-4I\right)^{2}p_{5}=\left(A-4I\right)p_{4}=0\\\left(A-4I\right)p_{6}&=p_{5}\quad \Rightarrow \quad \left(A-4I\right)^{3}p_{6}=\left(A-4I\right)^{2}p_{5}=\left(A-4I\right)p_{4}=0\end{aligned}}}$

${\displaystyle p_{1}}$, and are main vectors of the first order (i.e. eigenvectors), and main vectors of the second order and is a main vector of the third order. ${\displaystyle p_{2}}$ ${\displaystyle p_{4}}$ ${\displaystyle p_{3}}$ ${\displaystyle p_{5}}$ ${\displaystyle p_{6}}$

The cores of the mappings are spanned by the main vectors as follows: ${\displaystyle A-\lambda E}$

${\displaystyle {\begin{aligned}\operatorname {Ker} \left(A-2I\right)&=\left\langle p_{1},p_{2}\right\rangle \,,\quad \operatorname {Ker} \left(A-2I\right)^{n}=\left\langle p_{1},p_{2},p_{3}\right\rangle \ {\text{mit}}\ n\geq 2\,,\\\operatorname {Ker} \left(A-4I\right)&=\left\langle p_{4}\right\rangle \,,\quad \operatorname {Ker} \left(A-4I\right)^{2}=\left\langle p_{4},p_{5}\right\rangle \,,\quad \operatorname {Ker} \left(A-4I\right)^{n}=\left\langle p_{4},p_{5},p_{6}\right\rangle \ {\text{mit}}\ n\geq 3\end{aligned}}}$

The main spaces and eigenspaces for the two eigenvalues ​​are thus, with the eigenspaces being subspaces of the respective main spaces:

${\displaystyle {\begin{aligned}\operatorname {Hau} (A,2)=\operatorname {Ker} (A-2I)^{2}=\left\langle p_{1},p_{2},p_{3}\right\rangle &\supset \operatorname {E} (A,2)=\operatorname {Ker} (A-2I)=\left\langle p_{1},p_{2}\right\rangle \\\operatorname {Hau} (A,4)=\operatorname {Ker} (A-4I)^{3}=\left\langle p_{4},p_{5},p_{6}\right\rangle &\supset \operatorname {E} (A,4)=\operatorname {Ker} (A-4I)=\left\langle p_{4}\right\rangle \end{aligned}}}$

The dimensions of the main spaces agree with the multiples of the zeros of the characteristic polynomial, i.e. and . The main rooms form a direct decomposition of , i. H. . ${\displaystyle \dim \left(\operatorname {Hau} (A,2)\right)=3}$ ${\displaystyle \dim \left(\operatorname {Hau} (A,4)\right)=3}$ ${\displaystyle V=\mathbb {R} ^{6}}$ ${\displaystyle V=\operatorname {Hau} (A,2)\oplus \operatorname {Hau} (A,4)}$

The matrix has a decomposition , whereby it is diagonalizable and nilpotent: with ${\displaystyle A}$ ${\displaystyle A=A_{D}+A_{N}}$ ${\displaystyle A_{D}}$ ${\displaystyle A_{N}}$ ${\displaystyle P^{-1}(A_{D}+A_{N})P=J_{D}+J_{N}}$

${\displaystyle J_{D}={\begin{bmatrix}2&0&0&0&0&0\\0&2&0&0&0&0\\0&0&2&0&0&0\\0&0&0&4&0&0\\0&0&0&0&4&0\\0&0&0&0&0&4\end{bmatrix}}\ ,\quad J_{N}={\begin{bmatrix}0&0&0&0&0&0\\0&0&1&0&0&0\\0&0&0&0&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\0&0&0&0&0&0\end{bmatrix}}}$

literature

• Gerd Fischer: Lineare Algebra , Vieweg-Verlag, ISBN 3-528-97217-3 .