Diagonalizable matrix

In the mathematical field of linear algebra, a diagonalizable matrix is a square matrix that is similar to a diagonal matrix . It can be transformed into a diagonal matrix by means of a base change ( i.e. conjugation with a regular matrix ). The concept can be transferred to endomorphisms .

definition

Diagonal matrix

→ Main article : Diagonal matrix

A square matrix over a body , the elements of which are all zero, is called a diagonal matrix. Often you write for it ${\displaystyle D=(d_{ij})}$ ${\displaystyle K}$ ${\displaystyle d_{ij}\in K}$ ${\displaystyle i\neq j}$

${\displaystyle D=\operatorname {diag} (d_{1},d_{2},\dotsc ,d_{n}):={\begin{pmatrix}d_{1}&0&\cdots &0\\0&d_{2}&\ddots &\vdots \\\vdots &\ddots &\ddots &0\\0&\cdots &0&d_{n}\end{pmatrix}}}$.

Diagonalizable matrix

A square -dimensional matrix is called diagonalizable or diagonal-like if there is a diagonal matrix to which it is similar , that is, there is a regular matrix such that or . ${\displaystyle n}$ ${\displaystyle A}$ ${\displaystyle D_{A}}$ ${\displaystyle S}$ ${\displaystyle D_{A}=S^{-1}AS}$ ${\displaystyle SD_{A}=AS}$

An endomorphism over a finite-dimensional vector space is called diagonalizable if there is a basis of with respect to which the mapping matrix is ​​a diagonal matrix. ${\displaystyle f\colon V\to V}$ ${\displaystyle V}$ ${\displaystyle B}$ ${\displaystyle V}$ ${\displaystyle M_{B}(f)}$

Unitary diagonalizable matrix

A matrix is unitary diagonalizable if and only if a unitary transformation matrix exists such that is a diagonal matrix, where is the matrix to be adjoint . ${\displaystyle A\in \mathbb {C} ^{n\times n}}$ ${\displaystyle S\in \mathbb {C} ^{n\times n}}$ ${\displaystyle S^{H}AS=S^{-1}AS}$ ${\displaystyle S^{H}}$ ${\displaystyle S}$

From this follows for a real-valued matrix the unitary diagonalizability, if an orthogonal transformation matrix exists such that is a diagonal matrix, where is the matrix to be transposed . ${\displaystyle A}$ ${\displaystyle S\in \mathbb {R} ^{n\times n}}$ ${\displaystyle S^{T}AS=S^{-1}AS}$ ${\displaystyle S^{T}}$ ${\displaystyle S}$

In a finite-dimensional Prehilbert space , an endomorphism can be unitarily diagonalized if and only if there is an orthonormal basis of such that the mapping matrix is ​​a diagonal matrix. ${\displaystyle V}$ ${\displaystyle f\colon V\to V}$ ${\displaystyle B}$ ${\displaystyle V}$ ${\displaystyle M_{B}(f)}$

Further characterizations of the diagonalisability

Let be a -dimensional matrix with entries from a body . Each of the following six conditions is met if and only if is diagonalizable. ${\displaystyle A}$ ${\displaystyle n}$ ${\displaystyle K}$ ${\displaystyle A}$

1. The minimal polynomial completely breaks down into different linear factors in pairs: with ${\displaystyle m_{A}(\lambda )}$ ${\displaystyle m_{A}(\lambda )=\pm (\lambda -\lambda _{1})\cdot \dots \cdot (\lambda -\lambda _{n})}$ ${\displaystyle \lambda _{i}\in K}$
2. The characteristic polynomial completely breaks down into linear factors and the geometric multiplicity corresponds to the algebraic multiplicity for each eigenvalue . ${\displaystyle \chi _{A}(\lambda )}$ ${\displaystyle \lambda _{i}\in K}$
3. There is a basis for which consists of eigenvectors for . ${\displaystyle K^{n}}$ ${\displaystyle A}$
4. The sum of the dimensions of the respective natural spaces is equal to : where the spectrum designated. ${\displaystyle n}$ ${\displaystyle \sum _{\lambda \in \sigma (A)}\dim(E_{\lambda }(A))=n}$ ${\displaystyle \sigma (A)}$
5. ${\displaystyle K^{n}}$is the direct sum of the respective natural spaces: . ${\displaystyle K^{n}=\bigoplus _{\lambda \in \sigma (A)}E_{\lambda }(A)}$
6. All Jordan blocks of Jordan normal form have dimension 1. ${\displaystyle J_{A}}$

If and with the desired properties are found, then it holds that the diagonal entries of , namely , are eigenvalues ​​of too certain unit vectors . Then is . The so are eigenvectors of , in each case to the eigenvalue . ${\displaystyle S}$ ${\displaystyle D_{A}}$ ${\displaystyle D_{A}}$ ${\displaystyle \lambda _{i}}$ ${\displaystyle D_{A}}$ ${\displaystyle e_{i}}$ ${\displaystyle ASe_{i}=SD_{A}e_{i}=S\lambda _{i}e_{i}=\lambda _{i}Se_{i}}$ ${\displaystyle Se_{i}}$ ${\displaystyle A}$ ${\displaystyle \lambda _{i}}$

Since it should be invertible, it is also linearly independent. ${\displaystyle S}$ ${\displaystyle (Se_{1},\ldots ,Se_{n})}$

In summary, this results in the necessary condition that a -dimensional diagonalizable matrix must have linearly independent eigenvectors. The space on which it operates has a basis of eigenvectors of the matrix. However, this condition is also sufficient, because from found linearly independent eigenvectors of with the associated eigenvalues ​​suitable and very direct can be constructed. ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle A}$ ${\displaystyle D_{A}}$ ${\displaystyle S}$

The problem is reduced to finding linearly independent eigenvectors of . ${\displaystyle n}$ ${\displaystyle A}$

A necessary but not sufficient condition for diagonalisability is that the characteristic polynomial completely breaks down into linear factors: So it is not diagonalisable, although . A sufficient, but not necessary, condition for diagonalisability is that it completely breaks down into pairwise different linear factors: So is diagonalisable, although . ${\displaystyle \chi _{A}}$ ${\displaystyle A={\begin{pmatrix}0&1\\0&0\end{pmatrix}}}$ ${\displaystyle \chi _{A}(X)=X^{2}}$ ${\displaystyle \chi _{A}}$ ${\displaystyle A={\begin{pmatrix}1&0\\0&1\end{pmatrix}}}$ ${\displaystyle \chi _{A}(X)=(X-1)^{2}}$

Properties of a diagonalizable matrix

If a matrix is ​​diagonalizable, the geometric multiplicity of its eigenvalues ​​is equal to the respective algebraic multiplicity . This means that the dimension of the individual eigenspaces corresponds to the algebraic multiplicity of the corresponding eigenvalues ​​in the characteristic polynomial of the matrix.

The matrix power of a diagonalizable matrix can be calculated by ${\displaystyle A}$

${\displaystyle A^{n}=S\cdot D_{A}^{n}\cdot S^{-1}}$

The power of a diagonal matrix is ​​obtained by raising the diagonal elements to the power.

Diagonalization

If a matrix can be diagonalized, there is a diagonal matrix for which the similarity condition is fulfilled: ${\displaystyle A}$ ${\displaystyle D_{A}}$

${\displaystyle D_{A}=S^{-1}AS}$

To diagonalize this matrix, one calculates the diagonal matrix and a corresponding basis from eigenvectors. This is done in three steps: ${\displaystyle D_{A}}$

1. The eigenvalues ​​of the matrix are determined. (Individual eigenvalues ​​can occur more than once.) ${\displaystyle \lambda _{i}}$ ${\displaystyle A}$
2. The eigenspaces for all eigenvalues ​​are calculated, i.e. systems of equations of the following form are solved ${\displaystyle E\left(\lambda _{i}\right)}$ ${\displaystyle \lambda _{i}}$

   ${\displaystyle (A-\lambda _{i}I)\cdot {\begin{pmatrix}e_{1}\\\vdots \\e_{n}\end{pmatrix}}=0}$.

3. Because the geometric multiplicity is equal to the algebraic multiplicity of any eigenvalue, we can find a basis of for any maximal set of corresponding eigenvalues . ${\displaystyle \lambda _{i_{1}}=\ldots =\lambda _{i_{k}}}$ ${\displaystyle \left\{b_{i_{1}},\ldots ,b_{i_{k}}\right\}}$ ${\displaystyle E\left(\lambda _{i_{1}}\right)=\ldots =E(\lambda _{i_{k}})}$
4. Now the diagonal form of the matrix with respect to the base is : ${\displaystyle D_{A}}$ ${\displaystyle A}$ ${\displaystyle B=\left\{b_{1},\ldots ,b_{n}\right\}}$

   ${\displaystyle D_{A}=\operatorname {diag} (\lambda _{1},\lambda _{2},\dots ,\lambda _{n})}$

   ${\displaystyle S=\left(b_{1},\ldots ,b_{n}\right)}$

Simultaneous diagonalization

Occasionally you want to diagonalize two matrices with the same transformation . If that succeeds, then and and there and are diagonal matrices, ${\displaystyle A,B}$ ${\displaystyle S}$ ${\displaystyle S^{-1}AS=D_{1}}$ ${\displaystyle S^{-1}BS=D_{2}}$ ${\displaystyle D_{1}}$ ${\displaystyle D_{2}}$

${\displaystyle D_{1}\cdot D_{2}=D_{2}\cdot D_{1}\Rightarrow B\cdot A=SD_{2}S^{-1}\cdot SD_{1}S^{-1}=SD_{1}D_{2}S^{-1}=A\cdot B}$.

So the endomorphisms have to commute with one another. In fact, the reverse also applies: if two diagonalizable endomorphisms commutate, they can be diagonalized simultaneously. In quantum mechanics there is a basis of common eigenstates for two such operators.

example

Let be the matrix to be diagonalized. is (unitary) diagonalizable, since is symmetrical , d. H. it applies . ${\displaystyle A={\begin{pmatrix}1&0&1\\0&2&0\\1&0&1\end{pmatrix}}\in \mathbb {R} ^{3\times 3}}$ ${\displaystyle A}$ ${\displaystyle A}$ ${\displaystyle A=A^{T}}$

The eigenvalues of can be determined by the zeros of the characteristic polynomial : ${\displaystyle \lambda _{i}}$ ${\displaystyle A}$ ${\displaystyle \chi _{A}}$

${\displaystyle \chi _{A}(\lambda )=\det(\lambda E_{3}-A)=\det {\begin{pmatrix}\lambda -1&0&-1\\0&\lambda -2&0\\-1&0&\lambda -1\end{pmatrix}}=\lambda (\lambda -2)^{2}}$

So . The eigenvalue 2 has algebraic multiplicity because it is the double zero of the characteristic polynomial. ${\displaystyle \lambda _{1}=0,\lambda _{2}=2}$ ${\displaystyle a_{\lambda _{2}}=2}$

To determine the eigenspaces, insert the eigenvalues ​​in . ${\displaystyle E(\lambda )=\mathrm {Kern} (\lambda E_{3}-A)}$

All with get, we summarize the extended coefficient matrix as a system of linear equations on with infinite solutions. ${\displaystyle v\in \mathbb {R} ^{3}}$ ${\displaystyle (\lambda E_{3}-A)v=0}$ ${\displaystyle {\begin{pmatrix}\lambda E_{3}-A\mid 0\end{pmatrix}}}$

For we get , with the Gaussian elimination method we get and thus as a solution set the eigenspace: ${\displaystyle \lambda =0}$ ${\displaystyle {\begin{pmatrix}-A\mid 0\end{pmatrix}}={\begin{pmatrix}{\begin{array}{ccc|c}-1&0&-1&0\\0&-2&0&0\\-1&0&-1&0\end{array}}\end{pmatrix}}}$ ${\displaystyle {\begin{pmatrix}{\begin{array}{ccc|c}1&0&1&0\\0&1&0&0\\0&0&0&0\end{array}}\end{pmatrix}}}$

${\displaystyle E(\lambda _{1})=\left\{{\begin{pmatrix}\alpha \\0\\-\alpha \end{pmatrix}}:\alpha \in \mathbb {R} \right\}=\operatorname {Lin} {\begin{pmatrix}1\\0\\-1\end{pmatrix}}}$,

where denotes the linear envelope . ${\displaystyle \operatorname {Lin} }$

For we get , from this and thus as a solution set, the eigenspace: ${\displaystyle \lambda =2}$ ${\displaystyle {\begin{pmatrix}2\cdot E_{3}-A\mid 0\end{pmatrix}}={\begin{pmatrix}{\begin{array}{ccc|c}1&0&-1&0\\0&0&0&0\\-1&0&1&0\end{array}}\end{pmatrix}}}$ ${\displaystyle {\begin{pmatrix}{\begin{array}{ccc|c}1&0&-1&0\\0&0&0&0\\0&0&0&0\end{array}}\end{pmatrix}}}$

${\displaystyle E(\lambda _{2})=\left\{{\begin{pmatrix}\alpha \\\beta \\\alpha \end{pmatrix}}:\alpha ,\beta \in \mathbb {R} \right\}=\operatorname {Lin} \left\{{\begin{pmatrix}1\\0\\1\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}}\right\}}$.

We get the eigenvectors from the bases of the eigenspaces, they form a basis of . ${\displaystyle v_{1}={\begin{pmatrix}1\\0\\-1\end{pmatrix}},v_{2}={\begin{pmatrix}1\\0\\1\end{pmatrix}},v_{3}={\begin{pmatrix}0\\1\\0\end{pmatrix}}}$ ${\displaystyle \mathbb {R} ^{3}}$

If we normalize with and we get an orthonormal basis , since symmetric and the eigenvectors of the semi-simple eigenvalues ​​are orthogonal to each other (in this case ). ${\displaystyle v_{1},v_{2},v_{3}}$ ${\displaystyle b_{1}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1\\0\\-1\end{pmatrix}},b_{2}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1\\0\\1\end{pmatrix}},b_{3}={\begin{pmatrix}0\\1\\0\end{pmatrix}}}$ ${\displaystyle {\mathcal {B}}=\left\{b_{1},b_{2},b_{3}\right\}}$ ${\displaystyle A}$ ${\displaystyle v_{2}\perp v_{3}}$

So it applies . From this we get the inverse using the properties of orthonormal bases . ${\displaystyle S=\left(b_{1},b_{2},b_{3}\right)={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1&1&0\\0&0&{\sqrt {2}}\\-1&1&0\end{pmatrix}}}$ ${\displaystyle S^{-1}=S^{T}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1&0&-1\\1&0&1\\0&{\sqrt {2}}&0\end{pmatrix}}}$

${\displaystyle D_{A}}$determined by . ${\displaystyle D_{A}=\operatorname {diag} (\lambda _{1},\lambda _{2},\lambda _{3})=\operatorname {diag} (0,2,2)={\begin{pmatrix}0&0&0\\0&2&0\\0&0&2\end{pmatrix}}}$

So we get for ${\displaystyle D_{A}=S^{-1}AS}$

${\displaystyle {\begin{pmatrix}0&0&0\\0&2&0\\0&0&2\end{pmatrix}}={\frac {1}{2}}{\begin{pmatrix}1&0&-1\\1&0&1\\0&{\sqrt {2}}&0\end{pmatrix}}{\begin{pmatrix}1&0&1\\0&2&0\\1&0&1\end{pmatrix}}{\begin{pmatrix}1&1&0\\0&0&{\sqrt {2}}\\-1&1&0\end{pmatrix}}}$

and thus the diagonalization

${\displaystyle A=SD_{A}S^{-1}={\frac {1}{2}}{\begin{pmatrix}1&1&0\\0&0&{\sqrt {2}}\\-1&1&0\end{pmatrix}}{\begin{pmatrix}0&0&0\\0&2&0\\0&0&2\end{pmatrix}}{\begin{pmatrix}1&0&-1\\1&0&1\\0&{\sqrt {2}}&0\end{pmatrix}}}$.

See also

• Jordan normal form
• Frobenius normal form

Individual evidence

1. ↑ Uwe Storch , Hartmut Wiebe: Textbook of Mathematics, Volume 2: Lineare Algebra. BI-Wissenschafts-Verlag, Mannheim et al. 1990, ISBN 3-411-14101-8 .