Generalized Modal Matrix

In linear algebra, the modal matrix is used in the diagonalization process involving eigenvalues and eigenvectors.

Specifically the modal matrix $M$ for the matrix $A$ is the ''n'' × ''n'' matrix formed with the eigenvectors of $A$ as columns in $M$. It is utilized in the similarity transformation

: $D = M^AM,$

where $D$ is an ''n'' × ''n'' diagonal matrix with the eigenvalues of $A$ on the main diagonal of $D$ and zeros elsewhere. The matrix $D$ is called the spectral matrix for $A$. The eigenvalues must appear left to right, top to bottom in the same order as their corresponding eigenvectors are arranged left to right in $M$.

Example

The matrix

:$A = \begin 3 & 2 & 0 \\ 2 & 0 & 0 \\ 1 & 0 & 2 \end$

has eigenvalues and corresponding eigenvectors

:$\lambda_1 = -1, \quad \, \mathbf b_1 = \left( -3, 6, 1 \right) ,$
:$\lambda_2 = 2, \qquad \mathbf b_2 = \left( 0, 0, 1 \right) ,$
:$\lambda_3 = 4, \qquad \mathbf b_3 = \left( 2, 1, 1 \right) .$

A diagonal matrix $D$, similar to $A$ is

:$D = \begin -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end.$

One possible choice for an invertible matrix $M$ such that $D = M^AM,$ is

:$M = \begin -3 & 0 & 2 \\ 6 & 0 & 1 \\ 1 & 1 & 1 \end.$

Note that since eigenvectors themselves are not unique, and since the columns of both $M$ and $D$ may be interchanged, it follows that both $M$ and $D$ are not unique.

Generalized modal matrix

Let $A$ be an ''n'' × ''n'' matrix. A generalized modal matrix $M$ for $A$ is an ''n'' × ''n'' matrix whose columns, considered as vectors, form a canonical basis for $A$ and appear in $M$ according to the following rules:

* All Jordan chains consisting of one vector (that is, one vector in length) appear in the first columns of $M$.
* All vectors of one chain appear together in adjacent columns of $M$.
* Each chain appears in $M$ in order of increasing rank (that is, the generalized eigenvector of rank 1 appears before the generalized eigenvector of rank 2 of the same chain, which appears before the generalized eigenvector of rank 3 of the same chain, etc.).

One can show that

where $J$ is a matrix in Jordan normal form. By premultiplying by $M^$, we obtain

Note that when computing these matrices, equation () is the easiest of the two equations to verify, since it does not require inverting a matrix.

Example

This example illustrates a generalized modal matrix with four Jordan chains. Unfortunately, it is a little difficult to construct an interesting example of low order.
The matrix

:$A = \begin -1 & 0 & -1 & 1 & 1 & 3 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 2 & -1 & -1 & -6 & 0 \\ -2 & 0 & -1 & 2 & 1 & 3 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ -1 & -1 & 0 & 1 & 2 & 4 & 1 \end$

has a single eigenvalue $\lambda_1 = 1$ with algebraic multiplicity $\mu_1 = 7$. A canonical basis for $A$ will consist of one linearly independent generalized eigenvector of rank 3 (generalized eigenvector rank; see generalized eigenvector), two of rank 2 and four of rank 1; or equivalently, one chain of three vectors $\left\$, one chain of two vectors $\left\$, and two chains of one vector $\left\$, $\left\$.

An "almost diagonal" matrix $J$ in ''Jordan normal form'', similar to $A$ is obtained as follows:

:$M = \begin \mathbf z_1 & \mathbf w_1 & \mathbf x_1 & \mathbf x_2 & \mathbf x_3 & \mathbf y_1 & \mathbf y_2 \end = \begin 0 & 1 & -1 & 0 & 0 & -2 & 1 \\ 0 & 3 & 0 & 0 & 1 & 0 & 0 \\ -1 & 1 & 1 & 1 & 0 & 2 & 0 \\ -2 & 0 & -1 & 0 & 0 & -2 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & -1 & 0 \end,$

:$J = \begin 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \end,$

where $M$ is a generalized modal matrix for $A$, the columns of $M$ are a canonical basis for $A$, and $AM = MJ$. Note that since generalized eigenvectors themselves are not unique, and since some of the columns of both $M$ and $J$ may be interchanged, it follows that both $M$ and $J$ are not unique.

Notes

References

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* {{ citation , first1 = Evar D. , last1 = Nering , year = 1970 , title = Linear Algebra and Matrix Theory , edition = 2nd , publisher = Wiley , location = New York , lccn = 76091646

Matrices