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


 * $$ D = M^{-1}AM, $$

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{pmatrix}

3 & 2 & 0 \\ 2 & 0 & 0 \\ 1 & 0 & 2 \end{pmatrix}$$

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{pmatrix}

-1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix}.$$

One possible choice for an invertible matrix $$M$$ such that $$ D = M^{-1}AM, $$ is


 * $$M = \begin{pmatrix}

-3 & 0 & 2 \\ 6 & 0 & 1 \\ 1 & 1 & 1 \end{pmatrix}.$$

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 $$ M^{-1} $$, 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{pmatrix}

-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{pmatrix}$$

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 $$ \left\{ \mathbf x_3, \mathbf x_2, \mathbf x_1 \right\} $$, one chain of two vectors $$ \left\{ \mathbf y_2, \mathbf y_1 \right\} $$, and two chains of one vector $$ \left\{ \mathbf z_1 \right\} $$, $$ \left\{ \mathbf w_1 \right\} $$.

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



M = \begin{pmatrix} \mathbf z_1 & \mathbf w_1 & \mathbf x_1 & \mathbf x_2 & \mathbf x_3 & \mathbf y_1 & \mathbf y_2 \end{pmatrix} = \begin{pmatrix} 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{pmatrix},$$


 * $$J = \begin{pmatrix}

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{pmatrix}, $$

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.