Modal matrix

In linear algebra, the modal matrix is used in the diagonalization process involving eigenvalues and eigenvectors.^[1]

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 $M$.^[2]

Example

The matrix

$A = \begin{pmatrix} 3 & 2 & 0 \\ 2 & 0 & 0 \\ 1 & 0 & 2 \end{pmatrix}$

has eigenvalues and corresponding eigenvectors

$\lambda_1 = -1, \quad \, \bold b_1 = \left( -3, 6, 1 \right) ,$

$\lambda_2 = 2, \qquad \bold b_2 = \left( 0, 0, 1 \right) ,$

$\lambda_3 = 4, \qquad \bold b_3 = \left( 2, 1, 1 \right) .$

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

$M = \begin{pmatrix} -3 & 0 & 2 \\ 6 & 0 & 1 \\ 1 & 1 & 1 \end{pmatrix}.$^[3]

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.^[4]

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.).^[5]

One can show that

$AM = MJ,$

(1)

where $J$ is a matrix in Jordan normal form. By premultiplying by $M^{-1}$, we obtain

$J = M^{-1}AM.$

(2)

Note that when computing these matrices, equation (1) is the easiest of the two equations to verify, since it does not require inverting a matrix.^[6]

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.^[7] 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 $\left\{ \bold x_3, \bold x_2, \bold x_1 \right\}$, one chain of two vectors $\left\{ \bold y_2, \bold y_1 \right\}$, and two chains of one vector $\left\{ \bold z_1 \right\}$, $\left\{ \bold w_1 \right\}$.

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

$M = \begin{pmatrix} \bold z_1 & \bold w_1 & \bold x_1 & \bold x_2 & \bold x_3 & \bold y_1 & \bold y_2 \end{pmatrix} = \begin{pmatrix} 0 & 1 & -1 & 0 & 0 & -2 & 1 \\ 0 & 3 & 0 & 0 & 1 & 0 & 0 \\ -1 & 1 & 1 & 1 & 0 & 2 & 0 \\ -2 & 0 & -1 & 0 & 0 & -2 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & -1 & 0 \end{pmatrix},$

$J = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix},$

where $M$ is a generalized modal matrix for $A$, the columns of $M$ are a canonical basis for $A$, and $AM = MJ$.^[8] 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.^[9]

Notes

1. Bronson (1970, pp. 179-183)
2. Bronson (1970, p. 181)
3. Beauregard & Fraleigh (1973, pp. 271,272)
4. Bronson (1970, p. 181)
5. Bronson (1970, p. 205)
6. Bronson (1970, pp. 206-207)
7. Nering (1970, pp. 122,123)
8. Bronson (1970, pp. 208,209)
9. Bronson (1970, p. 206)

References

• Beauregard, Raymond A.; Fraleigh, John B. (1973), A First Course In Linear Algebra: with Optional Introduction to Groups, Rings, and Fields, Boston: Houghton Mifflin Co., ISBN 0-395-14017-X
• Bronson, Richard (1970), Matrix Methods: An Introduction, New York: Academic Press, LCCN 70097490
• Nering, Evar D. (1970), Linear Algebra and Matrix Theory (2nd ed.), New York: Wiley, LCCN 76091646