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

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

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

$AM=MJ,$ 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.

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:

$M=\left({\mathbf {z}}_{1},{\mathbf {w}}_{1},{\mathbf {x}}_{1},{\mathbf {x}}_{2},{\mathbf {x}}_{3},{\mathbf {y}}_{1},{\mathbf {y}}_{2}\right)={\begin{pmatrix}0&1&-1&0&0&-2&1\\0&3&0&0&1&0&0\\-1&1&1&1&0&2&0\\-2&0&-1&0&0&-2&0\\1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\0&0&0&-1&0&-1&0\end{pmatrix}},$

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

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

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