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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

$${\displaystyle D=M^{-1}AM,}$$

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

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

has eigenvalues and corresponding eigenvectors

$${\displaystyle \lambda _{1}=-1,\quad \,\mathbf {b} _{1}=\left(-3,6,1\right),}$$
$${\displaystyle \lambda _{2}=2,\qquad \mathbf {b} _{2}=\left(0,0,1\right),}$$
$${\displaystyle \lambda _{3}=4,\qquad \mathbf {b} _{3}=\left(2,1,1\right).}$$

A diagonal matrix D, similar to A is

$${\displaystyle D={\begin{pmatrix}-1&0&0\\0&2&0\\0&0&4\end{pmatrix}}.}$$

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

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

$${\displaystyle AM=MJ,}$$ (1)

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

$${\displaystyle J=M^{-1}AM.}$$ (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

$${\displaystyle A={\begin{pmatrix}-1&0&-1&1&1&3&0\\0&1&0&0&0&0&0\\2&1&2&-1&-1&-6&0\\-2&0&-1&2&1&3&0\\0&0&0&0&1&0&0\\0&0&0&0&0&1&0\\-1&-1&0&1&2&4&1\end{pmatrix}}}$$

has a single eigenvalue $$\lambda _{1}=1$$ with algebraic multiplicity $${\displaystyle \mu _{1}=7}$$. 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 $${\displaystyle \left\{\mathbf {x} _{3},\mathbf {x} _{2},\mathbf {x} _{1}\right\}}$$, one chain of two vectors $${\displaystyle \left\{\mathbf {y} _{2},\mathbf {y} _{1}\right\}}$$, and two chains of one vector $${\displaystyle \left\{\mathbf {z} _{1}\right\}}$$, $${\displaystyle \left\{\mathbf {w} _{1}\right\}}$$.

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

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

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

where M is a generalized modal matrix for A, the columns of M are a canonical basis for A, and A M = M J {\displaystyle AM=MJ} 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

Bronson (1970, pp. 179–183)
Bronson (1970, p. 181)
Beauregard & Fraleigh (1973, pp. 271,272)
Bronson (1970, p. 181)
Bronson (1970, p. 205)
Bronson (1970, pp. 206–207)
Nering (1970, pp. 122,123)
Bronson (1970, pp. 208,209)

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

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