Matrix differential equation for rectangular matrix

weetabixharry

Given a matrix differential equation (system of equations?) of the form:

[itex]\textbf{X}^{\prime}(t) = \textbf{AX}(t)[/itex]

(where X is a complex matrix, t is real scalar and A is always a square and normal real matrix) I am able to find (e.g. here) that a general solution for square [itex]\textbf{X}[/itex] is:

[itex]\textbf{X}(t) = \textbf{E}diag\{exp\{\underline{\lambda}t\}\}[/itex]

where [itex]\textbf{E}[/itex] is the matrix whose columns are the eigenvectors of A and [itex]\underline{\lambda}[/itex] the vector of corresponding eigenvalues. [itex]diag\{exp\{\underline{\lambda}t\}\}[/itex] is a diagonal matrix, with diagonal entries [itex]exp\{\underline{\lambda}t\}[/itex].

However, what do I do if [itex]\textbf{X}[/itex] is a "tall" rectangular matrix? (i.e. X is (MxN), where M>N)? Can I somehow select only N of the eigenvectors/values?

Thanks very much for any help!

pasmith

That cannot be correct; it does not include an arbitrary constant.

The solution of
[tex]X' = AX[/tex]
where X (and X') is MxN and A is MxM (required for the matrix multiplication to be defined) and constant is
[tex]
X(t) = \exp(At)X(0)
[/tex]
where
[tex]
\exp(A) = \sum_{n=0}^{\infty} \frac1{n!} A^n.
[/tex]

Now it is true that if [itex]A[/itex] is diagonalizable then one way to calculate [itex]\exp(At)[/itex] is to use the relation [itex]A^n = P^{-1}\Lambda^nP[/itex], where [itex]\Lambda[/itex] is diagonal, to obtain [itex]\exp(At) = P^{-1}\exp(\Lambda t)P[/itex]. It is then easily shown from the above definition that [itex]\exp(\mathrm{diag}(\lambda_1,\dots,\lambda_M)) = \mathrm{diag}(e^{\lambda_1}, \dots, e^{\lambda_M})[/itex], so that
[tex]
X(t) = P^{-1} \mathrm{diag}(e^{\lambda_1 t}, \dots, e^{\lambda_M t})PX(0)
[/tex]
where, in your notation, [itex]E = P^{-1}[/itex].

This is not a problem; the above solution works whether X is square or not.

grep6

It is a problem to take the exponential of a non-square matrix.
How can you calculate A^n when you cant multiply a non-square matrix with itself, its non conformable.

pasmith

A must be square; otherwise the matrix equation
[tex]
X' = AX
[/tex]
does not make sense.