Matrix differential equation for rectangular matrix

[weetabixharry]

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

$\textbf{X}^{\prime}(t) = \textbf{AX}(t)$

(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 $\textbf{X}$ is:

$\textbf{X}(t) = \textbf{E}diag\{exp\{\underline{\lambda}t\}\}$

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

However, what do I do if $\textbf{X}$ 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]

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

$\textbf{X}^{\prime}(t) = \textbf{AX}(t)$

(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 $\textbf{X}$ is:

$\textbf{X}(t) = \textbf{E}diag\{exp\{\underline{\lambda}t\}\}$

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

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

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

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

However, what do I do if $\textbf{X}$ is a "tall" rectangular matrix? (i.e. X is (MxN), where M>N)?

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 can't multiply a non-square matrix with itself, its non conformable.

 

[pasmith]

It is a problem to take the exponential of a non-square matrix.
How can you calculate A^n when you can't multiply a non-square matrix with itself, its non conformable.

A must be square; otherwise the matrix equation
$$X' = AX$$
does not make sense.

 

[blue_raver22]

I would like to provide a response to your question regarding the matrix differential equation for rectangular matrix. Firstly, I would like to clarify that the term "rectangular matrix" is not commonly used in mathematics, and it would be more accurate to refer to it as a non-square matrix.

Moving on to your question, the solution for a non-square matrix \textbf{X}(t) would be similar to the one for a square matrix, however, the matrix \textbf{E} would now be an (MxN) matrix, where each column represents an eigenvector of A. In this case, the diagonal matrix diag\{exp\{\underline{\lambda}t\}\} would be of size (NxN) since we are only considering N of the eigenvectors/values.

In order to select only N of the eigenvectors/values, we can use a technique called "truncation", where we only consider the N largest eigenvalues and their corresponding eigenvectors. This can be done by sorting the eigenvalues in descending order and selecting the first N eigenvalues and their eigenvectors.

I hope this helps in providing a solution for your problem. However, I would suggest consulting a mathematician or a numerical analyst for a more detailed and accurate explanation.