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Note: I posted my question on math.stackexchange but got no answer. That is why I am asking it here.

Let $A$ be a $n\times n$ square matrix such that the real part of all eigenvalues are negative. For each $i,j$, let $\exp(At)_{ij}$ be the element $(i,j)$ of the matrix. It is well known that: $$ \int_0^\infty \exp(At)_{ij}dt = -(A^{-1})_{ij}$$ Is it possible to simplify a similar expression where each element is squared: $$ \int_0^\infty (\exp(At)_{ij})^2 dt = ??$$

I am wondering if it is possible to simplify the above expression. If it helps, I can assume that $A$ is diagonalizable. Note that unless for one-dimensional matrices, $(\exp(At)_{ij})^2\ne\exp(2At)_{ij}$.

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    $\begingroup$ Note that OP wants an elementwise square, not the usual matrix-multiplication square. $\endgroup$ – Federico Poloni Nov 24 '16 at 8:19
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Inspired strongly by Anthony's answer, here is a formula that works for arbitrary $A$. Let $M$ be the $n^2 \times n^2$ square matrix given by $$M= A \otimes I_n + I_n \otimes A_n$$ i.e. in terms of indices

$$ M_{ij,kl}=A_{i,k} \delta_{j,l} + \delta_{i,k}A_{j,l}$$

Then because $A \otimes I_n$ and $I_n \otimes A$ commute, $$e^{Mt} = (e^{At} \otimes I_n) (I_n \otimes e^{At}) = e^{At} \otimes e^{At}$$ i.e. in indices $$(e^{M t})_{ij,kl} = (e^{A t})_{i,j} (e^{At})_{k,l}$$ so in particular $$(e^{At})_{i,j}^2 = (e^{Mt})_{ii,jj}$$ and $$\int_t (e^{At})_{i,j}^2 = \int_t (e^{Mt})_{ii,jj} = - (M^{-1})_{ii,jj}$$

(Here I am using commas to separate the two indices of a matrix entry)

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  • $\begingroup$ Thanks for this answer. I am not sure to understand why the two summands would commute and why $e^{Mt}_{ij,kl}$ can be expressed in terms of $e^{At}$ (with these indices, it seems false). $\endgroup$ – N. Gast Nov 24 '16 at 16:18
  • $\begingroup$ I think this is problematic because each of the summands is of rank $n$, so the sum is of rank at most $2n$. For $n>2$, $M$ will not be invertible. $\endgroup$ – Anthony Quas Nov 24 '16 at 16:39
  • $\begingroup$ @AnthonyQuas That was a stupid mistake, fixed now. $\endgroup$ – Will Sawin Nov 24 '16 at 18:09
  • $\begingroup$ @N.Gast Sorry, that was wrong, is this clearer? $\endgroup$ – Will Sawin Nov 24 '16 at 18:09
  • $\begingroup$ @WillSawin: does M^{-1} exist? $\endgroup$ – Anthony Quas Nov 24 '16 at 18:42
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You can do something. Here's a computation for diagonalizable $A$. Let $A=BDB^{-1}$ and let the elements of $D$ be $-\lambda_1,\ldots,-\lambda_d$. Then \begin{align*} \int_0^\infty (e^{At})_{ij}^2\,dt&= \int_0^\infty (Be^{Dt}B^{-1})_{ij}^2\,dt\\ &=\int_0^\infty \sum_{k,k'} B_{ik}e^{-\lambda_k t}B^{-1}_{kj} B_{ik'}e^{-\lambda_k' t}B^{-1}_{k'j}\,dt\\ &=\int_0^\infty \sum_{k,k'} B_{ik}B^{-1}_{kj} B_{ik'}B^{-1}_{k'j}e^{-(\lambda_k+\lambda_k') t} \,dt\\ &=\sum_{k,k'} \frac{1}{\lambda_k+\lambda_k'}B_{ik}B^{-1}_{kj} B_{ik'}B^{-1}_{k'j}. \end{align*}

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Given a Hurwitz matrix $\mathrm A \in \mathbb R^{n \times n}$, let

$$\Phi (t) := \exp(\mathrm A t)$$

be the state transition matrix, and let its $(i,j)$-th entry be denoted by

$$\varphi_{ij} (t) := \mathrm e_i^{\top} \Phi (t) \, \mathrm e_j$$

Hence,

$$\begin{array}{rl } \displaystyle\int_0^{\infty} \left( \varphi_{ij} (t) \right)^2 \, \mathrm d t &= \displaystyle\int_0^{\infty} \left( \mathrm e_i^{\top} \Phi (t) \, \mathrm e_j \right)^2 \, \mathrm d t\\\\ &= \displaystyle\int_0^{\infty} \mathrm e_i^{\top} \Phi (t) \, \mathrm e_j \mathrm e_j^{\top} \Phi^{\top} (t) \, \mathrm e_i \, \mathrm d t\\\\ &= \mathrm e_i^{\top} \underbrace{\left( \displaystyle\int_0^{\infty} \Phi (t) \, \mathrm e_j \mathrm e_j^{\top} \Phi^{\top} (t) \, \mathrm d t \right)}_{=: \mathrm W_c} \mathrm e_i = \mathrm e_i^{\top} \mathrm W_c \mathrm e_i\end{array}$$

where $\mathrm W_c$ is the controllability Gramian of the pair $(\mathrm A, \mathrm e_j)$ and is the solution to the following controllability Lyapunov equation

$$\boxed{\mathrm A \mathrm W_c + \mathrm W_c \mathrm A^{\top} + \mathrm e_j \mathrm e_j^{\top} = \mathrm O_n}$$

Thus, the $n$ columns of the integral of the entrywise product

$$\int_0^{\infty} \left( \Phi (t) \circ \Phi (t) \right) \mathrm d t$$

are the diagonals of $\mathrm W_c^{(1)}, \mathrm W_c^{(2)}, \dots, \mathrm W_c^{(n)}$, where $\mathrm W_c^{(1)}, \mathrm W_c^{(2)}, \dots, \mathrm W_c^{(n)}$ are the solutions to the following $n$ controllability Lyapunov equations

$$\begin{array}{cl} \mathrm A \mathrm W_c^{(1)} + \mathrm W_c^{(1)} \mathrm A^{\top} + \mathrm e_1 \mathrm e_1^{\top} &= \mathrm O_n\\ \mathrm A \mathrm W_c^{(2)} + \mathrm W_c^{(2)} \mathrm A^{\top} + \mathrm e_2 \mathrm e_2^{\top} &= \mathrm O_n\\ \vdots & \\ \mathrm A \mathrm W_c^{(n)} + \mathrm W_c^{(n)} \mathrm A^{\top} + \mathrm e_n \mathrm e_n^{\top} &= \mathrm O_n\end{array}$$

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  • $\begingroup$ In MATLAB, the Gramian can be computed using function gram. $\endgroup$ – Rodrigo de Azevedo Nov 25 '16 at 22:18

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