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In the expression
$$\begin{bmatrix}A & C \\ 0 & B\end{bmatrix}^n = \begin{bmatrix}A^n & * \\ 0 & B^n\end{bmatrix},$$ I wonder whether the term denoted by * can be expressed in a simple form when we assume the following: (1) $A$ has its eigenvalues on or inside the unit circle. Those on the unit circle are simple; (2) $B$ has its eigenvalues strictly inside the unit circle; (3) $A$ and $B$ may have different dimensions.

In fact, I am interested for the value of * as $n \rightarrow \infty$. It would be $C(I-B)^{-1}$ when $A=I$ but in a general case, $A^n$, though bounded, may not converge as $n \rightarrow \infty$. So, I wonder whether * can have a simple expression.

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1 Answer 1

Let

$$M (n) := \left[\begin{array}{cc} A & C\\ 0 & B\end{array}\right]^n = \left[\begin{array}{cc} A^n & U (n)\\ 0 & B^n\end{array}\right]$$

where I call $U (n)$ what you call $*$. Thus, we have that $M (n+1)$ is given by

$$M (n+1) = \left[\begin{array}{cc} A^{n+1} & U (n+1)\\ 0 & B^{n+1}\end{array}\right] = \left[\begin{array}{cc} A & C\\ 0 & B\end{array}\right] \left[\begin{array}{cc} A^n & U (n)\\ 0 & B^n\end{array}\right]$$

and, therefore, we obtain the following matrix difference equation

$$U (n+1) = A \, U (n) + C \, B^n$$

Note that $U (0) = 0$, $U (1) = C$, and $U (2) = A \, C + C \, B$. The value of $U (n)$ as $n$ goes to infinity, which we denote by $\bar{U}$, is given by $\bar{U} = A \, \bar{U} + C B^{\infty}$, which yields $\bar{U} = (I - A)^{-1} C B^{\infty}$. Since you say that $B$ "has its eigenvalues strictly inside the unit circle", we conclude that $B^{\infty} = 0$ and, hence, $\bar{U} = 0$. One can also easily obtain the general solution, which is

$$U (n) = A^n \, U (0) + \displaystyle \sum_{i=0}^{n-1} A^i \, C \, B^{n-1-i}$$

and since $U (0) = 0$, the "natural response" is zero and we are left with the "forced response"

$$U (n) = \displaystyle \sum_{i=0}^{n-1} A^i \, C \, B^{n-1-i}$$

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Rod, thanks for the insight, though I think finding $\bar{U}$ here solves no purpose since (I-A) may not be invertible (and I am having to deal with the case where all eigenvalues of A are on the unit circle). I actually wonder whether $U(n)$, as $n$ tends to infinity, could be written as a closed-form expression. I can see now that such an expression will probably not exist if it does not converge as $n\rightarrow \infty$. –  Gemmi Sep 9 '12 at 8:56
    
@ Gemmi: Even if the eigenvalues of $A$ are on the unit circle, $(I-A)^{-1}$ may exist. The Jordan normal form of $A$ is $A = P \, J \, P^{-1}$ and thus $I - A = P \, (I - J)\, P^{-1}$. You're in trouble only if $J$ has ones on the diagonal. In any case, since $B^{\infty} = 0$, we have $(I - A) \, \bar{U} = 0$, where the right-hand side is the zero matrix. This leads to several (as many as the number of columns of $A$) linear systems of equations of the form $(I - A) \, \bar{u}_k = 0$, where $\bar{u}_k$ is the $k$-th column of $\bar{U}$ and the right-hand side is the zero vector. –  Rod Carvalho Sep 9 '12 at 14:22
    
@ Gemmi: Note also that if $(I - A)$ is not invertible, then there will be infinitely many matrices $\bar{U}$ that satisfy the matrix equation $(I - A) \, \bar{U} = 0$. –  Rod Carvalho Sep 9 '12 at 14:30

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