Let $A \in \Bbb R^{n \times n}$ be the following block matrix:

$$A:= \begin{bmatrix} a^T & \alpha\\ I_{n-1} & 0_{n-1} \end{bmatrix}$$

where $a, 0_{n-1} \in \Bbb R^{n-1}$ are vectors and $\alpha$ is a scalar. Then, is there a closed form for

$$e_1^T A^k e_1,\quad k\in\Bbb N$$

i.e., the top left element of $A^k$?

By closed form, I mean something like the combination of sums ($\sum$) or products ($\prod$) or anything else that would make solving for $k$ the equation $$e_1^T A^k e_1= 1/2$$ easier. Such a $k$ is called the half life of an AR(p) process whose coefficients are the first row of $A$.

Explanation for the tag: This problem arises from computing the impulse response function, and therefore the half life, of a general AR process.

Edit: $A$ is diagonalisable, non singular, and has spectral radius < 1 for what it's worth. So one numerical way to solve the equation $e_1^T A^k e_1= 1/2$ I can think about is to first diagonalise it as $A=QDQ^{-1}$, then we can define for any real number $k$ the matrix power $A^k$ as $$A^k=QD^kQ^{-1}$$ And once we obtain $Q$, $D$ numerically, we can expect to get a polynomial in $\lambda_{1,\cdots, n}^k$ where $\lambda$ are eigenvalues of $A$, and thus is solvable by software.

  • $\begingroup$ Is $a$ a matrix ?? $\endgroup$ – Yves Daoust Mar 30 '19 at 16:24
  • $\begingroup$ @YvesDaoust No, it's a (n-1) vector. $\endgroup$ – Vim Mar 30 '19 at 16:24
  • $\begingroup$ It's much easier to compute $A^k$ if we have $$ \pmatrix{I&0\\a^T & \alpha} $$ for what that's worth $\endgroup$ – Omnomnomnom Mar 30 '19 at 16:26
  • $\begingroup$ Yes that would be triangular. But i dont think it's possible to have such a nice form for the AR(p) structure (tried and failed). $\endgroup$ – Vim Mar 30 '19 at 16:28
  • 1
    $\begingroup$ Isn't $A$ a companion matrix? $\endgroup$ – Rodrigo de Azevedo Mar 30 '19 at 17:10

If you are looking to implement a solution in software, you can do the following. You want $e_1^TA^ke_1$, which is the first entry of $A^ke_1$. If you write $$ A=\begin{bmatrix} a_1&a_2&\cdots&a_n&\alpha \\ 1&0&0&\cdots&0\\ 0&1&0&\cdots&0\\ \vdots&\vdots&\ddots&\ddots&0\\ 0&0&\cdots&1&0 \end{bmatrix}, $$ then $$ Ax=\begin{bmatrix} \sum_{j=1}^na_jx_j+\alpha x_{n+1}\\ x_1\\ \vdots\\ x_n \end{bmatrix}. $$ So if we write $z_k$ for the first entry of $A^ke_1$ (and $z_j=0$ when $j<0$) we have the recursion $$ z_{m}=\sum_{j=1}^{n} x_{j}z_{m-j}+\alpha z_{m-n-1}. $$ This recursion is very easy to implement in software. One could also attempt to solve the recursion explicitly, by looking at the characteristic polynomial (of the recursion).

  • $\begingroup$ Thanks! Actually my main challenge isn't how to compute this value, but rather how to solve the equation $|e_1^T A^k e_1|=1/2$ for k. One nasty thing is that $k$ is continuous here ($k\in\Bbb R_+$), which is why I mentioned diagonalising $A$ first to make the potentially fractional power $A^k$ well defined. $\endgroup$ – Vim Mar 31 '19 at 6:42
  • $\begingroup$ The "real power" thing will only work naturally if the eigenvalues of $A$ are positive. Do you have any condition that guarantees that? From the couple of simple examples I tried, non-positive and non-real eigenvalues pop up easily. $\endgroup$ – Martin Argerami Mar 31 '19 at 18:49

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