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Let $W$ be a $n$ dimensional real vector. Let $\lambda_i$ be a list of $m$ distinct complex numbers. Let $X$ be a real $n \times n$ matrix whose spectral radius $\rho(X)$ satisfies $\rho(X) \lambda_i < 1$ for all $i$.

What conditions can I impose on $X$ and $W$ to ensure that the vectors

\begin{align*} v_i = \bigg[\sum^{\infty}_{k=0} ( X \lambda_i )^k \bigg]W \end{align*}

are linearly independent?

Allen Hart
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  • Some necessary conditions are that $X$ must be non-derogatory and $W$ must be a cyclic vector. From there, I'm not sure what conditions would need to be imposed on the $\lambda_i$ – Ben Grossmann Jul 30 '20 at 11:24
  • Also, it is worth noting that given your conditions, we cannot guarantee that the infinite sum of matrices/vectors converges. We would need a uniform bound, i.e. something let $\rho(X) \lambda_i \leq \alpha$ for some $\alpha < 1$. – Ben Grossmann Jul 30 '20 at 11:26
  • I think you're right about the condition on $\rho(X) \lambda_i$, so I've changed it in the question. – Allen Hart Jul 30 '20 at 12:21
  • Actually, I was wrong: I misunderstood what the $\lambda_i$ were. There are finitely many, so saying that $\lambda_i < 1$ for all $i$ is equivalent. – Ben Grossmann Jul 30 '20 at 12:25
  • ahh I understand, you were concerned that $\lambda_i$ could form a sequence that converges to 1? I'll return the condition to it's original form. Thanks! – Allen Hart Jul 30 '20 at 12:26
  • Right, that's the idea. Or that $\rho(X) \lambda_i$ might converge to $1$ as $i \to \infty.$ – Ben Grossmann Jul 30 '20 at 13:18

1 Answers1

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As I state in my comment on the question, we require minimally that $X$ is non-derogatory and that $W$ is an $X$-cyclic vector.

With that all stated, we note that if the sum converges, it must satisfy $$ \sum^{\infty}_{k=0} ( X \lambda_i )^k = (I - \lambda_i X)^{-1}. $$ Now, there exist coefficients $a^i_k$ such that $$ (I - \lambda_i X)^{-1} = a^i_0 I + a^i_1 X + \cdots + a^i_{n-1}X^{n-1}. $$ Let $v_i = (a_0^i,a_1^i,\dots,a_{n-1}^i) \in \Bbb C^n$. Because the powers $I,X,\dots,X^{n-1}$ are linearly independent, we find that the matrix summation $\sum^{\infty}_{k=0} ( X \lambda_i )^k$ are linearly independent if and only if the vectors $v_i$ are linearly independent.

With that said: given that the matrices $\sum^{\infty}_{k=0} ( X \lambda_i )^k$ are linearly independent, it is not clear how one should choose $W$ to ensure that the vectors $\left[\sum^{\infty}_{k=0} ( X \lambda_i )^k\right]W$ are linearly independent.

Incorrect work:

Now, suppose that $X$ has characteristic polynomial $p(x) = \sum_{k=0}^n a_k x^k$, with $a_n = 1$. We note that $$ a_0 I + a_1 \left[\frac{I - (I - \lambda_i X)}{\lambda_i}\right] + \cdots + a_{n-1}\left[\frac{I - (I - \lambda_i X)}{\lambda_i}\right]^{n-1} + \left[\frac{I - (I - \lambda_i X)}{\lambda_i}\right]^n = 0. $$ Now, let $q_i(x) = \lambda_i^n p(\frac{1 - x}{\lambda_i}) = \sum_{k=0}^n a^i_k x^k$, with $a^i_n = 1$. We have $$ a_0^i I + a_1^i (I - \lambda_i X) + \cdots + a_{n-1}^i(I - \lambda_i X)^{n-1} + (I - \lambda_i X)^n = 0 \implies\\ (I - \lambda_i X) \cdot \frac{a_1^i I + \cdots + a_{n-1}^i(I - \lambda_i X)^{n-2} + (I - \lambda_i X)^{n-1}}{-a_0^i} = I \implies\\ (I - \lambda_i X)^{-1} = \frac{a_1^i I + \cdots + a_{n-1}^i(I - \lambda_i X)^{n-2} + (I - \lambda_i X)^{n-1}}{-a_0^i}. $$ Now, let $v_i = (a_1^i,\dots,a_{n-1}^i,1) \in \Bbb C^n$. Because the powers $I,X,\dots,X^{n-1}$ are linearly independent, we find that the matrix summation $\sum^{\infty}_{k=0} ( X \lambda_i )^k$ are linearly indepednent if and only if the vectors $v_i$ are linearly independent.

Ben Grossmann
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