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Let $X$ be a Hilbert space with ON-basis $\lbrace e_n : ~ n \in \mathbb{N} \rbrace$. Furthermore let $A, ~ \Gamma : X \to X$ be linear operators with $A e_n = \alpha_n e_n$ and $\Gamma e_n = \gamma_n e_n$ respectively. Let $c,\lambda \in \mathbb{C}$.

My professor stated, that

$\left[ \begin{array}{ll} 0 & \Gamma \\ \lambda-A & c \Gamma \end{array} \right]$

is similar to the operator of multiplication by the sequence of matrices

$\left[ \begin{array}{ll} 0 & \gamma_n \\ \lambda-\alpha_n & c \gamma_n \end{array} \right] \in \mathbb{C}^{2 \times 2}$

Why is that correct? And what is even ment by "the operator of multiplication by the sequence of matrices"?


If you start calculating with some $v =(v_1, v_2) \in X \times X$ you get

$\left[ \begin{array}{ll} 0 & \Gamma \\ \lambda-A & c \Gamma \end{array} \right]v = \left[ \begin{array}{l} \Gamma v_2 \\ (\lambda -A) v_1 - c \Gamma v_1 \end{array} \right]$

and if you let $v_1 = \sum_{n \in \mathbb{N}} r_n e_n$, $v_2 = \sum_{n \in \mathbb{N}} s_n e_n$ this yields

$\left[ \begin{array}{ll} 0 & \Gamma \\ \lambda-A & c \Gamma \end{array} \right]v = \left[ \begin{array}{l} \sum_{n \in \mathbb{N}} s_n \gamma_n e_n \\ \lambda \sum_{n \in \mathbb{N}} r_n e_n - \sum_{n \in \mathbb{N}} r_n \alpha_n e_n - c \sum_{n \in \mathbb{N}} s_n \gamma_n e_n \end{array} \right]$

and this yields

$\left[ \begin{array}{ll} 0 & \Gamma \\ \lambda-A & c \Gamma \end{array} \right]v = \sum_{ n \in \mathbb{N}} \left[ \begin{array}{l} s_n \gamma_n e_n \\ \lambda r_n e_n - r_n \alpha_n e_n - c s_n \gamma_n e_n \end{array} \right]$.

But this doesn't help me at all. Any hint?

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1 Answer 1

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My guess is that $A$ and $\Gamma$ share eigenvectors, $e_n$. The "big" matrix is written according to the product $X \times X$. (Bases of the two copies of $X$ need not be chosen here.) The small matrices (with $\alpha_n$ and $\gamma_n$) are only $2$-by-$2$ the way they are written, and they correspond to subspaces of $X \times X$ generated by $(e_n, 0)$ and $(0, e_n)$. They have to be extended to cover $X \times X$ by adding identity to properly act on $X \times X$. A formal definition may look like this:

Let $P_n:X\times X \to X\times X$ be defined as follows. Suppose $\left(u, v\right) = \left(\sum_{i=1}^\infty u_i e_i, \sum_{i=1}^\infty v_i e_i\right)$ and $(x, y) = \left(\sum_{i=1}^\infty x_i e_i, \sum_{i=1}^\infty y_i e_i\right) = P_n(u, v)$. Then \begin{align*} x_i & = \begin{cases} \gamma_i v_i & ; i = n\\ u_i & ; i \ne n \end{cases}\\ y_i & = \begin{cases} (\lambda - \alpha_i)u_i + c \gamma_i v_i & ; i = n \\ v_i & ; i \ne n \end{cases} \end{align*} These $P_n$ are operators defined by the small $2$-by-$2$ matrix you wrote above. The big operator should be the same as $\prod_{i=1}^\infty P_i$, which is at first glance informal but can be made formal by considering how it operates. Its action is exactly what you wrote.

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I edited the post according to what you wrote: indeed $A$ and $\Gamma$ share eigenvectors. Sorry I forgot that. I have to think a little about the rest of your post. –  mjb Jan 28 '13 at 14:01

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