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Suppose that I have a symmetric Toeplitz $n\times n$ matrix $\mathbf{A}=\left[\begin{array}{cccc}a_1&a_2&\cdots& a_n\\a_2&a_1&\cdots&a_{n-1}\\\vdots&\vdots&\ddots&\vdots\\a_n&a_{n-1}&\cdots&a_1\end{array}\right]$ such that $a_i\geq 0$, and a diagonal matrix $\mathbf{B}=\left[\begin{array}{cccc}b_1&0&\cdots& 0\\0&b_2&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&b_n\end{array}\right]$ where $b_i=c/\beta_i$ for some constant $c>0$ such that $\beta_i>0$.

Now let $\mathbf{M}=\mathbf{A}(\mathbf{A}+\mathbf{B})^{-1}\mathbf{A}$.

Can one express a partial derivative $\partial \operatorname{Tr}[\mathbf{M}]\bigg/{\partial\beta_i}$ in closed form, where $\operatorname{Tr}[\mathbf{M}]$ is the trace operator?

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1 Answer 1

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Expanding $\mathbf A(\mathbf A + \mathbf B + \mathbf E)^{-1}\mathbf A$ in $\mathbf E$ yields $\mathbf A(\mathbf A + \mathbf B)^{-1}\mathbf A-\mathbf A(\mathbf A + \mathbf B)^{-1}\mathbf E(\mathbf A + \mathbf B)^{-1}\mathbf A$ up to first order. Thus

$$ \begin{eqnarray} \frac{\partial\operatorname{Tr}[M]}{\partial\beta_i} &=& -\operatorname{Tr}\left[\mathbf A(\mathbf A + \mathbf B)^{-1}\frac{\partial\mathbf B}{\partial\beta_i}(\mathbf A + \mathbf B)^{-1}\mathbf A\right] \\ &=& -\operatorname{Tr}\left[\frac{\partial\mathbf B}{\partial\beta_i}(\mathbf A + \mathbf B)^{-1}\mathbf A\mathbf A(\mathbf A + \mathbf B)^{-1}\right] \\ &=& \frac c{\beta_i^2}\left((\mathbf A + \mathbf B)^{-1}\mathbf A\mathbf A(\mathbf A + \mathbf B)^{-1}\right)_{ii}\;. \end{eqnarray} $$

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Thank you! I really appreciate the answer, but could you please elaborate on what $\mathbf{E}$ is? Is it an arbitrary matrix? Being unfamiliar with matrix calculus, I would really like to learn how one does something like this (parallels to vector calculus would be greatly appreciated)... Again, thank you! –  M.B.M. Mar 21 '12 at 2:35
@Bullmoose: You're welcome. Perhaps I should have written $\Delta\mathbf B$ instead of $\mathbf E$. You can think of it as a Taylor expansion up to first order: expand $\mathbf B(\beta_i+\Delta\beta_i)$ as $\mathbf B(\beta_i+\Delta\beta_i)=\mathbf B(\beta_i)+\Delta\beta_i(\partial\mathbf B/\partial\beta_i)$, write $\Delta\mathbf B=\Delta\beta_i(\partial\mathbf B/\partial\beta_i)$, and expand the expression for small $\Delta\mathbf B$; then the linear term in $\Delta\mathbf B$ is the linear term $\Delta \operatorname{Tr}[M]=\Delta\beta_i(\partial\operatorname{Tr}[M]/\partial\beta_i)‌​$. –  joriki Mar 21 '12 at 2:56

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