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Suppose I have a function $f(x) = A^n$ where $A$ is a square matrix, $x$ is a positive real scalar, and $n$ is a natural number.

I would like to calculate the derivative of $f$ with respect to $x$ (each entry in $A$ is a function of $x$).

Is there a simple formula for this in general or do I need to know what $n$ is and use the product rule?

I found this, but I don't understand it (in particular I don't understand what $DS(A)$ or $S(A)$ means).

edit: Each entry in $A$ is differentiable.

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  • $\begingroup$ the derivative is zero since $f(x)$ is constant. $\endgroup$ – Tsemo Aristide Oct 9 '17 at 20:51
  • $\begingroup$ Why would $f(x)$ be constant? Each entry in $A$ is a function of $x$ $\endgroup$ – HXSP1947 Oct 9 '17 at 20:53
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    $\begingroup$ The product rule applies to matrix multiplication, with the caveat that you can't reorder products. Thus $\frac d{dx}(A^n)=\sum_{k=1}^n A^{k-1}(\frac d{dx}A)A^{n-k}$. $\endgroup$ – stewbasic Oct 9 '17 at 20:56
  • $\begingroup$ Possible duplicate of Is there a general form for the derivative of a matrix to a power? $\endgroup$ – mfl Oct 8 '18 at 19:20
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There are two cases.

Case 1. $A$ is diagonalizable and you know explicitly $D$ diagonal, $P$ invertible s.t. $A^n=PD^nP^{-1}$.

Then $(A^n)'=P'D^nP^{-1}+nPD^{n-1}D'P^{-1}-PD^nP^{-1}P'P^{-1}$.

Case 2. Otherwise

$(A^n)'=A'A^{n-1}+AA'A^{n-2}+\cdots+A^{n-1}A'$ (sum of $n$ matrices) where $A'=[{a_{i,j}}']$.

There are no simplifications.

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In general, taking square matrices to integer powers is done by finding diagonal matrix $D$ and matrices $P$ and $P^{-1}$ such that $A=P^{-1}DP$.

This is because $A^n=P^{-1}D^nP$, and because $D$ is diagonal, one can simply raise each element of $D$ to the $n$ to get $D^n$. This way, we can relatively easily find a closed form for $A^n$.

If all entries of $A$ are differentiable, it should be easy to differentiate once you have this closed form of $A^n$.

This said, it may be very hard to find the eigenvalues of this function-valued matrix to find $P$ and $D$.

If $A$ be an $m\times m$ matrix: To find $P$ and $D$, first find $m$ linearly independent eigenvalues $\lambda_1, \lambda_2, \cdots, \lambda_m$ and corresponding eigenvectors $v_1, v_2, \cdots, v_m$. $P$ is the matrix $[v_1, v_2, \cdots, v_m]$ and $D=I\begin{bmatrix}\lambda_1\\\lambda_2\\\vdots\\\lambda_m\end{bmatrix}$.

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    $\begingroup$ The issue here is that even when the matrix $A$ is diagonalizable, $P$ and $D$ are not pleasant functions of the entries of $A$, even in the simplest case when $A$ has a full set of distinct eigenvalues. @stewbasic's formula will be better in practice for moderate $n$. $\endgroup$ – user7530 Oct 10 '17 at 16:53

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