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I need to differentiate the function $u$ shown below with respect to a vector $\psi$: ($a, c$ and $f$ are constants)

$$u(\psi) =\left[\begin{array}{cccc} a & f & 0 & 0\\ c & a & f & 0\\ 0 & c & a & f\\ 0 & 0 & c & a \end{array}\right]\left[\begin{array}{c} \psi^{1}\\ \psi^{2}\\ \psi^{3}\\ \psi^{4} \end{array}\right]$$

I'm thinking that the answer would be: $\left[\begin{array}{cccc} a & f & 0 & 0\\ c & a & f & 0\\ 0 & c & a & f\\ 0 & 0 & c & a \end{array}\right]$ by working out the derivative of each term of the matrix with respect to its corresponding $\psi$. But I was hesitant that the function could be written as:

$u=\left[\begin{array}{c} a\psi^{1}+f\psi^{2}\\ c\psi^{1}+a\psi^{2}+f\psi^{3}\\ c\psi^{2}+a\psi^{3}+f\psi^{4}\\ c\psi^{3}+a\psi^{4} \end{array}\right]$ and hence its derivative will just be: $\left[\begin{array}{c} a\\ a\\ a\\ a \end{array}\right]$, if the rows were differentiated with respect to only the corresponding $\psi$. Please let me know which is correct. I think the first one, but need to be sure.

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2 Answers

up vote 3 down vote accepted

The answer does depend on what you mean by "derivative". Your function $u$ appears to be a nice linear function $u:\mathbb{R}^4 \rightarrow \mathbb{R}^4$. Most of the time when one says they want to take "the derivative" of a function, $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ at $\vec{x} = \vec{a} $ they mean they want the linear function $Df_{\vec{a}}: \mathbb{R}^n \rightarrow \mathbb{R}^m$ that satisfies the condition:

$$\lim_{\vec{x} \rightarrow \vec{a}} \frac{|f(\vec{x}) - f(\vec{a})-Df_{\vec{a}}(\vec{x})|}{|\vec{x} - \vec{a}|} = 0 $$

Assuming such a $Df$ exists it turns out to be given by the matrix with rows equal to the gradients of the components of $f$. So in your case, $u= (a \psi^1 + f \psi^2, c \psi^1 + a \psi^2 + f \psi^3,c \psi^2 + a \psi^3 + f \psi^4, c \psi^3 + a \psi^4) = (u_1,u_2,u_3,u_4)$ the derivative $Du$ is given by:

$$\left( \begin{array}{c} \nabla u_1\\ \nabla u_2 \\ \nabla u_3 \\ \nabla u_4 \end{array} \right) % =\left( \begin{array}{cccc} a & f & 0 & 0 \\ c & a & f & 0 \\ 0 & c & a & f \\ 0 & 0 & c & a \\ \end{array} \right) $$

Which was your first instinct! Go withyour gut!

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You have

$$u_i(\psi)=\sum_jA_{ij}\psi_j$$

and so

$$\partial_{\psi_k}u_i(\psi)=\sum_jA_{ij}\delta_{jk}=A_{ik}$$

that is your original matrix.

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