2

Find the Total derivative of

i)$f(A,B)=A+B$ ,

ii)$g(A,B)=AB$

iii)$h(A,B)=A^2$

where $f,g:M(n,\mathbb{R}) \times M(n,\mathbb{R}) \to M(n,\mathbb{R})$ and $h:M(n,\mathbb{R}) \to M(n,\mathbb{R})$

Let $A=[a_{ij}]$ , $B=[b_{ij}]$, where $ 1 \le i \le n, 1\le j \le n$

If I regard the $f$ to be a function in the $2n^2$ variables, then I believe $$Df= \begin{bmatrix} 1 & 0 & 0 & \cdots &0 &1 & \cdots &0\\ \vdots & \vdots & & \vdots &\vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots &1 &0 & \cdots &1\\ \end{bmatrix}$$ where $Df$ is a $n^2 \times 2n^2$ matrix, the second $1$ appears in the $n+1$th place and so on.

Similarly $$DG= \begin{bmatrix} B^t & 0_{n\times n} & 0_{n \times n} & \cdots &0_{n \times n} &a_{11}I_{n\times n} & \cdots &a_{1n}I_{n\times n}\\ 0_{n \times n} & B^{t}_{n\times n} & 0_{n \times n} & \cdots &0_{n \times n} &a_{21}I_{n\times n} & \cdots &a_{2n}I_{n\times n}\\ \vdots & \vdots & & \vdots &\vdots & \vdots & & \vdots \\ 0_{n\times n}& 0_{n\times n} & 0_{n \times n} & \cdots &B^{t}_{n \times n} &a_{n1}I_{n\times n} & \cdots &a_{nn}I_{n\times n}\\ \end{bmatrix}$$ where $Dg$ is a $n^2 \times 2n^2$ matrix, and $$DH=\begin{bmatrix} A^t+a_{11}I_{n \times n} &{ a_{12}}I_{n\times n} & {a_{13}}I_{n \times n} & \cdots &{a_{1n}}I_{n \times n}\\ a_{21}I_{n \times n} &A^t+{ a_{22}}I_{n\times n} & {a_{23}}I_{n \times n} & \cdots &{a_{2n}}I_{n \times n} \\ \vdots & \vdots & \vdots &\cdots &\vdots\\ a_{n1}I_{n \times n} &{ a_{n2}}I_{n\times n} & {a_{n3}}I_{n \times n} & \cdots &A^t+{a_{nn}}I_{n \times n}\\ \end{bmatrix}$$.

I think these are the derivative matrices. Thanks for the help!!

tattwamasi amrutam
  • 12,802
  • 5
  • 38
  • 73

1 Answers1

1

i) Linear. Therefore, $D_xf=f$ for all $x \in M(n,\mathbb{R}) \times M(n,\mathbb{R})$

ii) $g(x+h)=g(A+H_1,B+H_2)=(A+H_1)(B+H_2)=AB +AH_2+H_1B+H_1H_2 \implies $

$$\displaystyle D_xf=x_1.\left(pr_2\left(\cdot\right)\right) + \left(pr_1\left(\cdot\right)\right).x_2$$

iii)$h(x+h)=h(A+H_1,B+H_2)=(A+H_1)(A+H_1)=A^2+AH_1+H_1A+H_1^2 \implies$

$$\displaystyle D_xf=x_1.\left(pr_1\left(\cdot\right)\right) + \left(pr_1\left(\cdot\right)\right).x_1$$

(You could also see this as the composition of the map $A \mapsto (A,A)$ and $g$, which obviously yields the same result by the chain rule).

Aloizio Macedo
  • 34,292
  • Are my answers correct? ? – tattwamasi amrutam Aug 02 '15 at 18:52
  • Evaluate the canonical basis you are considering on the linear maps I provided. If the image of the $i$-th element equal the $i$-th collumn of your matrices, then your answer is correct. I advise you to grasp the concept as I told, though. Many times trying to find the matrix first can be not only troublesome but also not helpful at all for a given problem – Aloizio Macedo Aug 02 '15 at 18:55