Exercise 7.3.3

• 1.

  Take derivative on both sides of $A_{m\times n}{x}_{n\times 1}=b_{m\times 1}$ w.r.t. $b_j$, we have: $A\frac{\mathit{\partial x}}{\partial b_j}=\frac{\mathit{\partial b}}{\partial b_j}$ So we have $A\left[\begin{array}{c}\hfill \frac{\partial x_1}{\partial b_j}\hfill \\ \hfill \dots \hfill \\ \hfill \frac{\partial x_j}{\partial b_j}\hfill \\ \hfill \dots \hfill \\ \hfill \frac{\partial x_n}{\partial b_j}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill \dots \hfill \\ \hfill 1\hfill \\ \hfill \dots \hfill \\ \hfill 0\hfill \end{array}\right]$

Suppose the $\frac{\mathit{\partial x}}{\mathit{\partial b}}$ is the Jacobian matrix of $x$ w.r.t. $b$. So we have $A_{m\times n}{\frac{\mathit{\partial x}}{\mathit{\partial b}}}_{n\times m}=I_{m\times m}$. The entry $\frac{\partial x_i}{\partial b_j}$ is thus solution of the equation.

• 1.

  Let’s change the matrix $A_{m\times n}$ to a 1 by $\mathit{mn}$ vector, with $A_{11},\dots ,A_{1n},A_{21},\dots ,A_{2n},\dots ,A_{m1},\dots ,A_{\mathit{mn}}$ in the entries. So the Jacobian of $x$ w.r.t. $A$ has $n\times \mathit{mn}$ elements.

Take derivative of $x$ w.r.t. $A_{\mathit{jk}}$ on both sides of the equation $\mathit{Ax}=b$, we have

Write in matrix format we have

$A\left[\begin{array}{c}\hfill \frac{\partial x_1}{\partial A_{\mathit{jk}}}\hfill \\ \hfill \dots \hfill \\ \hfill \frac{\partial x_k}{\partial A_{\mathit{jk}}}\hfill \\ \hfill \dots \hfill \\ \hfill \frac{\partial x_n}{\partial A_{\mathit{jk}}}\hfill \end{array}\right]+\left[\begin{array}{c}\hfill 0\hfill \\ \hfill \dots \hfill \\ \hfill x_k\hfill \\ \hfill \dots \hfill \\ \hfill 0\hfill \end{array}\right]=0$

Collect all terms, we have

$A_{\mathit{mn}}{J}_{n(\mathit{mn})}+X_{m(\mathit{mn})}=0$ where we have $X_{j,(k-1)n+1:\mathit{kn}}=x^T$ and zeros on all other columns on the $j$th row.

For example, when $m=2,n=3$, we have $X=\left[\begin{array}{cccccc}\hfill x_1\hfill & \hfill x_2\hfill & \hfill x_3\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill x_1\hfill & \hfill x_2\hfill & \hfill x_3\hfill \end{array}\right]$

So the derivatives in $J$ are the solution of this equation.