Given a function $f:R^2\rightarrow R^2$ such that $f(x,y)=(xy, \cos xy)$, I need to compute the Jacobian matrix Df with respect to the basis $\{(1,0), (1,1)\}$. Not confident in my answer though. Please help me verify it. If it is not correct, please give me a hint. Here is my try:
With respect to the standard basis $\{1,0), (0,1)\}$, we could derive a Jacobian matrix
$$Df=
\begin{bmatrix}
\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\
\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial x}
\end{bmatrix}
=
\begin{bmatrix}
y & x\\
-y\sin xy & -x\sin xy
\end{bmatrix}
.$$
To compute the Jacobian matrix with respect to the nonstandard basis $\{(1,0), (1,1)\}$, I multiply $Df(x,y)$ by this basis and get
$$\overline{Df}(x,y)=
\begin{bmatrix}
y & x\\
-y\sin xy & -x\sin xy
\end{bmatrix}
\times
\begin{bmatrix}
1 & 1\\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
y & y+x\\
-y\sin xy & -y\sin xy-x\sin xy
\end{bmatrix}
.$$
I did it in this way, because my textbook [Hoffman] says, "the columns of the matrix relative tot he new basis will be the derivative $Df(x,y)$ applied to the new basis in $R^2$ with this image vector expressed in the new basis in $R^2$."
Anyone disagrees or agrees with my answer?
Best Answer
You can check this by going back to the definitions and properties of linear transformations.
Suppose $A:\mathbb R^2\to \mathbb R^2$ is a linear transformation such that $A((1,0))=a(1,0)+b(0,1)$ and $A((0,1))=c(1,0)+d(0,1)$. This determines $A$ uniquely and the matrix with respect to the basis $(1,0),(0,1)$ is
\begin{bmatrix} a& c\\ b & d \end{bmatrix}
Now,
$A((1,0))=a(1,0)+b(0,1)=a(1,0)+b((1,1)-(1,0))=(a-b)(1,0)+b(1,1)$
and
$A((1,1))=A((1,0))+A((0,1))=(a+c)(1,0)+(b+d)(0,1)=(a+c)(1,0)+(b+d)((1,1)-(1,0))=((a+c)-(b+d))(1,0)+(b+d)(1,1)$
Then, the matrix with respect to the basis $(1,0), (1,1)$ is
\begin{bmatrix} a-b& (a+c)-(b+d)\\ b & b+d \end{bmatrix}