for a linear transformation :

$$x'=ax+by$$ $$y'=cx+dy$$

is this the direct transformation matrix or the inverse transformation matrix?

$$T = \begin{bmatrix} a&b\\ c&d\\ \end{bmatrix} \tag{1}\label{1}$$

if its direct, I'm having confusion with contravariant components of a vector... a curvilinear transformation can be considered linear in the vicinity of a point and hence (if (1) is true) the direct transformation matrix would be :

$$T = \begin{bmatrix} \frac{\partial {x_1}'}{\partial x_1} &\frac{\partial {x_1}'}{\partial x_2} \\ \frac{\partial {x_2}'}{\partial x_1}&\frac{\partial {x_2}'}{\partial x_2}\\ \end{bmatrix} \tag{2}$$

3Blue1Brown series on linear algebra taught me that $\begin{bmatrix}a \\ c \\ \end{bmatrix}$ in (1) and $\begin{bmatrix}\frac{\partial {x_1}'}{\partial x_1} \\ \frac{\partial {x_2}'}{\partial x_1} \\ \end{bmatrix}$ in (2) represent the covariant basis vectors of new (primed) coordinate system.

So the covariant basis vectors are transforming following the direct matrix and that means the contravariant components of a vector would transform following the inverse transformation matrix:

$$g = \begin{bmatrix} \frac{\partial {x_1}}{\partial {x_1}'} &\frac{\partial {x_1}}{\partial {x_2}'} \\ \frac{\partial {x_2}}{\partial {x_1}'}&\frac{\partial {x_2}}{\partial {x_2}'}\\ \end{bmatrix} \tag{3}$$

Thus, (from (3)), in Index notation, contravariant components of a vector transforms like :

$$A'^{i}= \frac{\partial {x_j}}{\partial x_i}A^{j} $$

this is contrary to the definition of contravariance : $$A'^{i}= \frac{\partial {x_i}'}{\partial x_j}A^{j} $$

please clarify where I'm wrong

  • $\begingroup$ Is there any difficulty in understanding the question? $\endgroup$ Jul 14, 2021 at 7:22
  • $\begingroup$ man, I am thankful and ready to wide any other idea what is needed $\endgroup$
    – janmarqz
    Jul 21, 2021 at 3:23

2 Answers 2


Direct transformation: $$ dx_1' = \frac{\partial x_1'}{\partial x_1}dx_1 + \frac{\partial x_1'}{\partial x_2}dx_2 \\ dx_2' = \frac{\partial x_2'}{\partial x_1}dx_1 + \frac{\partial x_2'}{\partial x_2}dx_2 $$ $$ \begin{bmatrix} dx_1' \\ dx_2' \end{bmatrix} = \Large \begin{bmatrix} \frac{\partial x_1'}{\partial x_1} & \frac{\partial x_1'}{\partial x_2} \\ \frac{\partial x_2'}{\partial x_1} & \frac{\partial x_2'}{\partial x_2} \end{bmatrix} \normalsize \begin{bmatrix} dx_1 \\ dx_2 \end{bmatrix} $$ Inverse transformation: $$ dx_1 = \frac{\partial x_1}{\partial x_1'}dx_1' + \frac{\partial x_1}{\partial x_2'}dx_2' \\ dx_2 = \frac{\partial x_2}{\partial x_1'}dx_1' + \frac{\partial x_2}{\partial x_2'}dx_2' $$ $$ \begin{bmatrix} dx_1 \\ dx_2 \end{bmatrix} = \Large \begin{bmatrix} \frac{\partial x_1}{\partial x_1'} & \frac{\partial x_1}{\partial x_2'} \\ \frac{\partial x_2}{\partial x_1'} & \frac{\partial x_2}{\partial x_2'} \end{bmatrix} \normalsize \begin{bmatrix} dx_1' \\ dx_2' \end{bmatrix} $$ With the direct transformation, calculating the inverse results in: $$ \begin{bmatrix} dx_1 \\ dx_2 \end{bmatrix} = \Large \begin{bmatrix} \frac{\partial x_2'}{\partial x_2} & -\frac{\partial x_1'}{\partial x_2} \\ -\frac{\partial x_2'}{\partial x_1} & \frac{\partial x_1'}{\partial x_1} \end{bmatrix} / \normalsize \det \begin{bmatrix} dx_1' \\ dx_2' \end{bmatrix} \\ \mbox{with} \qquad \det = \frac{\partial x_1'}{\partial x_1}\frac{\partial x_2'}{\partial x_2} - \frac{\partial x_2'}{\partial x_1}\frac{\partial x_1'}{\partial x_2} $$ With the inverse transformation, calculating the inverse results in: $$ \begin{bmatrix} dx_1' \\ dx_2' \end{bmatrix} = \Large \begin{bmatrix} \frac{\partial x_2}{\partial x_2'} & -\frac{\partial x_1}{\partial x_2'} \\ -\frac{\partial x_2}{\partial x_1'} & \frac{\partial x_1}{\partial x_1'} \end{bmatrix} / \normalsize \det' \begin{bmatrix} dx_1 \\ dx_2 \end{bmatrix} \\ \mbox{with} \qquad \det' = \frac{\partial x_1}{\partial x_1'}\frac{\partial x_2}{\partial x_2'} - \frac{\partial x_2}{\partial x_1'}\frac{\partial x_1}{\partial x_2'} $$ Hope you can take it from here.

  • $\begingroup$ You have clarified which is direct and inverse but not the rest of my question. Why do I end up with wrong equation for how the components of a vector transform? Are you claiming that the last two equations in my question are equivalent? $\endgroup$ Jul 15, 2021 at 5:52

Your coordinated transformation can be written as $$\left( \begin{array}{c} x'\\y' \end{array} \right) = \left( \begin{array}{cc} a&b\\c&d \end{array} \right) \left( \begin{array}{c} x\\y \end{array} \right)$$ which has derivatives $$\frac{\partial x'}{\partial x}=a\quad ,\quad \frac{\partial x'}{\partial y}=b, $$ $$\frac{\partial y'}{\partial x}=c\quad ,\quad \frac{\partial y'}{\partial y}=d. $$

If you use the canonical base $e_1=\left(\begin{array}{c}1\\0\end{array}\right)$ and $e_2=\left(\begin{array}{c}0\\1\end{array}\right)$ you will get the components of new base by transforming the old via $$\left( \begin{array}{cc} a&b\\c&d \end{array} \right) \left( \begin{array}{c} 1\\0 \end{array} \right)=\left(\begin{array}{c}a\\c\end{array}\right) \quad ,\quad \left( \begin{array}{cc} a&b\\c&d \end{array} \right) \left( \begin{array}{c} 0\\1 \end{array} \right)=\left(\begin{array}{c}b\\d\end{array}\right).$$ You could write the same as $$e'_1=ae_1+ce_2,$$ $$e'_2=be_1+de_2.$$

Now, solving for $e_1, e_2$ will give $$e_1=\frac{d}{ad-bc}e'_1+\frac{-c}{ad-bc}e'_2,$$ $$e_2=\frac{-d}{ad-bc}e'_1+\frac{a}{ad-bc}e'_2.$$

Observe that this correspond to the inverse coordinated transformation $$\left( \begin{array}{c} x\\y \end{array} \right) = \left( \begin{array}{cc} \frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc} \end{array} \right) \left( \begin{array}{c} x'\\y' \end{array} \right),$$ or in other words $$x=\frac{d}{ad-bc}x'+\frac{-b}{ad-bc}y',$$ $$y=\frac{-c}{ad-bc}x'+\frac{a}{ad-bc}y'.$$ So $$ \frac{\partial x}{\partial x'}=\frac{d}{ad-bc}\quad ,\quad \frac{\partial x}{\partial y'}=\frac{-b}{ad-bc} $$ $$ \frac{\partial y}{\partial x'}=\frac{-c}{ad-bc}\quad ,\quad \frac{\partial y}{\partial y'}=\frac{a}{ad-bc} $$

Finally, if you has a vector $A^1e_1+A^2e_2$ written in old coordinates then you will get its new components by subbing the solve above to get \begin{eqnarray*} A^1e_1+A^2e_2 &=&A^1\left(\frac{d}{ad-bc}e'_1+\frac{-c}{ad-bc}e'_2\right)+ A^2\left(\frac{-b}{ad-bc}e'_1+\frac{a}{ad-bc}e'_2\right),\\ &=&\left(A^1\frac{d}{ad-bc}+A^2\frac{-b}{ad-bc}\right)e'_1+ \left(A^1\frac{-c}{ad-bc}+A^2\frac{a}{ad-bc}\right)e'_2. \end{eqnarray*}

These are $$A'^1=A^1\frac{d}{ad-bc}+A^2\frac{-b}{ad-bc}\quad {\rm and}\quad A'^2=A^1\frac{-c}{ad-bc}+A^2\frac{a}{ad-bc}.$$

That in terms of derivatives are $$A'^1=A^1\frac{\partial x}{\partial x'}+A^2\frac{\partial x}{\partial y'}\quad {\rm and}\quad A'^2=A^1\frac{\partial y}{\partial x'}+A^2\frac{\partial y}{\partial y'},$$ which coincide with the law you seek to grasp.

  • $\begingroup$ In the end, we get $${A'}^i=A^1\frac{\partial x^1}{\partial {x'}^i} + A^2 \frac{\partial {x'}^2}{\partial {x'}^i}$$. But the definition of contravariant components say : $${A'}^i = A^1\frac{\partial {x'}^i}{\partial {x}^1} + A^2 \frac{\partial {x'}^i}{\partial {x'}^i} $$ $\endgroup$ Jul 15, 2021 at 5:08
  • $\begingroup$ sorry...typo in definition : $${A'}^i = A^1\frac{\partial {x'}^i}{\partial {x}^1} + A^2 \frac{\partial {x'}^i}{\partial {x}^2}$$ $\endgroup$ Jul 15, 2021 at 5:58
  • $\begingroup$ @RishabNavaneet, I had corrected, I had a distraction but now all is on track $\endgroup$
    – janmarqz
    Jul 19, 2021 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.