I am trying to understand the step-by-step of visually calculating linear transformations following 3blue1brown chapter 4 (see youtube 3blue1brown ch 4.

Background: 3blue1brown describes how we can visually capture one transformation by simply recording where the base vectors ($\hat{i}, \hat{j}$) land in the original coordinate system. So a rotation will produce new base vecors:

$\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} \right]$

where the first column captures the new base for $\hat{i}$ and the second column is the new base for $\hat{j}$. Following one transformation visually like this is straightforward. Conducting two transformations should produce similar results. But I am not getting consistent results.

For example, take the two transformations "first do a rotation, then do a sheer transformation"--see the link above for visuals. I interpret this as firs do the rotation and draw this new coordinate system: the new x-axis points vertically and the y-axis points horizontally. Call the new base vectors $\hat{i_r}$ and $\hat{j_r}$. Now we do a sheer transformation in this new coordinate system and record where $\hat{i_r}$ and $\hat{j_r}$ land: call these new base vectors $\hat{i_s}$ and $\hat{j_s}$. I record where $\hat{i_s}$ and $\hat{j_s}$ land in the original (non-transformed) coordinate system. This would give me a new (composition) matrix:

$\left[\begin{array}{cc} 0 & -1 \\ 1 & 1 \\ \end{array} \right]$

This is obviously wrong since the composition matrix should be:

$\left[\begin{array}{cc} 1 & -1 \\ 1 & 0 \\ \end{array} \right]$

A sheer transformation is represented by:

$\left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right]$

A rotation transformation by,

$\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} \right]$

Why does this visual-way of doing it ste-by-step not adding up? Curiously, if you do these two linear transformations backward, you do in fact get the expected results. In other words, the visual calculation of "first do a rotation, then do a sheer transformation", will produce the correct answer if we do first a sheer, then do a rotation in the sheer coordinate system, then record where the base vectors land in the original (non-transsformed) coordinate system. But this is counterintuitive of that order matters in doing linear transformations.

Your input is appriciated.

  • $\begingroup$ What is your rotation and sheer? Can't verify your calculations without knowing what transformation you wish to represent $\endgroup$ – Brevan Ellefsen Feb 24 at 20:11
  • $\begingroup$ Thanks for your input. Clarified the rotation and sheer transformation. $\endgroup$ – Adel Feb 24 at 20:51

I think we can consider this question in this way.

The matrix which does the rotation is then

$$ T = \begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix} $$

Then we do the shear transform. In the rotated system, the coordinates of two base vectors become $\hat{i}=\begin{bmatrix}1\\-1\end{bmatrix}$ and $\hat{j}=\begin{bmatrix}0\\1\end{bmatrix}$. So the transform is

$$ S = \begin{bmatrix} 1 & 0\\ -1 & 1 \end{bmatrix} $$

But we are in the transformed system. If we see this shear in the original system, it would be like

$$ T\cdot{S} = \begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0\\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1\\ 1 & 0 \end{bmatrix} $$

which is the composition of two transforms.

Hope this will help.


When we talk about a vector, say, $v = \begin{bmatrix}3\\4\end{bmatrix}$, we talk about it in the standard basis. I.e., $v = 3\hat{i} + 4\hat{j}$ or $v=\begin{bmatrix}\hat{i}&\hat{j}\end{bmatrix}\begin{bmatrix}3\\4\end{bmatrix}$.

First, we do the rotation.

enter image description here

Base vectors change from $\hat{i}$, $\hat{j}$ to $\hat{i}' = \hat{j}$, $\hat{j}' = -\hat{i}$.

$$ T = \begin{bmatrix} \hat{i}' & \hat{j}' \end{bmatrix} = \begin{bmatrix} \hat{j} & -\hat{i} \end{bmatrix} $$

Then we do the shear transform.

enter image description here

Here for base vectors, $\hat{i}'' = \hat{i}' - \hat{j}'$, $\hat{j}'' = \hat{j}'$

$$ S = \begin{bmatrix} \hat{i}'' & \hat{j}'' \end{bmatrix} = \begin{bmatrix} \hat{i}' - \hat{j'} & \hat{j}' \end{bmatrix} $$

But as you can see our base vectors are based on $\hat{i}'$ and $\hat{j}'$ instead of $\hat{i}$ and $\hat{j}$. So this only describes the shear transform in terms of the rotated system. But we know the relationship between them from the rotation transform. So we can get the transform

$$ T'= \begin{bmatrix} \hat{i} + \hat{j} & -\hat{i} \end{bmatrix} $$

which describes a rotation and a shear directly.

  • $\begingroup$ Thanks @Snjór! I guess where I am stuck is the part of visually seeing that the two base vectors in the rotated system are actually i = (1,-1) and j = (0,1)--your S matrix. How do we actually realize this geometrically? I am led to see these as i = (1,1), j = (-1,1). I am missing some obvious part... $\endgroup$ – Adel Feb 24 at 22:43
  • $\begingroup$ @Adel I updated my answer. Hope it makes some sense. $\endgroup$ – Snjór Feb 25 at 23:01

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