From S.L Linear Algebra:

We can define a rotation in terms of matrices.

Indeed, we call a linear map $L: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ a rotation if its associated matrix can be written in the form:

$$\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \, \ \cos(\theta) \end{pmatrix}$$

The geometric justification for this definition comes from Fig. 1.

enter image description here

We see that:

$$L(E^1) = (\cos \theta)E^1 + (\sin \theta)E^2$$

$$L(E^2) = (-\sin \theta)E^1 + (\cos \theta)E^2$$

Thus our definition corresponds precisely to the picture. When the matrix of the rotation is as above, we say that the rotation is by an angle $\theta$.

For example, the matrix associated with a rotation by an angle $\frac{\pi}{2}$ is:

$$R(\frac{\pi}{2})=\begin{pmatrix} 0 & -1 \\ 1 & \, \, \, 0 \end{pmatrix}$$

Linear Transformation Perspective:

I think that $L(E^1)$ and $L(E^2)$ are basis for the column space of the matrix $A$ (hence the basis for image under linear transformation $L$).

It is known, that $L=AX$ where $A$ is the matrix associated with $L$ and $X=(x_1, x_2)$ is input of $L$'s definition. Also $AX=b$ where $b$ is the element of 2-dimensional image subspace (correct?).

On the basis thereof, I think we get:

$$\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$$

where $A=\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}=\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \, \ \cos(\theta) \end{pmatrix}$

For example, $\cos(\theta)x_1 + \sin(\theta)x_2=b_1$ which seems to equivalent of $L(E^1) = (\cos \theta)E^1 + (\sin \theta)E^2$.

Geometry Perspective (problem is here):

This is where it gets confusing for me, $E_1$ and $E_2$ from the figure 1 look like unit vectors in the $x$ and $y$ direction respectively. If so, is there a proof that $||E_1||=||x_1||=1$ and that $||E_2||=||x_2||=1$, if not, what do they represent?

Furthermore, I'm aware from basic trigonometry that sine function represents a vertical leg of triangle in the unit circle, whereas cosine represents a horizontal one, does this have to do anything with the figure 1?

In short:

Is there any deeper explanation of geometric justification above? I'm unable to understand it completely.

Thank you!

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    $\begingroup$ What are $x_1$ and $x_2$? Also, the picture does not seem to imply that $E_1$ and $E_2$ are unit vectors, just that they are perpendicular vectors of the same nonzero length. $\endgroup$
    – Servaes
    Commented Jan 21, 2019 at 18:36
  • $\begingroup$ @Servaes $x_1$ and $x_2$ are elements of input vector $X$ such that $L(X)=b$ $\endgroup$
    – ShellRox
    Commented Jan 21, 2019 at 18:45
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    $\begingroup$ That raises the question of what $X$ and $b$ are. What do you mean by $||E_1||=||x_1||=1$? What part of this should there be a proof of? Also, I do not know from which book this excerpt comes. Perhaps $E_1$ and $E_2$ denote standard basis vectors in $\Bbb{R}^2$? $\endgroup$
    – Servaes
    Commented Jan 21, 2019 at 18:46
  • $\begingroup$ I apologize for misunderstanding, I thought $x_1=E^1$. $X$ defines a domain, whereas $b$ shall define the image. But in essence, $b$ is also a column space which has basis $L(E_1)$ and $L(E_2)$ (since image=column space). But "basis" for domain should be $X=(x_1, x_2)=(E^1, E^2)$, correct? $\endgroup$
    – ShellRox
    Commented Jan 21, 2019 at 18:50
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    $\begingroup$ This is not making any sense. From the picture and the text $E^1$ and $E^2$ are both vectors, so $(E^1,E^2)$ does not make sense (it is not an element of $\Bbb{R}^2$). $\endgroup$
    – Servaes
    Commented Jan 21, 2019 at 18:51

2 Answers 2


I do not know which book the excerpt is from, so I do not know what exactly is meant by $E^1$ and $E^2$; the picture only suggests that $E^1$ and $E^2$ are perpendicular vectors of the same nonzero length, but perhaps in the context of the book $E^1$ and $E^2$ are the standard basis vectors for $\Bbb{R}^2$. I'll take a guess at what the geometric idea is:

Let $L:\ \Bbb{R}^2\ \longrightarrow\ \Bbb{R}^2$ be a linear map given by a matrix $\tbinom{\hphantom{-}\cos\theta\ \sin\theta}{-\sin\theta\ \cos\theta}$. Let $e_1$ and $e_2$ be the standard basis vectors of $\Bbb{R}^2$. Then \begin{eqnarray*} L(e_1)&=&\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \, \ \cos\theta \end{pmatrix}\begin{pmatrix}1 \\0 \end{pmatrix} =\begin{pmatrix} \cos\theta \\ \sin\theta \end{pmatrix},\\ L(e_2)&=&\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \, \ \cos\theta \end{pmatrix}\begin{pmatrix}0 \\1 \end{pmatrix} =\begin{pmatrix} -\sin\theta \\ \hphantom{-}\cos\theta \end{pmatrix}, \end{eqnarray*} and the picture shows, by elementary trigonometry, that these vectors are precisely the standard basis vectors rotated over an angle $\theta$ about the origin. Because rotations are linear maps, by extension every vector $X=(x_1,x_2)\in\Bbb{R}^2$ is rotated over an angle $\theta$ about the origin, and hence we call $L$ a rotation.

  • $\begingroup$ Well that makes sense, thank you for the answer. The book is called "Linear Algebra" (written by Serge Lang). By the way, on your second standard basis equation I assume you meant $L(e_2)$ instead of $L(e_1)$? $\endgroup$
    – ShellRox
    Commented Jan 21, 2019 at 19:23
  • $\begingroup$ Indeed, edited. And thanks for providing the reference. $\endgroup$
    – Servaes
    Commented Jan 21, 2019 at 19:24

This is where it gets confusing for me, $E^1$ and $E^2$ from the figure 1 look like unit vectors in the $x$ and $y$ direction respectively.

That's just an unfortunate choice of example vectors.

In general, we can write any 2D rotation matrix $\mathbf{R}$ as $$\bbox{ \mathbf{R} = \left [ \begin{matrix} r_{11} & r_{12} \\ r_{21} & r_{22} \end{matrix} \right ] }$$ where unit vectors $$\bbox{ \hat{e}_1 = \left [ \begin{matrix} r_{11} \\ r_{21} \end{matrix} \right ] } , \quad \bbox{ \hat{e}_2 = \left [ \begin{matrix} r_{12} \\ r_{22} \end{matrix} \right ] }$$ describe the basis vectors after rotation. (The corresponding basis vector before rotation are of course $\left[ \begin{matrix} 1 \\ 0 \end{matrix} \right]$ and $\left[ \begin{matrix} 0 \\ 1 \end{matrix} \right]$.) Because pure rotation matrices are orthonormal, $\mathbf{R}^{-1} = \mathbf{R}^T$, the unit vectors $$\bbox{ \hat{\epsilon}_1 = \left [ \begin{matrix} r_{11} \\ r_{12} \end{matrix} \right ] } , \quad \bbox{ \hat{\epsilon}_2 = \left [ \begin{matrix} r_{21} \\ r_{22} \end{matrix} \right ] }$$ describe the basis vectors after the inverse rotation.

If we look at the 2D counterclockwise rotation by $\varphi$, $$\bbox{ \mathbf{R} = \left [ \begin{matrix} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \end{matrix} \right ] }$$ where $$\bbox{ \hat{e}_1 = \left[ \begin{matrix} \cos\varphi \\ \sin\varphi \end{matrix} \right] } , \quad \bbox{ \hat{e}_2 = \left[ \begin{matrix} -\sin\varphi \\ \cos\varphi \end{matrix} \right] } , \quad \bbox{ \hat{\epsilon}_1 = \left[ \begin{matrix} \cos\varphi \\ -\sin\varphi \end{matrix} \right] } , \quad \bbox{ \hat{\epsilon}_2 = \left[ \begin{matrix} \sin\varphi \\ \cos\varphi \end{matrix} \right] }$$ we notice that $\hat{\epsilon}_1$ and $\hat{\epsilon}_2$ are equivalent to $\hat{e}_1$ and $\hat{e}_2$, respectively, if we negate $\varphi$; and that $\lVert\hat{e}_1\rVert = \lVert \hat{e}_2 \rVert = \lVert \hat{\epsilon}_1 \rVert = \lVert \hat{\epsilon}_2 \rVert = 1$ and $\hat{e}_1 \cdot \hat{e}_2 = \hat{\epsilon}_1 \cdot \hat{\epsilon}_2 = 0$.

If we look at the definition of orthogonal matrices, we have $$\bbox{ \mathbf{R}^T \mathbf{R} = \mathbf{R} \mathbf{R}^T = \mathbf{I} }$$ If we were to explore these properties, we'd find that the column vectors of $\mathbf{R}$ must form an orthonormal basis, as must the row vectors of $\mathbf{R}$. $\mathbf{R}$ must also always have a determinant of $+1$ or $-1$.

The final wrinkle is that only orthogonal matrices with determinant $+1$ are pure rotation matrices. Those that have determinant $-1$ correspond to matrices with a reflection. The above rotation matrix has determinant $(\cos\varphi)^2 + (\sin\varphi)^2 = 1$. If you negate $\hat{e}_1$, the determinant becomes $-(\cos\varphi)^2 - (\sin\varphi)^2 = -1$, as one would expect, as you essentially add reflection along the first basis vector after rotation to $\mathbf{R}$.

All of the above also applies to 3D rotation matrices. (For exploration on that, pick a random unit axis vector $\hat{a}$, and a rotation around it $\varphi$. The rotation matrix that corresponds to is shown in the Rotation matrix Wikipedia article. Versors, or unit quaternions, can be easily used to represent an orientation analogously to the axis-angle formalism. Quaternion algebra makes combining rotations very easy.)

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    $\begingroup$ @ShellRox: You're welcome! I am not a mathematician, but as I do quite a bit of 2D/3D visualization (of atomic models and such), I wanted to show the geometric properties of rotation matrices that are particularly useful for programmers. Basically, if you tack on the versor/unit quaternion stuff on top, then something on how to combine rotations (say, for skeleton models and such), you quickly get a pretty comprehensive toolkit with everything with a direct geometric justification/description. I thought it might be useful; but I do agree that Servaes' answer is the more proper one. $\endgroup$ Commented Jan 22, 2019 at 15:23
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    $\begingroup$ @ShellRox: I think it is a case of using the same variable to refer to different things. I see it often in math books, because authors assume readers will infer the meaning correctly from the context :(. In the context of this question, $E^1$ and $E^2$ are just some arbitrary vectors, that are being rotated by the linear transform $L$. In this notation, $L(E^1)$ and $L(E^2)$ are just those two vectors, rotated. The basis vectors before any rotation are $[1, 0]^T$ and $[0, 1]^T$. After rotation, they are $L([1,0]^T)$ and $L([0,1]^T)$, but these we can see directly from the matrix. $\endgroup$ Commented Jan 22, 2019 at 16:09
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    $\begingroup$ The reverse (meaning if we consider our basis vectors $[1, 0]^T$ and $[0, 1]^T$ after the rotation) is $L^{-1}([1,0]^T)$ and $L^{-1}([0,1]^T)$. It is an arbitrary choice, because the inverse of any rotation is a valid rotation too. In general, $L(\vec{v}) = \mathbf{R}\vec{v}$ and $L^{-1}(\vec{v}) = \mathbf{R}^{-1} \vec{v} = \mathbf{R}^T\vec{v}$, if that matters. $\endgroup$ Commented Jan 22, 2019 at 16:12
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    $\begingroup$ And that hence general inverse linear mapping is from properties of orthogonal matrix. Also I see that every linear map associated with orthogonal matrices is isomorphism, injective map? Interesting... $\endgroup$
    – ShellRox
    Commented Jan 22, 2019 at 17:09
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    $\begingroup$ @ShellRox: I personally find it useful to remember that the geometric interpretation of $\hat{e}_1$, $\hat{e}_2$, and $\hat{\epsilon}_1$ and $\hat{\epsilon}_2$ basis vectors also apply to all nondegenerate transformation matrices. That is, that if you know how the basis vectors are transformed, you immediately have the transformation matrix, and can (easily, algebraically!) find the inverse transformation as well. For me, this has been an indispendable tool for simplifying problems like 3D trilateration. $\endgroup$ Commented Jan 22, 2019 at 18:06

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