I'm new to Linear Algebra and I'm having a hard time wrapping my head around linear transformations, specifically a rotation.

From Anton's book (Elementary Linear Algebra, 11th Edition) he states:

$T(e_1) = T(1,0) = (\cos\theta, \sin\theta)$


$T(e_2) = T(0,1) = (-\sin\theta, \cos\theta)$


$A = [T(e_1) | T(e_2)] = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$

When I rotate a vector $\begin{bmatrix} x\\y \end{bmatrix}$ I get

$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} x \cdot\cos\theta \, - \, y \cdot\sin\theta \\ x \cdot\sin\theta \, + \, y \cdot\cos\theta \end{bmatrix}$

Correct me if I'm wrong, but I thought that column 1 of $A$ $\begin{bmatrix} \cos\theta \\ \sin\theta \end{bmatrix}$, holds the 'x' values and column 2 holds the 'y' values. What I'm confused about is why does $x'$ contain both an $x$ component and a $y$ component?

When I transform $\begin{bmatrix}1 \\ 0 \end{bmatrix}$ by a rotation $\theta$, the new $x$ value is just $\cos\theta$ while the new $y$ value is just $\sin\theta$. I don't understand why with the matrix transformation, $x'$ and $y'$ get both $x$ and $y$ components summed together.

I hope I'm making sense.

  • $\begingroup$ Look up MITs Gilbert Strang. ocw.mit.edu/courses/mathematics/… He hosts a Linear Algebra class at MIT, the videos of which are available online. You will find them very methodical and useful $\endgroup$
    – DWD
    May 3, 2017 at 19:08
  • $\begingroup$ You transformed $[1,0]^T$ correctly: note that even though the unrotated point has a y coordinate of 0, the rotated one has both coordinates non-zero (in general) - there is no mystery in that, is there? Now do $[0,1]^T$. Then do an arbitrary $[x,y]^T = x\cdot[1,0]^T + y\cdot [0,1]^T$. Does that clarify things? $\endgroup$
    – NickD
    May 3, 2017 at 19:50
  • $\begingroup$ @DWD, oh man, this is a treasure trove! $\endgroup$
    – mikeglaz
    Jun 21, 2017 at 12:06

5 Answers 5


Your typical point with polar coordinates $r$ and $\phi$ and has vector $$\begin{bmatrix} r\cos\phi\\r\sin\phi \end{bmatrix}.$$ Multiplying $A$ into this gives $$\begin{bmatrix} r\cos\theta\cos\phi-r\sin\theta\sin\phi\\r\sin\theta\cos\phi+r\cos\theta\sin\phi \end{bmatrix} =\begin{bmatrix} r\cos(\theta+\phi)\\r\sin(\theta+\phi) \end{bmatrix}.$$ It's at the same distance from the origin, but rotated anticlockwise by an angle $\theta$.


Any vector in $\Bbb{R}^2$ can be represented as follows

$$v=\left(\begin{array}{c} x\\ y \end{array}\right)=|| v|| \left(\begin{array}{c} \cos(\phi)\\ \sin(\phi) \end{array}\right).$$

If $$T=\left(\begin{array}{cc} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{array}\right) $$

Then $$Tv= || v||\left(\begin{array}{cc} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{array}\right) \left(\begin{array}{c} \cos(\phi)\\ \sin(\phi) \end{array}\right)=||v||\left(\begin{array}{c} \cos(\phi)\cos(\theta)-\sin(\phi)\sin(\theta)\\ \cos(\phi)\sin(\theta)+\sin(\phi)\cos(\theta) \end{array}\right)$$

Now, you will recognize this as the angle sum formula. That is

$$\left(\begin{array}{c} \cos(\phi)\cos(\theta)-\sin(\phi)\sin(\theta)\\ \cos(\phi)\sin(\theta)+\sin(\phi)\cos(\theta) \end{array}\right)=\left(\begin{array}{c} \cos(\phi+\theta)\\ \sin(\phi+\theta) \end{array} \right) $$ so ultimately you can see that this matrix is simply rotating by adding the appropriate angle, and this is why you see the original components of the vector after applying the linear transformation.

  • $\begingroup$ I understand why $x'$ contains $\cos\phi\cos\theta$. Because those are the x components. But why are the y components added to $x'$ as well, namely $\sin\phi\sin\theta$? $\endgroup$
    – mikeglaz
    May 3, 2017 at 19:22
  • $\begingroup$ This is due to how matrix multiplication is defined. Take $$\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\left(\begin{array}{c} x\\ y \end{array}\right) =x\left(\begin{array}{c} a\\ c \end{array}\right) +y\left(\begin{array}{c} b\\ d \end{array}\right).$$ $\endgroup$ May 3, 2017 at 19:23
  • $\begingroup$ I know this is how it is defined. But I want to understand, conceptually, why x and y get both components... $\endgroup$
    – mikeglaz
    May 3, 2017 at 19:38
  • $\begingroup$ I think your final answer is part of my answer. $\begin{bmatrix} \cos(\phi + \theta) \\ \sin(\phi + \theta)\end{bmatrix}$. In this version $x'$ gets only the x component and $y'$ gets only the y component. $\endgroup$
    – mikeglaz
    May 3, 2017 at 20:06
  • $\begingroup$ I suppose I don't understand what you mean by "gets." Would you elaborate? $\endgroup$ May 3, 2017 at 20:46

why does $x′$ contain both an $x$ component and a $y$ component?

This is true for any matrix transformation. If the transformation is represented by matrix $$ A:=\left[\begin{matrix}a&b\\ c&d\\ \end{matrix}\right] $$ then for an arbitrary vector $\begin{bmatrix}x \\ y \end{bmatrix}$ you get $$ \begin{align} \begin{bmatrix}x' \\ y' \end{bmatrix}&=T\begin{bmatrix}x \\ y \end{bmatrix}\\ &=T\left(x\begin{bmatrix}1 \\ 0 \end{bmatrix} + y\begin{bmatrix}0 \\ 1 \end{bmatrix}\right)\\ &=xT\left(\begin{bmatrix}1 \\ 0 \end{bmatrix}\right) + yT\left(\begin{bmatrix}0 \\ 1 \end{bmatrix}\right)\\ &=x\begin{bmatrix}a \\ c \end{bmatrix} + y\begin{bmatrix}b \\ d\end{bmatrix}\\ &=\begin{bmatrix}ax \\ cx \end{bmatrix} + \begin{bmatrix}by \\ dy\end{bmatrix}\\ &=\begin{bmatrix}ax+by \\ cx+dy \end{bmatrix}. \end{align} $$ When $\begin{bmatrix}x \\ y \end{bmatrix}$ is a unit vector, say $\begin{bmatrix}x \\ y \end{bmatrix}=\begin{bmatrix}1 \\ 0 \end{bmatrix}$, all but one of the coordinates is zero so the effect is to pick out just one column of $A$. Otherwise more than one column of $A$ comes into play.

  • $\begingroup$ Did you mean $T\bigg( x\begin{bmatrix} 1 \\ 0 \end{bmatrix} + y\begin{bmatrix} 0 \\ 1 \end{bmatrix}\bigg)$? $\endgroup$
    – mikeglaz
    May 3, 2017 at 19:31
  • $\begingroup$ @mikeglaz Oops, yep. Copy-paste error... $\endgroup$
    – grand_chat
    May 3, 2017 at 19:34
  • $\begingroup$ and $x \begin{bmatrix} a \\ c \end{bmatrix} + y \begin{bmatrix} b \\ d \end{bmatrix}$? $\endgroup$
    – mikeglaz
    May 3, 2017 at 19:36
  • $\begingroup$ @mikeglaz Yes. Sigh... :) $\endgroup$
    – grand_chat
    May 3, 2017 at 19:37

Column 1 contains the $x$ and $y$ values for the vector representing where the transformation takes $e_1$. For example, if the transformation is counterclockwise rotation by $\pi/4$ ($45$ degrees) that vector goes to $$ (\sqrt{2}/2,\sqrt{2}/2) = (\cos(\pi/4), \sin(\pi/4)). $$

It's the first row, not the first column, that contains the $x$ values. Thinking about the rows is generally less informative than thinking about the columns.

I think your problem is not conceptual, it's with the notation. You seem to understand what's going on.


I found the answer. I kept thinking about this and drew out the angle rotation using a ruler, compass, and protractor to see if I could find why the rotation formula is the way it is. I don't have the time to learn TikZ but basically this is where I got to:

enter image description here

We are originally at angle $\alpha$ and want to rotate $\beta$ degrees. In other words, we want to get from $\overline{AE}$ to $\overline{AF}$.

The $x$ coordinate of $\overline{AE} = \overline{AC}$. The $y$ coordinate of $\overline{AE} = \overline{CE}$.

After a rotation of $\beta$ degrees, the new $x = x' = \overline{AB}$ and the new $y = y' = \overline{FB}$.

$x' = \overline{AB} = \overline{AC} - \overline{BC}$

$\overline{BC} = \overline{DE}$

$\triangle ACE: \cos\alpha = \frac{\overline{AC}}{\cos\beta}$

$\overline{AC} = \cos\alpha\cos\beta$

$\angle DFE = \angle CAE$

$\triangle DEF: \sin\alpha = \frac{\overline{DE}}{\sin\beta}$

$\overline{DE} = \sin\alpha\sin\beta$

$x' = \cos\alpha\cos\beta - \sin\alpha\sin\beta$

$y' = \overline{FB} = \overline{FD} + \overline{DB}$

$\overline{DB} = \overline{CE}$

$\triangle DEF: \cos\alpha = \frac{\overline{FD}}{\sin\beta}$

$\overline{FD} = \cos\alpha\sin\beta$

$\triangle ACE: \sin\alpha = \frac{\overline{CE}}{\cos\beta}$

$\overline{CE} = \sin\alpha\cos\beta$

$y' = \cos\alpha\sin\beta + \sin\alpha\cos\beta$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .