First basis vector
\(R_\theta\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}\)
The first column is the rotated x-axis.
Lesson 5 of 9 · Linear algebra foundation
A matrix is a recipe for moving vectors.
Your win: multiply a 2D vector by the rotation matrix and recognize the same coordinate rules from complex multiplication.
Mission link: RoPE implementations use real-valued tensors. The matrix view expresses the same complex rotation using ordinary real arithmetic.
Rows produce output coordinates
Multiplying a \(2\times2\) matrix by a two-component column vector forms one dot product per row:
\[
\begin{bmatrix}a&b\\c&d\end{bmatrix}
\begin{bmatrix}x\\y\end{bmatrix}
=
\begin{bmatrix}ax+by\\cx+dy\end{bmatrix}.
\]
The rotation recipe
\[
\underbrace{\begin{bmatrix}
\cos\theta&-\sin\theta\\
\sin\theta& \cos\theta
\end{bmatrix}}_{R_\theta}
\begin{bmatrix}x\\y\end{bmatrix}
=
\begin{bmatrix}
x\cos\theta-y\sin\theta\\
x\sin\theta+y\cos\theta
\end{bmatrix}.
\]
The output is exactly the pair obtained from \((x+iy)e^{i\theta}\). Complex multiplication and matrix multiplication are two notations for the same motion.
A quarter turn
At \(\theta=\pi/2\), \(\cos\theta=0\) and \(\sin\theta=1\):
\[
R_{\pi/2}
=\begin{bmatrix}0&-1\\1&0\end{bmatrix},
\qquad
R_{\pi/2}\begin{bmatrix}x\\y\end{bmatrix}
=\begin{bmatrix}-y\\x\end{bmatrix}.
\]
A 90° counterclockwise turn maps \((x,y)\) to \((-y,x)\).
Numbers substituted: \(R_{\pi/2}(3,2)\)
\[
\begin{aligned}
R_{\pi/2}\begin{bmatrix}3\\2\end{bmatrix}
&=
\begin{bmatrix}0&-1\\1&0\end{bmatrix}
\begin{bmatrix}3\\2\end{bmatrix}\\
&=
\begin{bmatrix}0(3)+(-1)(2)\\1(3)+0(2)\end{bmatrix}
=
\begin{bmatrix}-2\\3\end{bmatrix}.
\end{aligned}
\]
Numbers substituted: \(R_{\pi/6}(2,0)\)
\[
\begin{aligned}
R_{\pi/6}\begin{bmatrix}2\\0\end{bmatrix}
&=
\begin{bmatrix}\sqrt3/2&-1/2\\1/2&\sqrt3/2\end{bmatrix}
\begin{bmatrix}2\\0\end{bmatrix}\\
&=\begin{bmatrix}\sqrt3\\1\end{bmatrix}.
\end{aligned}
\]
This matches the complex-number calculation in Lesson 4.
Numbers substituted: \(R_{\pi}(-1,4)\)
\[
\begin{bmatrix}-1&0\\0&-1\end{bmatrix}
\begin{bmatrix}-1\\4\end{bmatrix}
=
\begin{bmatrix}1\\-4\end{bmatrix}.
\]
A half turn negates both coordinates.
Columns show where the basis goes
Second basis vector
\(R_\theta\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}-\sin\theta\\\cos\theta\end{bmatrix}\)
The second column is the rotated y-axis.
Retrieval check
Which output is \(R_{\pi/2}(x,y)\)?
Practice before moving on
- Compute \(R_{\pi/2}\begin{bmatrix}4\\-1\end{bmatrix}\).
- Compute \(R_{\pi}\begin{bmatrix}2\\5\end{bmatrix}\).
- Compute \(R_{0}\begin{bmatrix}-3\\7\end{bmatrix}\).
- Use \(R_{\pi/4}\) to rotate \((\sqrt2,0)\).
- Verify by lengths that \((3,2)\) and its quarter-turn output \((-2,3)\) have equal magnitude.
- Explain what the two columns of \(R_{\pi/2}\) say geometrically.
Check solutions
- \(\begin{bmatrix}1\\4\end{bmatrix}\).
- \(\begin{bmatrix}-2\\-5\end{bmatrix}\).
- \(\begin{bmatrix}-3\\7\end{bmatrix}\).
- \(\begin{bmatrix}1\\1\end{bmatrix}\).
- Both squared magnitudes are \(13\).
- The x-axis basis goes to \((0,1)\), and the y-axis basis goes to \((-1,0)\).
Pairwise form: a full RoPE matrix is block diagonal: one \(2\times2\) rotation block per adjacent pair, with zeros between different pairs.
Primary source: MIT OCW: Linear Transformations and Their Matrices. Watch for the idea that a matrix is determined by where it sends the basis vectors.
Ask the teaching agent to give you a rotation at \(0\), \(\pi/2\), or \(\pi\) to calculate by hand.