Find the standard matrix of T : R 2 → R 2 T:\R^2\to \R^2 T : R 2 → R 2 rotates points (about the origin) through − π / 4 −π/4 − π /4 radians (clockwise).
Let T : R 2 → R 2 T:\R^2\to \R^2 T : R 2 → R 2 be a linear transformation given by rotating vectors in the counter-clockwise direction through an angle of θ . θ. θ .
Then the matrix A A A of T T T is given by
T ( ( 1 0 ) ) = ( cos θ sin θ ) T\bigg(\begin{pmatrix}
1 \\
0
\end{pmatrix}\bigg)=\begin{pmatrix}
\cos \theta \\
\sin \theta
\end{pmatrix} T ( ( 1 0 ) ) = ( cos θ sin θ )
T ( ( 0 1 ) ) = ( − sin θ cos θ ) T\bigg(\begin{pmatrix}
0 \\
1
\end{pmatrix}\bigg)=\begin{pmatrix}
- \sin \theta \\
\cos \theta
\end{pmatrix} T ( ( 0 1 ) ) = ( − sin θ cos θ )
A = ( cos θ − sin θ sin θ cos θ ) A=\begin{pmatrix}
\cos \theta & - \sin\theta \\
\sin \theta & \cos \theta
\end{pmatrix} A = ( cos θ sin θ − sin θ cos θ ) Given θ = π / 4. \theta=\pi/4. θ = π /4. Then
A = ( cos ( π / 4 ) − sin ( π / 4 ) sin ( π / 4 ) cos ( π / 4 ) ) = ( 2 / 2 − 2 / 2 2 / 2 2 / 2 ) A=\begin{pmatrix}
\cos (\pi/4) & - \sin (\pi/4) \\
\sin (\pi/4) & \cos(\pi/4)
\end{pmatrix}=\begin{pmatrix}
\sqrt{2}/2& -\sqrt{2}/2 \\
\sqrt{2}/2 & \sqrt{2}/2
\end{pmatrix} A = ( cos ( π /4 ) sin ( π /4 ) − sin ( π /4 ) cos ( π /4 ) ) = ( 2 /2 2 /2 − 2 /2 2 /2 )
Comments