Rotation Inverse
1. **Problem statement:** Determine which given statements about the linear transformation $T:\mathbb{R}^2 \to \mathbb{R}^2$ defined by rotation through angle $\theta$ are false.
2. **Matrix of $T$:**
$$
\begin{pmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{pmatrix}
$$
This matrix represents a rotation in $\mathbb{R}^2$.
3. **Check statement (a):**
Given inverse candidate matrix:
$$
\begin{pmatrix}
-\cos \theta & \sin \theta \\
-\sin \theta & -\cos \theta
\end{pmatrix}
$$
The true inverse of a rotation matrix $R$ is its transpose, which equals rotation by $-\theta$:
$$
R^{-1} = R^T = \begin{pmatrix}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{pmatrix}
$$
The candidate matrix in (a) differs by signs; to verify, multiply candidate inverse by original:
$$
\begin{pmatrix}
-\cos \theta & \sin \theta \\
-\sin \theta & -\cos \theta
\end{pmatrix}
\begin{pmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{pmatrix} =
\begin{pmatrix}
(-\cos \theta)(\cos \theta) + (\sin \theta)(\sin \theta) & (-\cos \theta)(-\sin \theta) + (\sin \theta)(\cos \theta) \\
(-\sin \theta)(\cos \theta) + (-\cos \theta)(\sin \theta) & (-\sin \theta)(-\sin \theta) + (-\cos \theta)(\cos \theta)
\end{pmatrix}
$$
Simplify entries:
- Top-left: $-\cos^2 \theta + \sin^2 \theta = -(\cos^2 \theta - \sin^2 \theta)$
- Top-right: $\cos \theta \sin \theta + \sin \theta \cos \theta = 2\sin \theta \cos \theta$
- Bottom-left: $-\sin \theta \cos \theta - \cos \theta \sin \theta = -2 \sin \theta \cos \theta$
- Bottom-right: $\sin^2 \theta - \cos^2 \theta = -(\cos^2 \theta - \sin^2 \theta)$
This matrix is not the identity matrix $I$, so (a) is **false**.
4. **Statement (b):** $\text{rank}(T) = 2$.
Since $T$ is a rotation, it is invertible and maps $\mathbb{R}^2$ onto itself with full rank. So rank is 2. (b) is **true**.
5. **Statement (c):** The inverse transformation does not exist.
Since $T$ is a rotation, it is invertible, so inverse exists. Hence (c) is **false**.
6. **Statement (d):** $\text{nullity}(T) = 1$.
By Rank-Nullity Theorem:
$$ \text{rank}(T) + \text{nullity}(T) = \dim(\mathbb{R}^2) = 2 $$
Since $\text{rank}(T) = 2$, nullity$=0$. So (d) is **false**.
7. **Statement (e):** $T$ is one-to-one.
An invertible linear transformation is one-to-one. So (e) is **true**.
**Summary of false statements:** (a), (c), (d).
**Final answer:** Statements (a), (c), and (d) are false.