Question

Let \( T : \mathbb{R}^2 \rightarrow \mathbb{R}^2 \) be the linear transformation given by \( T(x, y) = (-x, y). \) Then

• A. \( T^{2k} = T \) for all \( k \ge 1 \)
• B. \( T^{2k+1} = -T \) for all \( k \ge 1 \)
• C. The range of \( T^2 \) is a proper subspace of the range of \( T \)
• D. The range of \( T^2 \) is equal to the range of \( T \)

Hint:

For linear transformations, if \( T^2 = I \), then the range and domain remain identical since \( T \) is invertible.

Answer 1

The Correct Option is D

Step 1: Compute \( T^2(x, y). \) 
\( T(x, y) = (-x, y) \Rightarrow T^2(x, y) = T(T(x, y)) = T(-x, y) = (x, y). \) 
 

Step 2: Observe pattern. 
\( T^2 \) is identity, so \( T^2 = I. \) Hence, \( T^{2k} = I \) and \( T^{2k+1} = T. \) 
 

Step 3: Compare ranges. 
The range of \( T \) is \( \mathbb{R}^2 \), since \( T(x, y) = (-x, y) \) is onto. Similarly, \( T^2 = I \) also maps to all of \( \mathbb{R}^2 \). 
 

Step 4: Conclusion. 
Therefore, the range of \( T^2 \) is equal to the range of \( T. \)