Question

Let \( T: \mathbb{R}^3 \to \mathbb{R}^3 \) be a linear transformation such that \[ T \left( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, T^2 \left( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, T^2 \left( \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right) = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}. \] Then the rank of \( T \) is \(\underline{\hspace{2cm}}\) .

Hint:

To determine the rank of a linear transformation, examine the linear dependence of the vectors mapped by the transformation. The rank is the dimension of the image (range) of the transformation.

Answer 1

Correct Answer: 2

Given the transformation \( T: \mathbb{R}^3 \to \mathbb{R}^3 \), we are provided with the following information: - \( T \left( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \), - \( T^2 \left( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \), - \( T^2 \left( \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right) = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \). We need to determine the rank of \( T \). Step 1: Understanding the problem The rank of a linear transformation \( T \) is the dimension of the image (or range) of \( T \), which is the subspace of \( \mathbb{R}^3 \) spanned by the vectors that \( T \) maps. This is equal to the number of linearly independent vectors that are mapped by \( T \). Step 2: Exploring the behavior of \( T \) and \( T^2 \) From the given information, we observe the following: \\ - \( T \) maps \( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \) to \( \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \). \\ - \( T^2 \) maps both \( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \) and \( \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \) to the same vector, \( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \). \\ Step 3: Analyzing the rank of \( T \) \\ We now investigate the linear dependence of the vectors under \( T \). We know that \( T^2 \) maps two distinct vectors (namely \( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \) and \( \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \)) to the same vector, \( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \). \\ - The fact that \( T^2 \) maps two distinct vectors to the same result implies that the image of \( T^2 \) is one-dimensional, i.e., it spans a one-dimensional subspace of \( \mathbb{R}^3 \). \\ - Since \( T^2 \) has a one-dimensional image, the rank of \( T \) can be at most 2 (because the image of \( T \) is at least as large as the image of \( T^2 \)). \\ Step 4: Conclusion The rank of \( T \) is the dimension of its image. From the analysis above, we find that the image of \( T \) must be a 2-dimensional subspace because the image of \( T^2 \) is one-dimensional and the transformation \( T \) is not degenerate (i.e., it does not collapse the whole space to a point). Therefore, the rank of \( T \) is 2. Final Answer: \[ \boxed{2}. \]