91Ó°ÊÓ

When dealing with linear transformations, we often need a convenient way to express and manipulate them. That's where the matrix representation of a linear transformation comes into play. A matrix can encode all the information needed to transform vectors in a specific vector space.
In the problem, the transformation \( T \) moves vectors in the space defined by basis \( S \) to another set of vectors also within that space. The matrix \( A \) relative to the basis \( S \) represents how \( T \) acts on the vectors \( u_1 \) and \( u_2 \).
This is done by finding \( T(u_1) \) and \( T(u_2) \), then expressing the results as linear combinations of \( u_1 \) and \( u_2 \). Thus, matrix \( A \) becomes:

• First column: coefficients of \( T(u_1) = 3u_1 - 2u_2 \)
• Second column: coefficients of \( T(u_2) = u_1 + 4u_2 \)

This results in the matrix:\[ A = \begin{bmatrix} 3 & 1 \ -2 & 4 \end{bmatrix} \]Each column represents the image of the respective basis vector under the transformation. This matrix provides a clear and compact description of how the transformation affects the space.

Sometimes it's helpful or necessary to describe a linear transformation using a different basis. A change of basis adjusts the perspective from which we view transformations, providing a potentially simpler or more insightful form.
In the exercise, we consider two different bases: \( S \) and \( S' \). To move between these, we use a change of basis matrix \( P \), which transforms one basis into another. For example, we are given:

• \( w_1 = u_1 + u_2 \)
• \( w_2 = 2u_1 + 3u_2 \)

To find \( P \), consider how basis \( S' \) is built from \( S \). Each column of \( P \) corresponds to the coefficients of \( w_1 \) and \( w_2 \) written in terms of \( u_1 \) and \( u_2 \). This yields:\[ P = \begin{bmatrix} 1 & 2 \ 1 & 3 \end{bmatrix} \]By using \( P \), we can transform a representation of \( T \) from \( S \) to \( S' \) and vice versa by calculating \( B = P^{-1}AP \), reconciling the two views into a coherent picture.

The inverse of a matrix is an essential tool in linear algebra, providing a way to "reverse" a transformation. For a square matrix \( P \), its inverse \( P^{-1} \) satisfies the equation\[ PP^{-1} = P^{-1}P = I \]where \( I \) is the identity matrix. An identity matrix acts like multiplying by 1, keeping any vector unchanged.
To find the inverse, we first calculate the determinant. For matrix \( P \):\[ \det(P) = (1)(3) - (1)(2) = 1 \]When the determinant is non-zero, the matrix is invertible. The \( 2\times2 \) inverse can be found by swapping the diagonal elements and negating the off-diagonal ones:\[ \text{adj}(P) = \begin{bmatrix} 3 & -2 \ -1 & 1 \end{bmatrix} \]With a determinant of 1, the inverse simplifies directly to its adjugate. Thus:\[ P^{-1} = \begin{bmatrix} 3 & -2 \ -1 & 1 \end{bmatrix} \]Using \( P^{-1} \) allows for transformation between different views or basis systems. In our problem, it helps confirm that \( B \) is indeed the representation of \( T \) in the new basis \( S' \), ensuring the correctness and consistency of the transformation framework.