91Ó°ÊÓ

An eigenvalue is a special number associated with a linear transformation or matrix. It reveals important characteristics about the transformation's behavior. In simpler terms, when a matrix acts on an eigenvector, the outcome is simply that vector scaled by the eigenvalue. So, if we have a transformation \( T \) and a vector \( v \), then the equation \( T(v) =  \lambda  v \) holds, where \(  \lambda  \) is the eigenvalue.For our transformation \( \alpha \) represented by the matrix \( A = \begin{pmatrix} 3 & 4 \ -1 & -1 \end{pmatrix} \), we identified that 1 is an eigenvalue because the transformation can be described by \( (A - I)v = 0 \), leading us to the equation \( \begin{pmatrix} 2 & 4 \ -1 & -2 \end{pmatrix}\cdot\begin{pmatrix} x \ y \end{pmatrix} = 0 \). This confirms that 1 is an eigenvalue since this equation is satisfied.

Eigenvectors are special vectors associated with a linear transformation that change by only a scalar factor (the eigenvalue) when that transformation is applied. When a matrix \( A \) operates on an eigenvector \( v \), the resulting vector is a scaled version of \( v \): \( A(v) =  \lambda  v \).For our transformation \( \alpha \) with eigenvalue 1, we solve \( (A - I)v = 0 \) to find the eigenvectors. In this case, the equation \( \begin{pmatrix} 2 & 4 \ -1 & -2 \end{pmatrix}\cdot\begin{pmatrix} x \ y \end{pmatrix} = 0 \) simplifies to \( x = -2y \). Setting \( y = 1 \) gives us the eigenvector \( \begin{pmatrix} -2 \ 1 \end{pmatrix} \). Unfortunately, since there is only one eigenvector, it does not span the whole space \( \mathbb{R}^2 \), demonstrating that the eigenvectors of \( \alpha \) do not fully span the space.

The characteristic polynomial is a powerful tool used to find the eigenvalues of a matrix. You derive it from the determinant of the matrix \( A - I \lambda \), where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalue. Solving the polynomial equation gives you the eigenvalues.In our example, the matrix \( A \) representing the transformation \( \alpha \) is \( \begin{pmatrix} 3 & 4 \ -1 & -1 \end{pmatrix} \). To find its eigenvalues, we form the matrix \( A - \lambda I = \begin{pmatrix} 3 - \lambda & 4 \ -1 & -1 - \lambda \end{pmatrix} \) and find its determinant, solving \( \det(A - \lambda I) = 0 \). Simplifying the determinant leads us to the characteristic polynomial, which, when solved, confirms that the eigenvalue \( \lambda = 1 \) is valid.

A matrix is a way to represent linear transformations compactly. Every linear transformation in \( \mathbb{R}^2 \) can be written as a matrix, where the transformation's effect on the standard basis vectors \( e_1 \) and \( e_2 \) determines the matrix columns.In our exercise, the linear transformation \( \alpha \) has effects \( \alpha(e_1) = 3e_1 - e_2 \) and \( \alpha(e_2) = 4e_1 - e_2 \). These transformations give us the columns of matrix \( A = \begin{pmatrix} 3 & 4 \ -1 & -1 \end{pmatrix} \). This matrix form allows us to easily calculate the outcome of applying \( \alpha \) to any vector by simple matrix multiplication.