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Eigenvalues are a fundamental concept in linear algebra, especially in the process of matrix diagonalization. Simply put, eigenvalues are special numbers associated with a square matrix that provide insight into its properties. Let's dive deeper into how we find these.
To determine the eigenvalues of a matrix, you solve the characteristic equation. For a given matrix \( B \), this equation is obtained by setting the determinant of \( (B - \lambda I) \) to zero. Here, \( \lambda \) represents the eigenvalues, and \( I \) is the identity matrix of the same dimensions as \( B \).
In the exercise, the matrix \( B \) had the form:
\[B = \begin{pmatrix} 3 & 1 \ 0 & 2 \end{pmatrix}\]

• The characteristic equation is: \[ \text{det}\begin{pmatrix} 3 - \lambda & 1 \ 0 & 2 - \lambda \end{pmatrix} = 0 \]
• By expanding, we solve \((3-\lambda)(2-\lambda) = 0\), which gives two solutions: \( \lambda_1 = 3 \) and \( \lambda_2 = 2 \).

These solutions are the eigenvalues of \( B \), indicating the scalar factors by which their corresponding eigenvectors are stretched.

Once the eigenvalues are known, eigenvectors are their next logical step. An eigenvector is a non-zero vector that only changes by a scalar factor when a linear transformation is applied. In simple terms, if multiplying a matrix by a vector yields a result that is a scalar multiple of that vector, then that vector is an eigenvector of the matrix.
To find eigenvectors, solve the equation \((B - \lambda I)\mathbf{v} = 0\). This will provide the directional vectors associated with each eigenvalue.
For the exercise,

• For \( \lambda_1 = 3 \), solving \((B - 3I)\mathbf{v} = 0\) gives the system: \ \[\begin{pmatrix} 0 & 1 \ 0 & -1 \end{pmatrix} \mathbf{v} = \mathbf{0}\]
• The eigenvector \( \mathbf{v}_1 \) is determined from this system as: \ \[ \mathbf{v}_1 = \begin{pmatrix} 1 \ 0 \end{pmatrix} \]
• For \( \lambda_2 = 2 \), solve \((B - 2I)\mathbf{w} = 0\) resulting in: \[\begin{pmatrix} 1 & 1 \ 0 & 0 \end{pmatrix} \mathbf{w} = \mathbf{0}\]
• The eigenvector \( \mathbf{w}_2 \) is \ \[ \mathbf{w}_2 = \begin{pmatrix} -1 \ 1 \end{pmatrix} \]

These vectors are the eigenvectors corresponding to their respective eigenvalues, indicating directions in which transformations stretch or compress.

The characteristic equation is a crucial concept when discussing eigenvalues and matrix diagonalization. It essentially serves as the gateway to finding these eigenvalues, which are key to understanding how a matrix transforms space.
The characteristic equation is derived from the determinant of \( (B - \lambda I) \), set equal to zero. Here's a breakdown:

• The matrix \( B \) undergoes a transformation such that \( B - \lambda I \) forms a new matrix, where \( \lambda \) is an unknown scalar at this point.
• The determinant of this new matrix, \( \text{det}(B - \lambda I) = 0 \), forms a polynomial equation in terms of \( \lambda \).

For the problem given,

• The initial matrix was: \ \[B = \begin{pmatrix} 3 & 1 \ 0 & 2 \end{pmatrix}\]
• The characteristic equation derived is: \ \[\text{det}\begin{pmatrix} 3 - \lambda & 1 \ 0 & 2 - \lambda \end{pmatrix} = 0\]

Solving \((3 - \lambda)(2 - \lambda) = 0\) yielded the eigenvalues \( \lambda_1 = 3 \) and \( \lambda_2 = 2 \). This polynomial equation is key in determining how the matrix can be diagonalized.