91Ó°ÊÓ

Eigenvalues are special scalars associated with a matrix that play a central role in the concept of diagonalization. For a square matrix like matrix \( A \), an eigenvalue \( \lambda \) is found when there exists a non-zero vector \( \mathbf{v} \) such that \( A \mathbf{v} = \lambda \mathbf{v} \). This means that when matrix \( A \) acts on \( \mathbf{v} \), it simply scales it by a factor of \( \lambda \).
To find these eigenvalues, you solve the characteristic equation \( \det(A - \lambda I) = 0 \). This equation arises from the requirement that \( A - \lambda I \) must have a zero determinant for \( \mathbf{v} \) to exist. In our problem, the calculated eigenvalues of matrix \( A \) are \( \lambda_1 = 3 \), \( \lambda_2 = 1 \), and \( \lambda_3 = 0 \). These values are used to form the diagonal matrix \( D \), which is a key step in simplifying the matrix operations.

Eigenvectors are vectors associated with eigenvalues and give direction to the scaling effect described by eigenvalues. When you have a matrix \( A \) and an eigenvalue \( \lambda \), the corresponding eigenvector \( \mathbf{v} \) satisfies the equation \( (A - \lambda I)\mathbf{v} = 0 \). This means the matrix \( A \) acts on \( \mathbf{v} \) by stretching or compressing it but not rotating it.
In our solution, for each eigenvalue, we find the respective eigenvector. For \( \lambda_1 = 3 \), the eigenvector is \( \mathbf{v}_1 = \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} \). For \( \lambda_2 = 1 \), we get \( \mathbf{v}_2 = \begin{bmatrix} i \ 0 \ 1 \end{bmatrix} \), and for \( \lambda_3 = 0 \), the eigenvector is \( \mathbf{v}_3 = \begin{bmatrix} 1 \ 0 \ -i \end{bmatrix} \). These vectors are crucial as they form the columns of the unitary matrix \( U \), which is used in diagonalizing \( A \).

Diagonalization is the process of transforming a matrix into a diagonal matrix. A diagonal matrix, like \( D \), has all its non-diagonal elements as zero. This simplifies computations, especially in finding powers of matrices since diagonal matrices are much easier to handle.
To diagonalize a matrix \( A \), you find the unitary matrix \( U \) composed of the normalized eigenvectors of \( A \), and construct \( D \) using its eigenvalues. Diagonalization is achieved through the relation \( A = U D U^{-1} \), where \( U^{-1} = U^* \), the conjugate transpose, due to \( U \) being unitary. In the exercise, we found \( D = \begin{bmatrix} 3 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 0 \end{bmatrix} \). Verify this relationship by ensuring \( U D U^* = A \), confirming successful diagonalization.

Orthonormality refers to a set of vectors that are both orthogonal and normalized. Orthogonality means that vectors in the set are perpendicular to each other, and normalization means each vector has a unit length.
For a matrix \( U \) to be unitary, the columns must be orthonormal eigenvectors. In the exercise, upon forming \( U \) from the eigenvectors \( \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \), we ensure they are orthonormal by normalizing \( \mathbf{v}_2 \) and \( \mathbf{v}_3 \). Normalization was done by dividing each vector by its length. This orthonormality guarantees that \( U \) retains its unitary property, meaning \( U U^* = I \), the identity matrix.
The orthonormality condition ensures that transformations involving \( U \) preserve the geometric structure of the data, such as angles and lengths, critical for maintaining the integrity of matrix operations during diagonalization.