Diagonalizable matrices are key players in linear algebra. They're matrices that can be transformed into diagonal form, making calculations way easier. This transformation involves eigenvalues and eigenvectors, which are crucial concepts in this chapter.

Understanding when a matrix is diagonalizable is super important. It's all about the relationship between eigenvalues and their multiplicities. Knowing these conditions helps you quickly determine if a matrix can be diagonalized, saving you time and effort in many applications.

Diagonalizable matrices

Definition and properties

• A square matrix A is diagonalizable if it is similar to a diagonal matrix D
  • There exists an invertible matrix P such that $P^{-1}AP = D$
  • The diagonal entries of D are the eigenvalues of A ($\lambda_1, \lambda_2, ..., \lambda_n$)
  • The columns of P are the corresponding eigenvectors of A ($v_1, v_2, ..., v_n$)
• If A is diagonalizable, then $A = PDP^{-1}$
  • D is a diagonal matrix with the eigenvalues of A on its diagonal
  • P is a matrix whose columns are the corresponding eigenvectors of A
• Diagonalizable matrices have distinct eigenvalues or eigenvalues with geometric multiplicities equal to their algebraic multiplicities
  • Distinct eigenvalues (no repeated eigenvalues)
  • Eigenvalues with geometric multiplicity (dimension of eigenspace) equal to algebraic multiplicity (multiplicity as a root of the characteristic polynomial)

Significance and applications

• Diagonalization simplifies matrix operations and calculations
  • Powers of a diagonalizable matrix A can be computed as $A^k = PD^kP^{-1}$
  • Exponential of a diagonalizable matrix A can be computed as $e^A = Pe^DP^{-1}$
• Diagonalization is used in various applications
  • Principal component analysis (PCA) in data science and machine learning
  • Quantum mechanics to represent observables and states
  • Systems of differential equations to decouple and solve them independently

Conditions for diagonalizability

Necessary and sufficient conditions

• A matrix A is diagonalizable if and only if A has n linearly independent eigenvectors, where n is the size of the matrix
  • The eigenvectors form a basis for the vector space $\mathbb{R}^n$
  • Example: A 3x3 matrix A is diagonalizable if it has 3 linearly independent eigenvectors
• A matrix A is diagonalizable if and only if the sum of the geometric multiplicities of its eigenvalues is equal to n
  • The geometric multiplicity of an eigenvalue is the dimension of its eigenspace
  • Example: If a 4x4 matrix A has eigenvalues with geometric multiplicities 2, 1, and 1, then A is diagonalizable (2+1+1 = 4)
• A matrix A is diagonalizable if and only if the minimal polynomial of A has no repeated factors
  • The minimal polynomial is the monic polynomial of the lowest degree that annihilates A
  • The minimal polynomial of a diagonalizable matrix is a product of distinct linear factors $(\lambda_1 - x)(\lambda_2 - x)...(\lambda_k - x)$
• A matrix A is diagonalizable if and only if there exists a basis for $\mathbb{R}^n$ consisting entirely of eigenvectors of A
  • The basis vectors are the columns of the matrix P in the diagonalization $A = PDP^{-1}$
• A matrix A is diagonalizable if and only if A is similar to a diagonal matrix
  • Similarity is defined by the relation $P^{-1}AP = D$, where P is an invertible matrix and D is a diagonal matrix

Sufficient conditions

• If a matrix A has n distinct eigenvalues, then A is diagonalizable
  • Each eigenvalue has a geometric multiplicity of 1, summing up to n
  • Example: A 2x2 matrix with eigenvalues 1 and 2 is diagonalizable
• If a matrix A is symmetric (A = A^T), then A is diagonalizable
  • Symmetric matrices have orthogonal eigenvectors that form a basis for $\mathbb{R}^n$
  • Example: The matrix $\begin{pmatrix} 1 & 2 \ 2 & 1 \end{pmatrix}$ is symmetric and diagonalizable

Diagonalizability of matrices

Determining diagonalizability

• Find the eigenvalues of the matrix A by solving the characteristic equation $\det(A - \lambda I) = 0$
  • The roots of the characteristic polynomial are the eigenvalues
  • Example: For $A = \begin{pmatrix} 3 & 1 \ 0 & 2 \end{pmatrix}$, the characteristic equation is $(\lambda - 3)(\lambda - 2) = 0$, giving eigenvalues 3 and 2
• For each distinct eigenvalue $\lambda_i$, find the corresponding eigenvectors by solving the equation $(A - \lambda_iI)v = 0$
  • The non-zero solutions of this equation are the eigenvectors associated with $\lambda_i$
  • Example: For the eigenvalue 3, solve $\begin{pmatrix} 0 & 1 \ 0 & -1 \end{pmatrix}v = 0$, giving eigenvector $\begin{pmatrix} 1 \ 0 \end{pmatrix}$
• Check if the eigenvectors associated with each eigenvalue span the entire vector space $\mathbb{R}^n$
  • If the eigenvectors form a basis for $\mathbb{R}^n$, the matrix is diagonalizable
  • Example: If a 3x3 matrix has eigenvectors $\begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix}$, it is diagonalizable
• Alternatively, check if the geometric multiplicity of each eigenvalue is equal to its algebraic multiplicity
  • If this condition holds for all eigenvalues, the matrix is diagonalizable
  • Example: A matrix with eigenvalues 1 (algebraic multiplicity 2) and 2 (algebraic multiplicity 1) is diagonalizable if the eigenspace of 1 has dimension 2

Diagonalization process

• Form the matrix P by using the eigenvectors of A as its columns
  • Normalize the eigenvectors to ensure P is invertible
  • Example: If the eigenvectors are $\begin{pmatrix} 1 \ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \ 1 \end{pmatrix}$, then $P = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$
• Form the diagonal matrix D by placing the eigenvalues of A on its diagonal
  • The order of the eigenvalues should match the order of the corresponding eigenvectors in P
  • Example: If the eigenvalues are 3 and 2, then $D = \begin{pmatrix} 3 & 0 \ 0 & 2 \end{pmatrix}$
• Express A as $A = PDP^{-1}$
  • This is the diagonalization of A
  • Example: $\begin{pmatrix} 3 & 1 \ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 3 & 0 \ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & -1 \ 0 & 1 \end{pmatrix}$

Diagonalizability vs Geometric multiplicity

• The geometric multiplicity of an eigenvalue $\lambda$ is the dimension of the eigenspace associated with $\lambda$
  • The eigenspace is the null space of the matrix $A - \lambda I$
  • The geometric multiplicity is the number of linearly independent eigenvectors corresponding to $\lambda$
• The algebraic multiplicity of an eigenvalue $\lambda$ is the multiplicity of $\lambda$ as a root of the characteristic polynomial
  • It is the power of the factor $(\lambda - x)$ in the characteristic polynomial
  • Example: If the characteristic polynomial is $(\lambda - 1)^2(\lambda - 2)$, the algebraic multiplicities are 2 for $\lambda = 1$ and 1 for $\lambda = 2$
• For a matrix to be diagonalizable, the geometric multiplicity of each eigenvalue must be equal to its algebraic multiplicity
  • If the geometric multiplicity is less than the algebraic multiplicity for any eigenvalue, the matrix is not diagonalizable
  • Example: A 3x3 matrix with eigenvalues 1 (algebraic multiplicity 2) and 2 (algebraic multiplicity 1) is diagonalizable if the eigenspace of 1 has dimension 2
• The sum of the geometric multiplicities of all eigenvalues must be equal to the size of the matrix for it to be diagonalizable
  • If the sum is less than the size of the matrix, there are not enough eigenvectors to form a basis for $\mathbb{R}^n$
  • Example: A 4x4 matrix with eigenvalues 1 (geometric multiplicity 2), 2 (geometric multiplicity 1), and 3 (geometric multiplicity 1) is diagonalizable (2+1+1 = 4)