Matrix Diagonalization

The diagonalization condition

1.A square matrix is said to be diagonalizable when it is possible to find a basis of the underlying vector space consisting entirely of eigenvectors of that matrix. When such a basis exists, the matrix can be written as a diagonal matrix whose entries are the eigenvalues. In this representation the computation of powers, exponentials, and solutions of linear differential equations reduces to operations on the diagonal entries.

2.Let $A$ be a square matrix of order $n$ with entries in $\mathbb{R}$ or $\mathbb{C}.$ The matrix $A$ is diagonalizable if and only if there exists an invertible matrix $P$ and a diagonal matrix $D$ such that the following relation holds:

$$A = P D P^{-1}$$

3.The columns of $P$ are eigenvectors of $A,$ and the diagonal entries of $D$ are the corresponding eigenvalues. Equivalently, the relation above can be rewritten as follows:

$$P^{-1} A P = D$$

4.This formulation makes clear that $P$ is a change-of-basis matrix: it transforms the standard representation of $A$ into the diagonal representation $D$ expressed in the eigenvector basis.

5.

The matrix $D$ is unique up to the ordering of the eigenvalues along the diagonal, while $P$ is not unique, since each eigenvector may be replaced by any nonzero scalar multiple.

Similar matrices

6.The relation $P^{-1} A P = D$ is an instance of a broader relation between square matrices. Two square matrices $A$ and $B$ of order $n$ are called similar, written $A \sim B,$ when there exists an invertible matrix $P$ such that:

$$B = P^{-1} A P$$

7.Similar matrices represent the same linear map in different bases, with $P$ the change-of-basis matrix between them. Quantities that do not depend on the choice of basis are therefore shared by similar matrices: $A$ and $B$ have the same characteristic polynomial, hence the same eigenvalues with the same algebraic multiplicities, and in particular the same determinant, the same trace, and the same rank.

8.In this language a matrix is diagonalizable exactly when it is similar to a diagonal matrix. The diagonal entries of that matrix are the eigenvalues of $A,$ and the columns of $P$ are the eigenvectors that form the new basis.

Eigenvalues and eigenvectors

9.The construction of the diagonalization relies entirely on the eigenstructure of $A.$ Recall that a scalar $\lambda$ is an eigenvalue of $A$ if there exists a nonzero vector $\mathbf{v}$ satisfying the following equation:

$$A\mathbf{v} = \lambda\mathbf{v}$$

10.Such a vector $\mathbf{v}$ is called an eigenvector of $A$ associated with $\lambda.$ The eigenvalues are determined by solving the characteristic equation, which is obtained by requiring that the matrix $A - \lambda I$ be singular:

$$\det(A - \lambda I) = 0$$

11.The left-hand side of this equation is a polynomial of degree $n$ in $\lambda,$ called the characteristic polynomial of $A.$ Its roots, counted with multiplicity, are the eigenvalues of $A.$ Once an eigenvalue $\lambda_k$ has been determined, the associated eigenvectors are the nonzero solutions of the homogeneous linear system:

$$(A - \lambda_k I)\mathbf{v} = \mathbf{0}$$

12.The set of all solutions, including the zero vector, constitutes a subspace of $\mathbb{R}^n$ (or $\mathbb{C}^n$), called the eigenspace associated with $\lambda_k.$

Algebraic and geometric multiplicity

13.Each eigenvalue $\lambda_k$ carries two distinct notions of multiplicity that play a central role in determining whether $A$ is diagonalizable. The algebraic multiplicity of $\lambda_k,$ denoted $m_a(\lambda_k),$ is the multiplicity of $\lambda_k$ as a root of the characteristic polynomial. The geometric multiplicity of $\lambda_k,$ denoted $m_g(\lambda_k),$ is the dimension of the corresponding eigenspace, that is:

$$m_g(\lambda_k) = \dim \ker(A - \lambda_k I)$$

14.It can be shown that for every eigenvalue the geometric multiplicity does not exceed the algebraic multiplicity:

$$1 \leq m_g(\lambda_k) \leq m_a(\lambda_k)$$

15.The matrix $A$ is diagonalizable if and only if, for every eigenvalue $\lambda_k,$ the geometric multiplicity equals the algebraic multiplicity. In particular, a matrix with $n$ distinct eigenvalues is always diagonalizable, since in that case both multiplicities are equal to one for every eigenvalue.

The diagonalization procedure

16.The practical construction of the matrices $P$ and $D$ follows a well-defined sequence of steps.

18.Once $P$ has been assembled, one verifies that it is invertible and computes $P^{-1},$ thereby completing the factorization $A = P D P^{-1}.$ The computation of $P^{-1}$ relies on the techniques described in the entry on the inverse matrix.

Example 1

19.Consider the following matrix:

$$A = \begin{pmatrix} 3 & 1 \\[6pt] 0 & 2 \end{pmatrix}$$

20.To find the eigenvalues, one computes the determinant of $A - \lambda I$:

$$\det(A - \lambda I) = \det \begin{pmatrix} 3 - \lambda & 1 \\[6pt] 0 & 2 - \lambda \end{pmatrix} = (3 - \lambda)(2 - \lambda)$$

21.The characteristic equation is therefore the following:

$$(3 - \lambda)(2 - \lambda) = 0$$

22.The two roots are $\lambda_1 = 2$ and $\lambda_2 = 3,$ both simple, so the matrix is diagonalizable. For $\lambda_1 = 2,$ one solves the system $(A - 2I)\mathbf{v} = \mathbf{0}$:

$$\begin{pmatrix} 1 & 1 \\[6pt] 0 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\[6pt] v_2 \end{pmatrix} = \begin{pmatrix} 0 \\[6pt] 0 \end{pmatrix}$$

23.The first row gives $v_1 + v_2 = 0,$ so that $v_1 = -v_2.$ Choosing $v_2 = 1,$ one obtains the eigenvector $\mathbf{v}_1 = (-1, 1)^{\mathrm{T}}.$ For $\lambda_2 = 3,$ one solves $(A - 3I)\mathbf{v} = \mathbf{0}$:

$$\begin{pmatrix} 0 & 1 \\[6pt] 0 & -1 \end{pmatrix} \begin{pmatrix} v_1 \\[6pt] v_2 \end{pmatrix} = \begin{pmatrix} 0 \\[6pt] 0 \end{pmatrix}$$

24.Both rows give $v_2 = 0,$ leaving $v_1$ free. Choosing $v_1 = 1,$ one obtains the eigenvector $\mathbf{v}_2 = (1, 0)^{\mathrm{T}}.$ The matrix $P$ is formed by placing these eigenvectors as columns:

$$P = \begin{pmatrix} -1 & 1 \\[6pt] 1 & 0 \end{pmatrix}$$

25.Its inverse is computed directly:

$$P^{-1} = \begin{pmatrix} 0 & 1 \\[6pt] 1 & 1 \end{pmatrix}$$

26.The diagonal matrix collects the eigenvalues in the corresponding order:

$$D = \begin{pmatrix} 2 & 0 \\[6pt] 0 & 3 \end{pmatrix}$$

27.One may verify that $A = P D P^{-1}$ holds by direct multiplication. The matrix $A$ is therefore diagonalizable, and its diagonalization is given by the factorization above, with $P$ and $D$ as constructed.

Example 2

28.Consider a matrix with a repeated eigenvalue. Let $A$ be the following $3 \times 3$ matrix:

$$ A = \begin{pmatrix} 4 & 1 & 0 \\[6pt] 0 & 4 & 0 \\[6pt] 0 & 0 & 2 \end{pmatrix} $$

29.The characteristic polynomial is obtained by expanding the determinant of $A - \lambda I$:

$$ \det(A - \lambda I) = \det \begin{pmatrix} 4 - \lambda & 1 & 0 \\[6pt] 0 & 4 - \lambda & 0 \\[6pt] 0 & 0 & 2 - \lambda \end{pmatrix} = (4 - \lambda)^2 (2 - \lambda) $$

30.The eigenvalues are therefore $\lambda_1 = 4,$ with algebraic multiplicity two, and $\lambda_2 = 2,$ with algebraic multiplicity one. The simple eigenvalue $\lambda_2 = 2$ presents no difficulty. Solving $(A - 2I)\mathbf{v} = \mathbf{0}$ gives the system:

$$ \begin{pmatrix} 2 & 1 & 0 \\[6pt] 0 & 2 & 0 \\[6pt] 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\[6pt] v_2 \\[6pt] v_3 \end{pmatrix} = \begin{pmatrix} 0 \\[6pt] 0 \\[6pt] 0 \end{pmatrix} $$

31.The second row gives $v_2 = 0,$ and substituting into the first row gives $v_1 = 0,$ while $v_3$ remains free. The eigenspace is therefore one-dimensional, spanned by the vector $\mathbf{v}_3 = (0, 0, 1)^{\mathrm{T}}.$ The repeated eigenvalue $\lambda_1 = 4$ is the critical case. Solving $(A - 4I)\mathbf{v} = \mathbf{0}$ gives the system:

$$ \begin{pmatrix} 0 & 1 & 0 \\[6pt] 0 & 0 & 0 \\[6pt] 0 & 0 & -2 \end{pmatrix} \begin{pmatrix} v_1 \\[6pt] v_2 \\[6pt] v_3 \end{pmatrix} = \begin{pmatrix} 0 \\[6pt] 0 \\[6pt] 0 \end{pmatrix} $$

32.The first row gives $v_2 = 0,$ and the third row gives $v_3 = 0,$ while $v_1$ remains free. The eigenspace associated with $\lambda_1 = 4$ is therefore one-dimensional, spanned by $\mathbf{v}_1 = (1, 0, 0)^{\mathrm{T}}.$ Since the geometric multiplicity of $\lambda_1 = 4$ is one, strictly less than its algebraic multiplicity of two, the three eigenvectors found are not sufficient to form a basis of $\mathbb{R}^3,$ and the matrix $A$ is not diagonalizable.

33.

This example illustrates that the presence of a repeated eigenvalue does not by itself prevent diagonalization: what matters is whether the corresponding eigenspace has dimension equal to the algebraic multiplicity. A matrix with a repeated eigenvalue may or may not be diagonalizable depending on the rank of $A - \lambda I.$

When diagonalization fails

34.Not every square matrix is diagonalizable. A matrix fails to be diagonalizable precisely when, for at least one eigenvalue, the geometric multiplicity is strictly less than the algebraic multiplicity. In such cases, the eigenspace associated with that eigenvalue is too small to provide a sufficient number of linearly independent eigenvectors. A standard example is the matrix:

$$B = \begin{pmatrix} 2 & 1 \\[6pt] 0 & 2 \end{pmatrix}$$

35.Its characteristic polynomial is $(2 - \lambda)^2,$ so $\lambda = 2$ is the only eigenvalue with algebraic multiplicity two. Solving $(B - 2I)\mathbf{v} = \mathbf{0}$ yields:

$$\begin{pmatrix} 0 & 1 \\[6pt] 0 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\[6pt] v_2 \end{pmatrix} = \begin{pmatrix} 0 \\[6pt] 0 \end{pmatrix}$$

36.The only condition is $v_2 = 0,$ so the eigenspace is one-dimensional, spanned by $(1, 0)^{\mathrm{T}}.$ Since the geometric multiplicity is one while the algebraic multiplicity is two, it is not possible to assemble a basis of eigenvectors for $\mathbb{R}^2,$ and the matrix $B$ is not diagonalizable.

37.

Matrices of this type are studied within the framework of Jordan normal form, which provides the canonical representation for non-diagonalizable matrices through the introduction of Jordan blocks.

Dependence on the field

38.Whether a matrix is diagonalizable can depend on the field of scalars, because the eigenvalues are the roots of the characteristic polynomial and those roots need not lie in the base field. The rotation matrix

$$A = \begin{pmatrix} 0 & 1 \\[6pt] -1 & 0 \end{pmatrix}$$

39.has characteristic polynomial $\lambda^2 + 1,$ which has no real roots, so over $\mathbb{R}$ the matrix has no eigenvectors and is not diagonalizable. Over $\mathbb{C}$ the same matrix has the two distinct eigenvalues $i$ and $-i,$ so it has two independent eigenvectors and is diagonalizable. A real matrix that is not diagonalizable over $\mathbb{R}$ may therefore become diagonalizable once it is regarded as a complex matrix.

Diagonalization of symmetric matrices

40.One class of matrices is diagonalizable without any computation. Every real symmetric matrix is diagonalizable over $\mathbb{R},$ and its eigenvectors can be chosen to form an orthonormal basis, a result known as the spectral theorem. The eigenvalues of a real symmetric matrix are all real, and even when an eigenvalue is repeated, its geometric multiplicity equals its algebraic multiplicity.

41.Consider the symmetric matrix:

$$ A = \begin{pmatrix} 0 & 1 & 1 \\[6pt] 1 & 0 & 1 \\[6pt] 1 & 1 & 0 \end{pmatrix} $$

42.Its characteristic polynomial is $-(\lambda + 1)^2(\lambda - 2),$ so the eigenvalues are $\lambda_1 = -1$ with algebraic multiplicity two and $\lambda_2 = 2$ with algebraic multiplicity one. Solving $(A + I)\mathbf{v} = \mathbf{0}$ gives the single condition $v_1 + v_2 + v_3 = 0,$ whose solution space is two-dimensional and spanned by $(1, -1, 0)^{\mathrm{T}}$ and $(1, 0, -1)^{\mathrm{T}}.$ The repeated eigenvalue therefore has geometric multiplicity two, matching its algebraic multiplicity, and solving $(A - 2I)\mathbf{v} = \mathbf{0}$ gives the eigenvector $(1, 1, 1)^{\mathrm{T}}.$ The three eigenvectors form a basis of $\mathbb{R}^3,$ so $A$ is diagonalizable, as the spectral theorem guarantees.

Powers of a diagonalizable matrix

43.One of the most immediate applications of diagonalization concerns the computation of integer powers of a matrix. For a diagonalizable matrix $A = P D P^{-1},$ the $k$-th power admits the following compact expression:

$$A^k = P D^k P^{-1}$$

44.This identity follows from the observation that $A^2 = (PDP^{-1})(PDP^{-1}) = PD^2P^{-1},$ and by induction the pattern extends to any positive integer $k.$ Since $D$ is diagonal, its $k$-th power is simply obtained by raising each diagonal entry to the $k$-th power:

$$ D^k = \begin{pmatrix} \lambda_1^k & & \\[6pt] & \ddots & \\[6pt] & & \lambda_n^k \end{pmatrix} $$

45.This observation transforms an otherwise laborious matrix multiplication into a straightforward scalar computation.

The Cayley-Hamilton theorem

46.Every square matrix satisfies its own characteristic equation. If $p_A(\lambda) = \det(A - \lambda I)$ is the characteristic polynomial of a matrix $A$ of order $n,$ then substituting $A$ for $\lambda$ and reading the constant term as a multiple of the identity gives the zero matrix:

$$p_A(A) = O$$

47.This is the Cayley-Hamilton theorem, and it holds for every square matrix, diagonalizable or not. For a matrix of order 2 the characteristic polynomial is $\lambda^2 - \mathrm{tr}(A)\lambda + \det(A),$ so the theorem reads:

$$A^2 - \mathrm{tr}(A) A + \det(A) I = O$$

48.Consider the matrix:

$$A = \begin{pmatrix} 2 & 1 \\[6pt] 1 & 3 \end{pmatrix}$$

49.Here $\mathrm{tr}(A) = 5$ and $\det(A) = 5,$ so the characteristic polynomial is $\lambda^2 - 5\lambda + 5.$ A direct computation confirms that $A^2 - 5A + 5I = O.$

50.The relation gives a way to invert a matrix without computing the adjugate. Rearranging $A^2 - 5A + 5I = O$ gives $A(5I - A) = 5I,$ so:

$$A^{-1} = \frac{1}{5}(5I - A) = \frac{1}{5}\begin{pmatrix} 3 & -1 \\[6pt] -1 & 2 \end{pmatrix}$$

51.The same idea applies in any order, since the theorem expresses $A^n,$ and through it every higher power and the inverse of an invertible matrix, in terms of $I, A, \ldots, A^{n-1}.$