Matrix Diagonalization · aiwiki.page

Matrix diagonalization is a procedure in linear algebra that represents a square matrix as

A=PDP^{-1},

where $P$ is invertible and $D$ is a diagonal matrix, with all off-diagonal entries equal to zero. Equivalently, $P^{-1}AP=D$ . The columns of $P$ are eigenvectors of $A$ , and the corresponding diagonal entries of $D$ are their eigenvalues. A matrix admitting this representation is called diagonalizable. The procedure replaces a potentially coupled transformation with independent scalar multiplications in suitable coordinates. (math.purdue.edu)

Definition and geometric meaning

Let $A$ be an $n\times n$ matrix over a field $F$ . Diagonalizability over $F$ requires both $P$ and $D$ to have entries in $F$ . The defining equation is equivalent to

AP=PD.

Writing $P=[v_1\ \cdots\ v_n]$ and $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$ gives $Av_i=\lambda_i v_i$ for every column. Since $P$ is invertible, these vectors constitute a basis of $F^n$ . (people.math.osu.edu)

Geometrically, a linear map on a finite-dimensional vector space is diagonalizable when it has a basis consisting entirely of eigenvectors. If $x=Py$ , then $Ax=PDy$ : in the eigenvector coordinates $y$ , each coordinate is multiplied by its associated eigenvalue. This is a similarity transformation, representing the same map in a different basis rather than changing the map itself. (math.purdue.edu)

Criteria for diagonalizability

The central criterion is that $A$ must possess $n$ linearly independent eigenvectors. For an eigenvalue $\lambda$ , its eigenspace is

E_\lambda=\ker(A-\lambda I),

where $I$ denotes the identity matrix and the kernel is the null space. Its dimension is the eigenvalue’s geometric multiplicity. Its algebraic multiplicity is its multiplicity as a root of the characteristic polynomial

\chi_A(t)=\det(tI-A).

Geometric multiplicity never exceeds algebraic multiplicity. (ximera.osu.edu)

A matrix is diagonalizable over $F$ exactly when its characteristic polynomial splits into linear factors over $F$ and every eigenvalue has equal geometric and algebraic multiplicities. Consequently, $n$ distinct eigenvalues in $F$ guarantee diagonalizability, but distinctness is not necessary: the identity matrix has only one eigenvalue and is already diagonal. Conversely, invertibility alone does not guarantee diagonalizability. (textbooks.math.gatech.edu)

The underlying field matters. For example,

R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}

has eigenvalues $i$ and $-i$ . It is therefore diagonalizable over the complex numbers, but not over the real numbers, where it has no eigenvectors. Even over the complex numbers, however, a matrix may lack enough independent eigenvectors. (people.math.osu.edu)

Construction and examples

In exact arithmetic, diagonalization can be constructed by finding the roots of $\chi_A(t)$ , solving $(A-\lambda I)v=0$ for each eigenvalue, and selecting a combined eigenvector basis. These homogeneous systems of linear equations can be solved using Gaussian elimination. The selected vectors become the columns of $P$ , and their eigenvalues enter $D$ in the same order. If their total number is less than $n$ , diagonalization is impossible. (textbooks.math.gatech.edu)

For a directly verifiable example, take

A=\begin{pmatrix}2&1\\0&3\end{pmatrix},\qquad P=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad D=\begin{pmatrix}2&0\\0&3\end{pmatrix}.

The columns $(1,0)^T$ and $(1,1)^T$ have eigenvalues $2$ and $3$ , respectively, and multiplication confirms $AP=PD$ .

By contrast,

J=\begin{pmatrix}1&1\\0&1\end{pmatrix}

has characteristic polynomial $(t-1)^2$ , but its eigenspace consists only of vectors $(a,0)^T$ . Its geometric multiplicity is one rather than two, so it is not diagonalizable. Matrices lacking an eigenvector basis are called defective. (textbooks.math.gatech.edu)

Orthogonal and unitary diagonalization

The spectral theorem gives stronger results for important matrix classes. Every real symmetric matrix has an orthonormal eigenvector basis and can be written

A=QDQ^T,

where $Q$ is an orthogonal matrix and $D$ is real diagonal. The transpose $Q^T$ therefore serves as its inverse. (ocw.mit.edu)

Over the complex numbers, a matrix is diagonalizable by a unitary matrix exactly when it is normal, meaning $A^*A=AA^*$ , where $A^*$ is the conjugate transpose. Thus $A=UDU^*$ . Hermitian matrices, satisfying $A=A^*$ , are normal and have real eigenvalues. Ordinary diagonalizability is weaker: it does not require orthogonal eigenvectors. (ocw.mit.edu)

Applications and numerical considerations

Diagonalization simplifies repeated matrix multiplication:

A^k=PD^kP^{-1},\qquad D^k=\operatorname{diag}(\lambda_1^k,\ldots,\lambda_n^k)

for nonnegative integers $k$ . This converts matrix powers into scalar powers. (textbooks.math.gatech.edu)

It also simplifies a matrix exponential:

e^{tA}=P\operatorname{diag}(e^{t\lambda_1},\ldots,e^{t\lambda_n})P^{-1}.

For the constant-coefficient differential equation $x'(t)=Ax(t)$ , changing variables to $x=Py$ yields independent equations $y_i'=\lambda_i y_i$ . (webhome.auburn.edu) In principal component analysis, diagonalizing a covariance matrix identifies orthogonal directions of variation; its eigenvalues give the corresponding variances. (cs357.cs.illinois.edu)

In numerical linear algebra, theoretical diagonalizability does not ensure a reliable computation. Nearly dependent eigenvectors can make $P$ ill-conditioned, increasing sensitivity to perturbations. General-purpose eigensolvers commonly compute a Schur decomposition first, using unitary or orthogonal transformations, and then recover eigenvectors when needed. Unlike diagonalization, complex Schur decomposition exists for every square complex matrix, with an upper-triangular factor rather than necessarily a diagonal one. (netlib.org)