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A matrix is defective when an eigenvalue has geometric multiplicity smaller than its algebraic multiplicity. Over a field in which the characteristic polynomial splits, this is precisely the obstruction to an eigenbasis: at least one Jordan block has size greater than one, and generalized eigenvectors are needed to complete a basis. The shear mapping in diagonalizable matrix is the simplest example.

For a matrix over a field $F$, diagonalizability requires an eigenbasis with entries in $F$ and eigenvalues in $F$. Passing to a larger field can supply missing roots of the characteristic polynomial. For example,is not a diagonalizable matrix over the rational numbers, since the quadratic factor has no rational root. It is a diagonalizable matrix over the real numbers and complex numbers, since its three eigenvalues are distinct. More generally, diagonalizability is equivalent to the minimal polynomial splitting into distinct linear factors over the specified field.

For an integer matrix, reduction modulo a prime number is a different operation from extension of its coefficient field, and can merge eigenvalues. In the same example modulo $5$, ${\chi }_A(t)=(t-1)(t-3{)}^2$, while $\ker (A-3I)$ is spanned by $(1,3,4{)}^T$. Thus the geometric multiplicity of $3$ is one and its algebraic multiplicity is two, preventing diagonalizability even though all eigenvalues belong to the finite field.

A finite-dimensional linear operator is diagonalizable over a specified field when it has an eigenbasis. Equivalently, its matrix $A$ admitswhere $P$ is an invertible matrix whose columns are eigenvectors, and $D$ is a diagonal matrix of the corresponding eigenvalues. This change of basis reveals a linear map as independent one-dimensional scalings. The matrix may look coupled in the original coordinates, but each coordinate in the eigenbasis evolves on its own.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ computed in eigenvector, takeThen $AP=PD$, hence $P^{-1}AP=D$. The orthonormal basis version is $Q=P/\sqrt{2}$, an orthogonal matrix satisfying $Q^{T}AQ=D$. Every real symmetric matrix has an orthonormal eigenbasis by the spectral theorem; a general diagonalizable matrix need not have orthogonal eigenvectors.

The immediate payoff is repeated application: adjacent factors $P^{-1}P$ cancel, giving $A^n=P{D}^n{P}^{-1}$. In this example,The component in $E_3$ grows while the component in $E_1$ stays fixed. After division by its Euclidean norm, this particular sequence approaches $(1,1{)}^T/\sqrt{2}$. This is the mechanism behind the power method; an initial vector entirely in $E_1$ would stay there and never find the dominant eigenvalue $3$.

Two different obstructions deserve comparison. The planar rotation $R$ in eigenvector is diagonalizable over $\mathbb{C}$ but not over $\mathbb{R}$, because its eigenvalues are $i$ and $-i$. By contrast, the shear mappinghas only $\lambda =1$, of algebraic multiplicity two, and $E_1=\mathrm{span}\{(1,0{)}^T\}$, of geometric multiplicity one. There is no second independent eigenvector, even over $\mathbb{C}$. This is a defective matrix, already a two-dimensional Jordan block. Its powers areThus even though both eigenvalues are one, repeated application can produce growth. Eigenvalues alone miss the coupling recorded by a Jordan block. A repeated eigenvalue is harmless when it has enough independent eigenvectors, as $I$ demonstrates. More generally, a matrix is diagonalizable exactly when its characteristic polynomial splits over the chosen field and every eigenvalue has equal geometric multiplicity and algebraic multiplicity. The proof outline is to join bases of the distinct eigenspaces; their direct sum fills the space exactly under these conditions.

A physical motivation is the separation of normal modes. Consider two equal masses $m$ between three identical springs of stiffness $k$, with the outer ends fixed. By Hooke's law and Newton's second law, the displacement vector $q=(q_1,q_2{)}^T$ obeysThe eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $1$ and $3$: the masses move together in the first normal mode and oppositely in the second. Writing $q=a(t)(1,1{)}^T+b(t)(1,-1{)}^T$ separates the coupled equations into two harmonic oscillators,Their angular frequencies are $\sqrt{k/m}$ and $\sqrt{3k/m}$; the eigenvalues give squared angular frequencies after multiplication by $k/m$. This is the beauty of diagonalization: finding the right directions turns interacting motions into independent patterns. For a physical treatment, see Yen-Jie Lee's "Coupled Oscillators, Normal Modes" (MIT 8.03SC, 2016), and the local article normal modes of two equal masses between three springs.^[ref]

On $0<x<L_x$, $0<y<L_y$ with positive side lengths and zero Dirichlet boundary conditions, separation of variables for $-\Delta u=\lambda u$ giveswith the displayed eigenvalues. The one-dimensional endpoint conditions exclude constant and hyperbolic factors and force the positive integer sine frequencies. The product Fourier sine series basis is complete in the square-integrable functions on the rectangle. Consequently an eigenfunction with prescribed $\lambda$ is a linear combination of exactly those modes with ${\lambda }_{mn}=\lambda$; if none match, only the zero solution exists. Distinct pairs with the same sum give an eigenvalue with geometric multiplicity greater than one. On a square of side $\pi$, for instance, $\lambda =25$ has precisely the pairs $(3,4)$ and $(4,3)$, so its eigenspace has dimension two.

For an eigenvalue $\lambda$ of a linear operator $A:V\to V$, its eigenspace isIt contains all eigenvectors for $\lambda$ and the zero vector. Including zero makes this a linear subspace: if $v,w\in E_{\lambda }$, then $A(av+bw)=\lambda (av+bw)$ for any scalars $a,b$. Every nonzero vector in this linear subspace has the same scaling under $A$. Its dimension is the geometric multiplicity of the eigenvalue. If $\lambda$ is not an eigenvalue, the shifted kernel is only $\{0\}$.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ from eigenvector,Each eigenspace is a line. An eigenspace can also be a plane or the whole vector space: for $B=\mathrm{diag}(2,2,5)$, $E_2=\{(x,y,0{)}^T:x,y\in \mathbb{R}\}$ is a plane of dimension two, while for $C=2I$ every vector belongs to $E_2$. A repeated eigenvalue therefore need not prevent diagonalizability.

The useful unit of information is the whole eigenspace, rather than a particular choice of eigenvectors inside it. On $E_{\lambda }$, $A$ is simply multiplication by $\lambda$. Distinct eigenspaces fit together as a direct sum, and a finite-dimensional linear operator is diagonalizable exactly when they fill the whole vector space. For the example above, every vector splits uniquely asThis decomposition makes the independent scalings visible without multiplying out the original matrix again.

Under $B=S^{-1}AS$, an eigenvector in old coordinates becomes $S^{-1}v$ in new coordinates. Indeed,Since $S^{-1}$ is an invertible matrix, the new coordinate vector is nonzero. The inverse conversion $S$ proves that the two eigenspaces have the same dimension, hence the same geometric multiplicity.

For $A=\mathrm{diag}(2,5)$, the eigenvector $e_2=(0,1{)}^T$ has eigenvalue $5$. In the oblique basis of the opening example, its coordinate vector isThe tuple changes from $(0,1)$ to $(-1,1)$, but reconstruction gives $-b_1+b_2=e_2$: the physical vertical direction is unchanged. Calling the eigenvectors unchanged is accurate only for the underlying vectors, not for their coordinate vectors.

In the finite field ${\mathbb{F}}_5=\mathbb{Z}/5\mathbb{Z}$, the characteristic polynomial becomesbecause $(\lambda -3{)}^2={\lambda }^2-\lambda -1$ modulo $5$. Splitting into linear factors does not by itself guarantee diagonalizability: the repeated eigenvalue needs enough eigenvectors.

Solving $(A-3I)(x,y,z{)}^T=0$ givesso $y=3x$ and $z=4x$. Hencehas dimension one, although $3$ has algebraic multiplicity two. The geometric multiplicity is too small, and $A$ is not diagonalizable over ${\mathbb{F}}_5$. This illustrates diagonalizability depends on the field: characteristic zero gave three distinct eigenvalues, while reduction modulo $5$ merges two of them.

If $\varphi =T^{\prime }\psi$ and $v\in \ker T$, then $\varphi (v)=\psi (Tv)=0$. Thus $\mathrm{im}{T}^{\prime }\subseteq (\ker T{)}^{\circ }$, using the annihilator of a vector subspace.

Conversely, let $\varphi$ vanish on the kernel of a linear map $\ker T$. Define a linear functional on $\mathrm{im}T$ by ${\psi }_0(Tv)=\varphi (v)$. This is well-defined: $Tv=Tw$ implies $v-w\in \ker T$, so $\varphi (v)=\varphi (w)$. Extend ${\psi }_0$ to a linear functional $\psi$ on $W$, using the extension argument from part (i). Then $T^{\prime }\psi =\varphi$, provingThe dimension formula from part (i) and the rank-nullity theorem now giveTo deduce the eigenvalue statement, apply this equality of rank of a linear map to $T-\lambda I$. Its dual map is $T^{\prime }-\lambda I$, so their kernel of a linear map dimensions agree. Thus the same $\lambda$ are eigenvalues, with the same geometric multiplicities.

For algebraic multiplicities as well, part (a)(iii) givesso equality of rank applies to every power. HenceAfter extending to a field where the characteristic polynomials split, take $r\ge \dim V$. These kernel of a linear map spaces are the generalized eigenspaces; the primary decomposition theorem for linear maps identifies their dimensions with the algebraic multiplicities. The same proof works over the original field for every eigenvalue in it. We obtainIn fact the argument also includes the zero eigenvalue.

Let $Hv=\lambda v$ with $v\ne 0$ and $H^{\ast }=H$. Then $v^{\ast }Hv=\lambda v^{\ast }v$, while taking the complex conjugate gives $v^{\ast }Hv=\overline{\lambda }{v}^{\ast }v$. Since $v^{\ast }v>0$, every eigenvalue of a Hermitian matrix is real.

If $Hv=\lambda v$ and $Hw=\mu w$, thenDistinct eigenvalues therefore force $v^{\ast }w=0$: their eigenvectors are orthogonal in the Hermitian inner product. A degenerate eigenvalue occurs more than once in the spectrum, counted with multiplicity. For a Hermitian matrix, its algebraic multiplicity equals its geometric multiplicity, so degeneracy gives an eigenspace of dimension greater than one; an orthonormal basis can be chosen within it.

For $t\ne 0$, factor the matrix asThe determinant of a scalar multiple of an $n\times n$ matrix gains the scalar's $n$th power. ThusThe right side extends to a polynomial at $t=0$: multiplying ${\chi }_A(t^{-1})$ by $t^n$ removes all negative powers. Its value there is $(\det A{)}^{-1}$, since the leading coefficient of ${\chi }_A(s)=\det (A-sI)$ is $(-1{)}^n$. This equals ${\chi }_{A^{-1}}(0)$, so the identity holds at zero with that interpretation.

For the relation between eigenvalues and the characteristic polynomial,The first equivalence is the definition of an eigenvector; the others use the kernel of a linear map and the criterion for an invertible matrix to have nonzero determinant. Since ${\chi }_A$ has degree $n$, the fundamental theorem of algebra gives $n$ roots over $\mathbb{C}$ counted with multiplicity. The multiplicity of a root is its algebraic multiplicity, and the eigenspace is exactly $\ker (A-\lambda I)$; its dimension is the geometric multiplicity, which need not equal the algebraic multiplicity.

Finally, factor $p(s)=s^4-\mu$ over $\mathbb{C}$ as $p(s)={\prod }_{j=1}^4(s-{\lambda }_j)$, including repeated roots when $\mu =0$. Substituting $A$ into this polynomial is valid because the resulting factors commute. If $\mu$ is an eigenvalue of $A^4$, thenAt least one factor must be zero, so its ${\lambda }_j$ is an eigenvalue of $A$ and satisfies ${\lambda }_j^4=\mu$. Thus the reverse direction of this instance of the spectral mapping theorem isThis argument also covers $\mu =0$, when the conclusion is that $A$ is singular and has eigenvalue zero.

Expansion along the first column gives the characteristic polynomialThus the eigenvalue $2$ has algebraic multiplicity two. Solving $(M-2I)(x,y,z{)}^T=0$ gives $y=0$ and then $z=0$, leaving $x$ arbitrary. Its eigenspace is $\mathrm{span}\{(1,0,0{)}^T\}$ and has geometric multiplicity one. A diagonalisable matrix needs an eigenvector basis, so $M$ is not diagonalisable.

If $B=S^{-1}AS$ with $S$ invertible, then $tI-B=S^{-1}(tI-A)S$. Multiplicativity of the determinant givesThe roots of the characteristic polynomial, counted with their multiplicities, are the eigenvalues with their algebraic multiplicities. Thus similar matrices have the same eigenvalues with the same algebraic multiplicities.

The converse is false. The matricesboth have characteristic polynomial $(t-1{)}^2$, but every matrix similar to $I_2$ is $S^{-1}{I}_{2}S=I_2$, whereas $J\ne I_2$. Alternatively, their geometric multiplicities for $1$ differ.

The Jordan normal form theorem says that every complex square matrix is similar to a direct sum of Jordan blocks, uniquely up to their order. For an eigenvalue $\lambda$, if its blocks have sizes $n_1,\dots ,n_r$, then its algebraic multiplicity and geometric multiplicity areand its contribution to the minimal polynomial is $(t-\lambda {)}^{{\max }_j{n}_j}$. Thuswhere $s_{\lambda }$ is the largest $\lambda$-block size.

For the given matrix,Moreover,HenceThe eigenvalue $-1$ therefore has one Jordan block of size two, so

The decomposition into $\mathrm{span}\{u\}$ and the zero-sum vector subspace gives characteristic polynomial $t^{n-1}(t-n)$. For $n\ge 2$, $n$ has algebraic multiplicity and geometric multiplicity one; zero has both multiplicities $n-1$. At $n=1$, only the eigenvalue 1 occurs, with both multiplicities one. Referring to a multiplicity zero for the absent root zero is merely a bookkeeping convention, not an assertion that it is an eigenvalue.

For $u=(1,\dots ,1{)}^T$, $J_n=u{u}^T$ maps $v$ to $({\sum }_i{v}_i)u$. Therefore $u$ is an eigenvector with eigenvalue $n$, while the zero-sum vector subspace is its kernel. For $n\ge 2$ a basis of the kernel is $e_1-e_n,\dots ,e_{n-1}-e_n$. These vectors and $u$ form an eigenbasis, proving diagonalization and characteristic polynomial $t^{n-1}(t-n)$. Both algebraic multiplicity and geometric multiplicity are one at $n$ and $n-1$ at zero. At $n=1$ the sole eigenvalue is 1; the zero-sum space has dimension zero and contributes no additional root.