A linear map can be simple in the wrong coordinates and opaque in the coordinates we first choose. A diagonal matrix such as $A=\operatorname{diag}(2,5)$ acts by stretching one coordinate axis by $2$ and the other by $5$. The same transformation written in a skew basis may have off-diagonal entries, mixed coordinates, and no visible axes of action. Eigenvalues answer the coordinate-free question hiding behind the matrix: which directions does the transformation preserve, and by what factors does it scale them? The first surprise is that a linear map need not send most vectors to multiples of themselves. If $A$ rotates the real plane through $90$ degrees, then no nonzero real vector stays on its own line. The vector is not merely moved; its direction is changed. Eigenvalues detect the exceptional invariant lines when they exist, and the failure of their existence tells us something equally important about the geometry of the operator and the field over which we are working. [example: A Rotation with No Real Eigenvalue] Let $V=\mathbb{R}^2$ and let $R:V\to V$ be represented in the standard basis by \begin{align*} R=\begin{pmatrix}0&-1\cr 1&0\end{pmatrix}. \end{align*} For $v=(x,y)$, multiplication by $R$ gives \begin{align*} R(x,y)=(-y,x). \end{align*} We show that no nonzero real vector $v$ satisfies $Rv=\lambda v$ for any real scalar $\lambda$. If such a vector existed, then \begin{align*} (-y,x)=\lambda(x,y)=(\lambda x,\lambda y). \end{align*} Equality of coordinates gives \begin{align*} -y=\lambda x \end{align*} and \begin{align*} x=\lambda y. \end{align*} Substituting $y=-\lambda x$ from the first equation into the second gives \begin{align*} x=\lambda(-\lambda x)=-\lambda^2x. \end{align*} Hence \begin{align*} (1+\lambda^2)x=0. \end{align*} Also, substituting $x=\lambda y$ into the first equation gives \begin{align*} -y=\lambda(\lambda y)=\lambda^2y, \end{align*} so \begin{align*} (1+\lambda^2)y=0. \end{align*} For real $\lambda$, we have $1+\lambda^2>0$, so both equations force $x=0$ and $y=0$. This contradicts $v\neq 0$, and therefore $R$ has no real eigenvalue. Over $\mathbb{C}$, the obstruction disappears because $1+\lambda^2=0$ has the two solutions $\lambda=i$ and $\lambda=-i$. For $\lambda=i$, the equation $-y=ix$ is satisfied by $x=1$ and $y=-i$, giving the eigenvector $(1,-i)$. For $\lambda=-i$, the equation $-y=-ix$ is satisfied by $x=1$ and $y=i$, giving the eigenvector $(1,i)$. Thus the same matrix has no eigenvalues over $\mathbb{R}$ but has eigenvalues $i$ and $-i$ over $\mathbb{C}$, so the scalar field is part of the eigenvalue question. [/example] The example shows two themes that will run through the whole page. First, eigenvalues are not merely numbers attached to a matrix; they are scalars attached to invariant directions of a linear operator. Second, existence and factorisation depend on the field. Over $\mathbb{C}$ every complex matrix has at least one eigenvalue, while over $\mathbb{R}$ rotations already show that a real operator can have none. ## Eigenvalues and Invariant Lines The central question is not how a matrix happens to look in one basis, but whether the transformation preserves any one-dimensional directions. The relevant maps are endomorphisms: linear maps from a vector space back to itself. That same-domain condition is essential because the equation below compares the transformed vector with the original vector. If a nonzero vector is sent to a scalar multiple of itself, the whole line it spans is invariant, and the scalar records the stretching, shrinking, sign reversal, or collapse along that line. [definition: Eigenvalue] Let $F$ be a field, let $V$ be a vector space over $F$, and let $T: V \to V$ be a linear map from $V$ to itself. A scalar $\lambda \in F$ is an eigenvalue of $T$ if there exists a nonzero vector $v \in V$ such that \begin{align*} T(v) = \lambda v. \end{align*} [/definition] The definition identifies the scalar, not the whole geometry around it. To use eigenvalues as structural information, we also need names for the operator being studied, the vectors that witness the scalar equation, the subspace of all such witnesses, and the set of all eigenvalues. ## Operators, Eigenvectors, and Spectra Every eigenvalue problem is a problem about a transformation that can be applied repeatedly to the same kind of vector. Rectangular matrices may encode useful linear maps, but they do not have eigenvalues in this sense because their outputs live in a different space from their inputs. The square-matrix situation is better understood intrinsically as a linear operator. [definition: Linear Operator] Let $F$ be a field and let $V$ be a vector space over $F$. A linear operator on $V$ is a linear map \begin{align*} T: V \to V. \end{align*} [/definition] This terminology separates the coordinate-free transformation from any particular matrix representing it. Once an eigenvalue is known, the next object to isolate is the vector that witnesses it. The zero vector satisfies $T(0)=\lambda 0$ for every scalar $\lambda$, so it cannot carry information about an invariant direction; the useful witnesses must be nonzero. [definition: Eigenvector] Let $F$ be a field, let $V$ be a vector space over $F$, let $T: V \to V$ be a linear operator, and let $\lambda \in F$ be an eigenvalue of $T$. A nonzero vector $v \in V$ is an eigenvector of $T$ with eigenvalue $\lambda$ if \begin{align*} T(v) = \lambda v. \end{align*} [/definition] A single eigenvector is only one representative of an invariant line, and several independent invariant directions can share the same scaling factor. To solve systems and count dimensions, we need all vectors associated to a fixed scalar collected into one linear object. [definition: Eigenspace] Let $F$ be a field, let $V$ be a vector space over $F$, let $T: V \to V$ be a linear operator, and let $\lambda \in F$. The eigenspace of $T$ associated to $\lambda$ is \begin{align*} E_\lambda(T) = \ker(T - \lambda I), \end{align*} where $I: V \to V$ is the identity operator. [/definition] If $\lambda$ is an eigenvalue, then $E_\lambda(T)$ contains nonzero vectors. If $\lambda$ is not an eigenvalue, then $E_\lambda(T)=\{0\}$. When studying one operator, repeatedly listing individual scalars is awkward: the real object of interest is the set of all scalars for which the kernel of $T-\lambda I$ becomes nonzero. Naming that set also prevents confusion later, where spectral theory contains scalars that are not eigenvalues. The next object packages this scalar-level information separately from the eigenspaces themselves. It answers the question of which scalars occur before asking how large the corresponding eigenspaces are. [definition: Point Spectrum] Let $F$ be a field, let $V$ be a vector space over $F$, and let $T: V \to V$ be a linear operator. The point spectrum of $T$ over $F$ is \begin{align*} \sigma_p(T) = \{\lambda \in F : \lambda \text{ is an eigenvalue of } T\}. \end{align*} [/definition] For finite-dimensional vector spaces, many authors call $\sigma_p(T)$ simply the spectrum when no infinite-dimensional issues are present. In functional analysis, the spectrum also includes scalars for which $T-\lambda I$ fails to have a bounded inverse, so the point spectrum is the eigenvalue part of a wider story. ## Characteristic Polynomials and Existence ### From Determinants to Roots The definition is geometric, but computation needs an equation. If $V$ is finite-dimensional and a basis is chosen, $T(v)=\lambda v$ becomes $(A-\lambda I)x=0$. A nonzero solution exists exactly when the matrix $A-\lambda I$ is singular, which leads to a polynomial whose roots are the eigenvalues. Here a matrix with entries in a field $F$ is written as an element of $F^{n\times n}$, $I_n$ denotes the $n\times n$ identity matrix, and $X$ is a formal variable rather than a particular scalar. [definition: Characteristic Polynomial] Let $F$ be a field, let $V$ be an $n$-dimensional vector space over $F$, and let $T: V \to V$ be a linear operator. If $A \in F^{n \times n}$ is the matrix of $T$ in an ordered basis of $V$, the characteristic polynomial of $T$ is \begin{align*} \chi_T(X) = \det(XI_n - A) \in F[X]. \end{align*} The notation $F[X]$ means the ring of polynomials in the formal variable $X$ with coefficients in $F$. [/definition] The definition mentions a basis, so there is an immediate danger: perhaps choosing a different basis changes the polynomial and therefore changes the computed eigenvalues. Unless this ambiguity is removed, the characteristic polynomial would be only a coordinate artifact rather than an invariant of the transformation. The obstruction is that one linear operator has many matrix representatives. Changing basis replaces $A$ by a similar matrix, so the determinant formula must give the same polynomial for all matrices that represent the same operator. In the invariant statement below, $\mathrm{Mat}_n(F)$ denotes the set of $n\times n$ matrices over $F$, $\mathrm{End}(V)$ denotes the vector space of linear maps from $V$ to itself, and notation such as $k[t]$ similarly means a polynomial ring over the field $k$. With that notation in place, the result resolves the ambiguity by showing that the characteristic polynomial survives change of coordinates. [quotetheorem:402] This theorem justifies the usual computational habit of choosing whatever basis makes $T$ easiest to write. Once the polynomial is well-defined, the key question is whether its roots exactly match the scalars that produce nonzero kernels of $T-\lambda I$. [quotetheorem:7911] This equivalence is the main bridge between geometry and algebra. A preserved direction becomes a vanishing determinant, and a determinant becomes a polynomial equation. [example: Eigenvalues of a Triangular Matrix] Let $A=\begin{pmatrix}3&7&-1\cr 0&3&4\cr 0&0&-2\end{pmatrix} \in \mathbb{R}^{3 \times 3}$. We compute its characteristic polynomial from $\chi_A(X)=\det(XI-A)$. First, \begin{align*} XI-A=\begin{pmatrix}X-3&-7&1\cr 0&X-3&-4\cr 0&0&X+2\end{pmatrix}. \end{align*} Expanding the determinant along the first row gives \begin{align*} \det(XI-A)=(X-3)\det\begin{pmatrix}X-3&-4\cr 0&X+2\end{pmatrix}-(-7)\det\begin{pmatrix}0&-4\cr 0&X+2\end{pmatrix}+1\det\begin{pmatrix}0&X-3\cr 0&0\end{pmatrix}. \end{align*} The three $2 \times 2$ determinants are \begin{align*} \det\begin{pmatrix}X-3&-4\cr 0&X+2\end{pmatrix}=(X-3)(X+2)-(-4)0=(X-3)(X+2), \end{align*} \begin{align*} \det\begin{pmatrix}0&-4\cr 0&X+2\end{pmatrix}=0(X+2)-(-4)0=0, \end{align*} and \begin{align*} \det\begin{pmatrix}0&X-3\cr 0&0\end{pmatrix}=0\cdot 0-(X-3)0=0. \end{align*} Therefore \begin{align*} \chi_A(X)=(X-3)(X-3)(X+2)+7\cdot 0+0=(X-3)^2(X+2). \end{align*} The characteristic equation is \begin{align*} (X-3)^2(X+2)=0, \end{align*} so the eigenvalues over $\mathbb{R}$ are $3$ and $-2$, with algebraic multiplicities $2$ and $1$ respectively. The off-diagonal entries do not change these roots, but they do change the eigenspaces. For $\lambda=3$, \begin{align*} A-3I=\begin{pmatrix}0&7&-1\cr 0&0&4\cr 0&0&-5\end{pmatrix}, \end{align*} so $(A-3I)(x,y,z)=0$ gives $7y-z=0$, $4z=0$, and $-5z=0$. Hence $z=0$, then $y=0$, while $x$ is free, so \begin{align*} E_3(A)=\operatorname{span}\{(1,0,0)\}. \end{align*} For $\lambda=-2$, \begin{align*} A+2I=\begin{pmatrix}5&7&-1\cr 0&5&4\cr 0&0&0\end{pmatrix}. \end{align*} The equation $(A+2I)(x,y,z)=0$ gives $5y+4z=0$, hence $y=-\frac{4}{5}z$, and then \begin{align*} 5x+7\left(-\frac{4}{5}z\right)-z=0. \end{align*} Thus \begin{align*} 5x-\frac{28}{5}z-z=5x-\frac{33}{5}z=0, \end{align*} so $x=\frac{33}{25}z$. Taking $z=25$ gives an eigenvector $(33,-20,25)$, and \begin{align*} E_{-2}(A)=\operatorname{span}\{(33,-20,25)\}. \end{align*} Thus the diagonal entries determine the eigenvalues of this triangular matrix, while the off-diagonal entries are still visible in the eigenvectors. [/example] ### Field Dependence The field still matters. A polynomial may fail to split over the field where the vector space was first defined, and then the determinant equation has fewer roots available inside that field. To know when at least one eigenvalue must exist, we need a field where nonconstant polynomials have roots. [quotetheorem:8486] The theorem is not a statement about real vector spaces. A real matrix may have no real eigenvalues, but after extending scalars to $\mathbb{C}$ its characteristic polynomial splits into linear factors and the complex eigenvalues appear. ## Eigenspaces and Multiplicity ### Counting Roots and Directions Finding the eigenvalues is only the first step. The real structure of an operator is often controlled by the dimensions of its eigenspaces: how many independent directions are preserved for each scalar, and whether these directions together span the whole space. The determinant equation records roots with repetition. The eigenspace records independent eigenvectors. To compare these two kinds of counting, we first need a name for the multiplicity that comes from the polynomial itself. [definition: Algebraic Multiplicity] Let $F$ be a field, let $V$ be a finite-dimensional vector space over $F$, let $T: V \to V$ be a linear operator, and let $\lambda \in F$ be a root of $\chi_T(X)$. The algebraic multiplicity of $\lambda$ is the largest integer $m \ge 1$ such that \begin{align*} (X-\lambda)^m \text{ divides } \chi_T(X) \text{ in } F[X]. \end{align*} [/definition] Algebraic multiplicity comes from the polynomial before solving for eigenvectors. That count alone does not say how many directions the operator actually preserves. A repeated root may give several independent eigenvectors, or it may give only one; to measure the usable supply of eigendirections, we count the dimension of the corresponding eigenspace. [definition: Geometric Multiplicity] Let $F$ be a field, let $V$ be a finite-dimensional vector space over $F$, let $T: V \to V$ be a linear operator, and let $\lambda \in F$ be an eigenvalue of $T$. The geometric multiplicity of $\lambda$ is \begin{align*} \dim E_\lambda(T). \end{align*} [/definition] Geometric multiplicity measures the number of independent eigenvector directions associated to $\lambda$. It is always at least $1$ for an eigenvalue, but it cannot grow without regard to the polynomial count. The obstruction appears when a root is repeated: the characteristic polynomial records how often the scalar occurs algebraically, while the eigenspace records how many independent directions are actually available. These two counts need a general comparison, because diagonalization and basis-building arguments fail if one assumes that repeated roots automatically supply enough eigenvectors. The bound below gives the required limit on the size of an eigenspace. The theorem also mentions the minimal polynomial. For a linear operator $\alpha$, its minimal polynomial $M_\alpha$ is the monic polynomial of least degree such that $M_\alpha(\alpha)=0$. If $\lambda$ is a root of $M_\alpha$, the number $c_\lambda$ denotes its multiplicity as a root of $M_\alpha$; it measures how long the generalized-eigenvector chains at $\lambda$ may need to be. [quotetheorem:409] The inequality explains a common source of disappointment. A repeated root of the characteristic polynomial may or may not provide several independent eigenvectors. [example: A Repeated Eigenvalue with Too Few Eigenvectors] Let $A=\begin{pmatrix}1&1\cr 0&1\end{pmatrix} \in \mathbb{R}^{2 \times 2}$. We compute the characteristic polynomial from $\chi_A(X)=\det(XI-A)$. Since \begin{align*} XI-A=\begin{pmatrix}X&0\cr 0&X\end{pmatrix}-\begin{pmatrix}1&1\cr 0&1\end{pmatrix}=\begin{pmatrix}X-1&-1\cr 0&X-1\end{pmatrix}, \end{align*} the $2 \times 2$ determinant formula gives \begin{align*} \chi_A(X)=\det\begin{pmatrix}X-1&-1\cr 0&X-1\end{pmatrix}=(X-1)(X-1)-(-1)0=(X-1)^2. \end{align*} Thus $1$ is a root of $\chi_A(X)$ with algebraic multiplicity $2$. Now compute the eigenspace for $\lambda=1$. We have \begin{align*} A-I=\begin{pmatrix}1&1\cr 0&1\end{pmatrix}-\begin{pmatrix}1&0\cr 0&1\end{pmatrix}=\begin{pmatrix}0&1\cr 0&0\end{pmatrix}. \end{align*} For a vector $(x,y)\in \mathbb{R}^2$, \begin{align*} (A-I)(x,y)=\begin{pmatrix}0&1\cr 0&0\end{pmatrix}\begin{pmatrix}x\cr y\end{pmatrix}=\begin{pmatrix}y\cr 0\end{pmatrix}. \end{align*} Therefore $(A-I)(x,y)=0$ is equivalent to \begin{align*} \begin{pmatrix}y\cr 0\end{pmatrix}=\begin{pmatrix}0\cr 0\end{pmatrix}, \end{align*} so $y=0$ and $x$ is free. Hence every vector in the eigenspace has the form \begin{align*} (x,0)=x(1,0), \end{align*} and therefore \begin{align*} E_1(A)=\operatorname{span}\{(1,0)\}. \end{align*} So the algebraic multiplicity of $1$ is $2$, but its geometric multiplicity is $\dim E_1(A)=1$; the repeated root supplies only one independent eigenvector direction. [/example] ### Independence of Distinct Eigenvalues Repeated eigenvalues can be subtle, but distinct eigenvalues behave with remarkable independence. The following theorem is needed because if different scalars preserve different directions, those directions cannot collapse into a nontrivial linear relation; this turns a list of eigenvalues into a supply of basis vectors. [quotetheorem:8488] This theorem is often the quickest way to build a basis. If an $n$-dimensional operator has $n$ distinct eigenvalues over the field, the corresponding eigenvectors already form a basis. ## Diagonalisation The most useful operators are those for which eigenvectors give coordinates. In such a basis the operator stops mixing coordinates and becomes a diagonal matrix. This is the best possible situation for computing powers, solving recurrences, and understanding long-term behaviour. [definition: Diagonalizable Operator] Let $F$ be a field, let $V$ be a finite-dimensional vector space over $F$, and let $T: V \to V$ be a linear operator. The operator $T$ is diagonalizable over $F$ if there exists an ordered basis of $V$ consisting entirely of eigenvectors of $T$. [/definition] Diagonalisation is not a property of an isolated matrix display; it is a property of the operator over the chosen field. To decide whether diagonalisation occurs, we need a theorem that compares the polynomial multiplicities with the dimensions of the corresponding eigenspaces. [quotetheorem:8491] The criterion says that diagonalisation fails in exactly two ways: the characteristic polynomial may not split over the field, or an eigenvalue may not have enough independent eigenvectors. [example: Diagonalising a Symmetric Matrix] Let $A=\begin{pmatrix}2&1\cr 1&2\end{pmatrix} \in \mathbb{R}^{2 \times 2}$. We compute its characteristic polynomial from $\chi_A(X)=\det(XI-A)$. Since \begin{align*} XI-A=\begin{pmatrix}X-2&-1\cr -1&X-2\end{pmatrix}, \end{align*} the $2 \times 2$ determinant formula gives \begin{align*} \chi_A(X)=(X-2)(X-2)-(-1)(-1). \end{align*} Thus \begin{align*} \chi_A(X)=(X-2)^2-1=X^2-4X+3=(X-1)(X-3). \end{align*} So the eigenvalues are $1$ and $3$. For $\lambda=3$, \begin{align*} A-3I=\begin{pmatrix}-1&1\cr 1&-1\end{pmatrix}. \end{align*} For $(x,y)\in \mathbb{R}^2$, \begin{align*} (A-3I)(x,y)=(-x+y,x-y). \end{align*} The equation $(A-3I)(x,y)=0$ is therefore equivalent to $-x+y=0$ and $x-y=0$, so $y=x$. Taking $x=1$ gives the eigenvector \begin{align*} v_3=(1,1). \end{align*} For $\lambda=1$, \begin{align*} A-I=\begin{pmatrix}1&1\cr 1&1\end{pmatrix}. \end{align*} For $(x,y)\in \mathbb{R}^2$, \begin{align*} (A-I)(x,y)=(x+y,x+y). \end{align*} The equation $(A-I)(x,y)=0$ is equivalent to $x+y=0$, so $y=-x$. Taking $x=1$ gives the eigenvector \begin{align*} v_1=(1,-1). \end{align*} Now put these eigenvectors as the columns of \begin{align*} P=\begin{pmatrix}1&1\cr 1&-1\end{pmatrix}. \end{align*} Its determinant is \begin{align*} \det P=1(-1)-1\cdot 1=-2, \end{align*} so $P$ is invertible, with \begin{align*} P^{-1}=\frac{1}{-2}\begin{pmatrix}-1&-1\cr -1&1\end{pmatrix}=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}\cr \frac{1}{2}&-\frac{1}{2}\end{pmatrix}. \end{align*} Also, \begin{align*} AP=\begin{pmatrix}2&1\cr 1&2\end{pmatrix}\begin{pmatrix}1&1\cr 1&-1\end{pmatrix}=\begin{pmatrix}3&1\cr 3&-1\end{pmatrix}. \end{align*} Therefore \begin{align*} P^{-1}AP=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}\cr \frac{1}{2}&-\frac{1}{2}\end{pmatrix}\begin{pmatrix}3&1\cr 3&-1\end{pmatrix}=\begin{pmatrix}3&0\cr 0&1\end{pmatrix}. \end{align*} Finally, \begin{align*} v_3\cdot v_1=(1,1)\cdot(1,-1)=1-1=0, \end{align*} so the two eigendirections are perpendicular. This is the first visible sign of the special geometry behind real symmetric matrices: they diagonalize along orthogonal axes. [/example] Diagonal form turns iteration into scalar arithmetic. If $T(v_i)=\lambda_i v_i$ and $x=\sum_i c_i v_i$, then \begin{align*} T^k(x) = \sum_i c_i \lambda_i^k v_i. \end{align*} This is why eigenvalues appear in difference equations, Markov chains, numerical methods, and stability questions. [example: Powers of a Diagonalizable Matrix] Let $A \in \mathbb{R}^{2 \times 2}$ be diagonalizable with eigenvectors $v_1=(1,1)$ and $v_2=(1,-1)$, with eigenvalues $3$ and $1$ respectively. Thus \begin{align*} Av_1=3v_1 \end{align*} and \begin{align*} Av_2=v_2. \end{align*} For $x=(4,2)$, the claimed decomposition is \begin{align*} 3v_1+v_2=3(1,1)+(1,-1)=(3,3)+(1,-1)=(4,2)=x. \end{align*} For every integer $k\ge 0$, induction from $Av_1=3v_1$ gives $A^k v_1=3^k v_1$: the step is \begin{align*} A^{k+1}v_1=A(3^k v_1)=3^k Av_1=3^k(3v_1)=3^{k+1}v_1. \end{align*} Similarly, since $Av_2=v_2$, we have $A^k v_2=1^k v_2=v_2$. Using linearity of $A^k$ and the decomposition $x=3v_1+v_2$, \begin{align*} A^k x=A^k(3v_1+v_2)=3A^k v_1+A^k v_2. \end{align*} Substituting the two eigenvector formulas gives \begin{align*} A^k x=3(3^k v_1)+1^k v_2. \end{align*} Hence \begin{align*} A^k x=3^{k+1}(1,1)+(1,-1). \end{align*} The component in the $3$-eigenspace is multiplied by $3^k$, while the component in the $1$-eigenspace is unchanged, so the $3$-eigenspace controls the long-term growth. [/example] ## Generalized Eigenvectors and Jordan Behaviour When diagonalisation fails, eigenvalues still organise the operator. The missing eigenvectors are replaced by chains of vectors that become eigenvectors after applying $T-\lambda I$ enough times. This is the first step toward Jordan normal form. [definition: Generalized Eigenvector] Let $F$ be a field, let $V$ be a finite-dimensional vector space over $F$, let $T: V \to V$ be a linear operator, and let $\lambda \in F$. A nonzero vector $v \in V$ is a generalized eigenvector of $T$ with eigenvalue $\lambda$ if there exists an integer $k \ge 1$ such that \begin{align*} (T-\lambda I)^k v = 0. \end{align*} [/definition] Every eigenvector is a generalized eigenvector with $k=1$. The new vectors record the directions that are not fixed immediately by $T-\lambda I$ but eventually collapse into the eigenspace. To use them structurally, the next definition collects all such vectors for a fixed eigenvalue into a subspace with a uniform exponent. [definition: Generalized Eigenspace] Let $F$ be a field, let $V$ be a finite-dimensional vector space over $F$, let $T: V \to V$ be a linear operator, and let $\lambda \in F$. The generalized eigenspace of $T$ associated to $\lambda$ is \begin{align*} G_\lambda(T) = \ker((T-\lambda I)^n), \end{align*} where $n=\dim V$. [/definition] The exponent $n$ gives a uniform stopping point in finite dimension. It packages all generalized eigenvectors associated to $\lambda$ into one subspace. The remaining issue is whether this larger subspace has the right size: if generalized eigenspaces are to repair the shortage of ordinary eigenvectors, their dimensions must match the multiplicities coming from the characteristic polynomial. The decomposition theorem organizes this comparison through the factorization of the minimal polynomial: \begin{align*} M_\alpha(t)=\prod_i (t-\lambda_i)^{c_i}. \end{align*} Here the distinct roots $\lambda_i$ are the eigenvalues being separated, and $c_i$ is the multiplicity of $\lambda_i$ as a root of $M_\alpha$. These exponents determine how far one must iterate $\alpha-\lambda_i I$ to capture the corresponding generalized eigenspace. [quotetheorem:411] This theorem explains why generalized eigenspaces are not a cosmetic addition. Even when eigenspaces are too small, generalized eigenspaces recover the full dimension predicted by the characteristic polynomial. [example: Generalized Eigenvectors in a Jordan Block] Let \begin{align*} A=\begin{pmatrix}1&1\cr 0&1\end{pmatrix} \end{align*} and consider the eigenvalue $\lambda=1$. Since \begin{align*} I=\begin{pmatrix}1&0\cr 0&1\end{pmatrix}, \end{align*} we have \begin{align*} A-I=\begin{pmatrix}1&1\cr 0&1\end{pmatrix}-\begin{pmatrix}1&0\cr 0&1\end{pmatrix}=\begin{pmatrix}0&1\cr 0&0\end{pmatrix}. \end{align*} Squaring this matrix gives \begin{align*} (A-I)^2=\begin{pmatrix}0&1\cr 0&0\end{pmatrix}\begin{pmatrix}0&1\cr 0&0\end{pmatrix}=\begin{pmatrix}0\cdot 0+1\cdot 0&0\cdot 1+1\cdot 0\cr 0\cdot 0+0\cdot 0&0\cdot 1+0\cdot 0\end{pmatrix}=\begin{pmatrix}0&0\cr 0&0\end{pmatrix}. \end{align*} Now let $e_1=(1,0)$ and $e_2=(0,1)$. For $e_1$, \begin{align*} (A-I)e_1=\begin{pmatrix}0&1\cr 0&0\end{pmatrix}\begin{pmatrix}1\cr 0\end{pmatrix}=\begin{pmatrix}0\cdot 1+1\cdot 0\cr 0\cdot 1+0\cdot 0\end{pmatrix}=\begin{pmatrix}0\cr 0\end{pmatrix}, \end{align*} so $Ae_1=e_1$, and $e_1$ is an eigenvector with eigenvalue $1$. For $e_2$, \begin{align*} (A-I)e_2=\begin{pmatrix}0&1\cr 0&0\end{pmatrix}\begin{pmatrix}0\cr 1\end{pmatrix}=\begin{pmatrix}0\cdot 0+1\cdot 1\cr 0\cdot 0+0\cdot 1\end{pmatrix}=\begin{pmatrix}1\cr 0\end{pmatrix}=e_1. \end{align*} Thus $(A-I)e_2\neq 0$, so $Ae_2\neq e_2$ and $e_2$ is not an eigenvector for $\lambda=1$. However, \begin{align*} (A-I)^2e_2=\begin{pmatrix}0&0\cr 0&0\end{pmatrix}\begin{pmatrix}0\cr 1\end{pmatrix}=\begin{pmatrix}0\cr 0\end{pmatrix}, \end{align*} so $e_2$ is a generalized eigenvector with eigenvalue $1$. Since $(A-I)^2$ is the zero map on $\mathbb{R}^2$, every vector lies in $\ker((A-I)^2)$, and therefore the generalized eigenspace for $\lambda=1$ is all of $\mathbb{R}^2$. [/example] The Jordan point of view separates the scalar part from the nilpotent part. On a Jordan block with eigenvalue $\lambda$, the operator acts like $\lambda I+N$, where $N$ shifts vectors along a finite chain and eventually vanishes. ## Spectral Meaning in Geometry and Dynamics ### Geometry of Symmetric Operators Eigenvalues are not only algebraic roots. They describe geometry: stretching factors, invariant axes, rotations after complexification, stability of equilibria, and natural frequencies of systems. A symmetric matrix is the cleanest geometric case. It has real eigenvalues, perpendicular eigenspaces, and an orthonormal eigenbasis. This turns an arbitrary-looking quadratic expression into independent stretching along perpendicular axes. [quotetheorem:925] The spectral theorem is why symmetric matrices feel more geometric than arbitrary matrices. Their eigenvectors form perpendicular axes, and the operator stretches each axis by a real factor. [example: Quadratic Form from Eigenvalues] Let $A=\begin{pmatrix}4&0\cr 0&-1\end{pmatrix}$ and define $q:\mathbb{R}^2\to\mathbb{R}$ by $q(x)=x^\top A x$. For $x=(x_1,x_2)$, write $x$ as a column vector when multiplying matrices. Then \begin{align*} Ax=\begin{pmatrix}4&0\cr 0&-1\end{pmatrix}\begin{pmatrix}x_1\cr x_2\end{pmatrix}=\begin{pmatrix}4x_1+0x_2\cr 0x_1-x_2\end{pmatrix}=\begin{pmatrix}4x_1\cr -x_2\end{pmatrix}. \end{align*} Therefore \begin{align*} q(x_1,x_2)=x^\top Ax=\begin{pmatrix}x_1&x_2\end{pmatrix}\begin{pmatrix}4x_1\cr -x_2\end{pmatrix}=x_1(4x_1)+x_2(-x_2)=4x_1^2-x_2^2. \end{align*} The coordinate axes are eigenvector directions. For $e_1=(1,0)$, \begin{align*} Ae_1=\begin{pmatrix}4&0\cr 0&-1\end{pmatrix}\begin{pmatrix}1\cr 0\end{pmatrix}=\begin{pmatrix}4\cr 0\end{pmatrix}=4e_1. \end{align*} Along this direction, \begin{align*} q(te_1)=q(t,0)=4t^2-0^2=4t^2, \end{align*} which is positive for every $t\neq 0$. For $e_2=(0,1)$, \begin{align*} Ae_2=\begin{pmatrix}4&0\cr 0&-1\end{pmatrix}\begin{pmatrix}0\cr 1\end{pmatrix}=\begin{pmatrix}0\cr -1\end{pmatrix}=-e_2. \end{align*} Along this direction, \begin{align*} q(te_2)=q(0,t)=4\cdot 0^2-t^2=-t^2, \end{align*} which is negative for every $t\neq 0$. Thus the eigenvalue $4$ gives a positive quadratic contribution along the first axis, while the eigenvalue $-1$ gives a negative quadratic contribution along the second axis; because the form takes both positive and negative values, it is indefinite. [/example] ### Dynamical Interpretation In differential equations, eigenvalues control growth and decay. The same algebraic equation that finds invariant directions also predicts whether solutions approach or leave an equilibrium. [example: Eigenvalues in a Linear Differential Equation] Let $x:\mathbb{R}\to\mathbb{R}^2$ solve \begin{align*} \frac{dx}{dt}=Ax,\qquad A=\begin{pmatrix}-1&0\cr 0&2\end{pmatrix}. \end{align*} Let $e_1=(1,0)$ and $e_2=(0,1)$. The coordinate axes are eigendirections because \begin{align*} Ae_1=\begin{pmatrix}-1&0\cr 0&2\end{pmatrix}\begin{pmatrix}1\cr 0\end{pmatrix}=\begin{pmatrix}-1\cr 0\end{pmatrix}=-e_1 \end{align*} and \begin{align*} Ae_2=\begin{pmatrix}-1&0\cr 0&2\end{pmatrix}\begin{pmatrix}0\cr 1\end{pmatrix}=\begin{pmatrix}0\cr 2\end{pmatrix}=2e_2. \end{align*} Thus $e_1$ has eigenvalue $-1$, while $e_2$ has eigenvalue $2$. Suppose \begin{align*} x(0)=c_1e_1+c_2e_2. \end{align*} We claim that \begin{align*} x(t)=c_1e^{-t}e_1+c_2e^{2t}e_2. \end{align*} First, at $t=0$ this gives \begin{align*} x(0)=c_1e^0e_1+c_2e^0e_2=c_1e_1+c_2e_2, \end{align*} so the initial condition is satisfied. Differentiating each scalar coefficient gives \begin{align*} \frac{dx}{dt}=c_1(-e^{-t})e_1+c_2(2e^{2t})e_2=-c_1e^{-t}e_1+2c_2e^{2t}e_2. \end{align*} On the other hand, using linearity of matrix multiplication and the two eigenvector equations above, \begin{align*} Ax(t)=A(c_1e^{-t}e_1+c_2e^{2t}e_2)=c_1e^{-t}Ae_1+c_2e^{2t}Ae_2. \end{align*} Substituting $Ae_1=-e_1$ and $Ae_2=2e_2$ gives \begin{align*} Ax(t)=-c_1e^{-t}e_1+2c_2e^{2t}e_2. \end{align*} Therefore $\frac{dx}{dt}=Ax$, as required. The eigenvalue $-1$ contributes the factor $e^{-t}$, which tends to $0$ as $t\to\infty$, while the eigenvalue $2$ contributes the factor $e^{2t}$, which grows without bound when $c_2\neq 0$. Thus negative eigenvalues give decaying modes in this diagonal system, positive eigenvalues give growing modes, and in general linear systems the sign of the real part of an eigenvalue is the stability signal. [/example] This dynamical reading is the reason eigenvalues appear whenever a nonlinear problem is linearised. Near an equilibrium $x^*$ of $\dot{x}=f(x)$, the Jacobian matrix $Jf_{x^*}$ is the linear approximation whose eigenvalues describe the first-order behaviour. [remark: Field and Interpretation] Complex eigenvalues of a real matrix are not a defect. If $A \in \mathbb{R}^{n \times n}$ has a non-real complex eigenvalue $a+ib$ with $b \neq 0$ and complex eigenvector $z=u+iw$, then the real plane $\operatorname{span}\{u,w\}$ is $A$-invariant. With respect to the ordered basis $(u,w)$ of that plane, the restriction of $A$ has real block form \begin{align*} \begin{pmatrix}a&-b\cr b&a\end{pmatrix}. \end{align*} For the operator itself this combines rotation with dilation factor $|a+ib|$; in the flow $\dot{x}=Ax$, the real part $a$ controls exponential growth or decay while $b$ controls oscillation. [/remark] ## Computation and Common Pitfalls The practical algorithm for eigenvalues is short: form $\chi_A(X)$, factor it, and solve $(A-\lambda I)x=0$ for each root. The conceptual hazards are also short: forgetting the field, confusing algebraic with geometric multiplicity, and treating a matrix representation as more intrinsic than the operator it represents. [example: The Same Linear Map in a Better Basis] Let $T:\mathbb{R}^2\to\mathbb{R}^2$ have eigenvectors $v_1=(1,1)$ and $v_2=(1,-1)$ with eigenvalues $3$ and $1$. In the ordered basis $\mathcal{B}=(v_1,v_2)$, the first basis vector is sent to \begin{align*} T(v_1)=3v_1=3v_1+0v_2, \end{align*} and the second basis vector is sent to \begin{align*} T(v_2)=v_2=0v_1+1v_2. \end{align*} Therefore the coordinate columns of $T(v_1)$ and $T(v_2)$ in the basis $\mathcal{B}$ are $(3,0)$ and $(0,1)$, so \begin{align*} [T]_{\mathcal{B}}=\begin{pmatrix}3&0\cr 0&1\end{pmatrix}. \end{align*} Now compare this with the standard-basis matrix. Put the basis vectors into the columns of the change-of-basis matrix \begin{align*} P=\begin{pmatrix}1&1\cr 1&-1\end{pmatrix}. \end{align*} Its determinant is \begin{align*} \det P=1(-1)-1\cdot 1=-2, \end{align*} so \begin{align*} P^{-1}=\frac{1}{-2}\begin{pmatrix}-1&-1\cr -1&1\end{pmatrix}=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}\cr \frac{1}{2}&-\frac{1}{2}\end{pmatrix}. \end{align*} If $D=\begin{pmatrix}3&0\cr 0&1\end{pmatrix}$, then the standard-basis matrix is \begin{align*} A=PDP^{-1}. \end{align*} First, \begin{align*} PD=\begin{pmatrix}1&1\cr 1&-1\end{pmatrix}\begin{pmatrix}3&0\cr 0&1\end{pmatrix}=\begin{pmatrix}3&1\cr 3&-1\end{pmatrix}. \end{align*} Hence \begin{align*} A=\begin{pmatrix}3&1\cr 3&-1\end{pmatrix}\begin{pmatrix}\frac{1}{2}&\frac{1}{2}\cr \frac{1}{2}&-\frac{1}{2}\end{pmatrix}=\begin{pmatrix}2&1\cr 1&2\end{pmatrix}. \end{align*} Thus the same operator that is diagonal in the eigenvector basis has off-diagonal entries in the standard basis. The eigenvalues are still visible from the standard matrix. Indeed, \begin{align*} \chi_A(X)=\det\begin{pmatrix}X-2&-1\cr -1&X-2\end{pmatrix}=(X-2)(X-2)-(-1)(-1). \end{align*} Therefore \begin{align*} \chi_A(X)=(X-2)^2-1=X^2-4X+3=(X-3)(X-1). \end{align*} So changing from the eigenvector basis to the standard basis changes the matrix entries, but it does not change the invariant scaling factors $3$ and $1$. [/example] Changing coordinates should not change the eigenvalues, but without a precise relation between the old and new matrices it is unclear which matrix changes are just coordinate changes and which represent genuinely different operators. The coordinate-change formula isolates the harmless changes: conjugating by an invertible matrix rewrites the same operator in a different ordered basis. This motivates a name for matrices related by an invertible change of coordinates. Once that relation is named, we can state invariance results for all such matrix representatives at once. [definition: Similar Matrices] Let $F$ be a field. Matrices $A,B \in F^{n \times n}$ are similar if there exists an invertible matrix $P \in F^{n \times n}$ such that \begin{align*} B = P^{-1}AP. \end{align*} [/definition] Similarity is the matrix version of choosing a different basis for the same linear operator. If the characteristic polynomial changed under similarity, then eigenvalue computations would depend on the coordinates rather than on the operator. The practical problem is that computations often replace a matrix by a simpler similar one, but that shortcut is valid only if the characteristic polynomial is unchanged by the replacement $A \mapsto P^{-1}AP$. The invariant needed here is therefore not just a comment about the definition of similarity; it is the fact that determinant-based eigenvalue calculations are stable under this coordinate change. [quotetheorem:402] The theorem does not say that similar matrices have the same eigenvectors as coordinate columns. Eigenvectors transform with the basis. The invariant statement is about eigenspaces of the operator, or about how coordinate vectors change under the change-of-basis matrix. [example: A Determinant Trap] Let $A=\begin{pmatrix}0&1\cr -2&3\end{pmatrix}$. We compute the eigenvalues from the characteristic polynomial $\chi_A(X)=\det(XI-A)$, and then compute the eigenspaces separately. First, \begin{align*} XI=\begin{pmatrix}X&0\cr 0&X\end{pmatrix}. \end{align*} Therefore \begin{align*} XI-A=\begin{pmatrix}X&0\cr 0&X\end{pmatrix}-\begin{pmatrix}0&1\cr -2&3\end{pmatrix}=\begin{pmatrix}X&-1\cr 2&X-3\end{pmatrix}. \end{align*} Using the $2\times 2$ determinant formula, \begin{align*} \chi_A(X)=\det\begin{pmatrix}X&-1\cr 2&X-3\end{pmatrix}=X(X-3)-(-1)2. \end{align*} Thus \begin{align*} \chi_A(X)=X^2-3X+2. \end{align*} Since \begin{align*} (X-1)(X-2)=X^2-2X-X+2=X^2-3X+2, \end{align*} we have \begin{align*} \chi_A(X)=(X-1)(X-2). \end{align*} So the possible eigenvalues are $1$ and $2$. For $\lambda=1$, \begin{align*} A-I=\begin{pmatrix}0&1\cr -2&3\end{pmatrix}-\begin{pmatrix}1&0\cr 0&1\end{pmatrix}=\begin{pmatrix}-1&1\cr -2&2\end{pmatrix}. \end{align*} For a vector $(x,y)$, \begin{align*} (A-I)(x,y)=(-x+y,-2x+2y). \end{align*} The equation $(A-I)(x,y)=0$ is therefore equivalent to \begin{align*} -x+y=0 \end{align*} and \begin{align*} -2x+2y=0. \end{align*} The first equation gives $y=x$, and the second equation gives $2(y-x)=0$, which is the same condition. Hence every vector in the $1$-eigenspace has the form \begin{align*} (x,x)=x(1,1). \end{align*} Taking $x=1$ gives the eigenvector $(1,1)$. For $\lambda=2$, \begin{align*} A-2I=\begin{pmatrix}0&1\cr -2&3\end{pmatrix}-\begin{pmatrix}2&0\cr 0&2\end{pmatrix}=\begin{pmatrix}-2&1\cr -2&1\end{pmatrix}. \end{align*} For a vector $(x,y)$, \begin{align*} (A-2I)(x,y)=(-2x+y,-2x+y). \end{align*} The equation $(A-2I)(x,y)=0$ is therefore equivalent to \begin{align*} -2x+y=0. \end{align*} Thus $y=2x$, so every vector in the $2$-eigenspace has the form \begin{align*} (x,2x)=x(1,2). \end{align*} Taking $x=1$ gives the eigenvector $(1,2)$. The determinant calculation finds the scalar factors $1$ and $2$, but the nullspace calculations are what find the corresponding invariant directions $\operatorname{span}\{(1,1)\}$ and $\operatorname{span}\{(1,2)\}$. [/example] A final pitfall is to assume that eigenvalues alone determine the operator. They do not. Multiplicities, eigenspaces, and generalized eigenspaces carry additional structure. [example: Same Eigenvalues, Different Diagonalisation] Let \begin{align*} A=\begin{pmatrix}1&0\cr 0&1\end{pmatrix} \end{align*} and \begin{align*} B=\begin{pmatrix}1&1\cr 0&1\end{pmatrix}. \end{align*} We compute the characteristic polynomial and the eigenspace for $\lambda=1$ for each matrix. For $A$, \begin{align*} XI-A=\begin{pmatrix}X&0\cr 0&X\end{pmatrix}-\begin{pmatrix}1&0\cr 0&1\end{pmatrix}=\begin{pmatrix}X-1&0\cr 0&X-1\end{pmatrix}. \end{align*} Using the $2\times 2$ determinant formula, \begin{align*} \chi_A(X)=\det(XI-A)=(X-1)(X-1)-0\cdot 0=(X-1)^2. \end{align*} Also, \begin{align*} A-I=\begin{pmatrix}1&0\cr 0&1\end{pmatrix}-\begin{pmatrix}1&0\cr 0&1\end{pmatrix}=\begin{pmatrix}0&0\cr 0&0\end{pmatrix}. \end{align*} Hence for every $(x,y)\in\mathbb{R}^2$, \begin{align*} (A-I)(x,y)=(0,0). \end{align*} Therefore \begin{align*} E_1(A)=\ker(A-I)=\mathbb{R}^2, \end{align*} so $\dim E_1(A)=2$. Since the standard basis vectors $(1,0)$ and $(0,1)$ are both eigenvectors for $\lambda=1$, they form a basis of $\mathbb{R}^2$ consisting of eigenvectors, and $A$ is diagonalizable. For $B$, \begin{align*} XI-B=\begin{pmatrix}X&0\cr 0&X\end{pmatrix}-\begin{pmatrix}1&1\cr 0&1\end{pmatrix}=\begin{pmatrix}X-1&-1\cr 0&X-1\end{pmatrix}. \end{align*} Again using the $2\times 2$ determinant formula, \begin{align*} \chi_B(X)=\det(XI-B)=(X-1)(X-1)-(-1)0=(X-1)^2. \end{align*} Thus $A$ and $B$ have the same characteristic polynomial. But their eigenspaces are different. For $B$, \begin{align*} B-I=\begin{pmatrix}1&1\cr 0&1\end{pmatrix}-\begin{pmatrix}1&0\cr 0&1\end{pmatrix}=\begin{pmatrix}0&1\cr 0&0\end{pmatrix}. \end{align*} For $(x,y)\in\mathbb{R}^2$, \begin{align*} (B-I)(x,y)=\begin{pmatrix}0&1\cr 0&0\end{pmatrix}\begin{pmatrix}x\cr y\end{pmatrix}=\begin{pmatrix}y\cr 0\end{pmatrix}. \end{align*} The equation $(B-I)(x,y)=0$ is therefore equivalent to \begin{align*} \begin{pmatrix}y\cr 0\end{pmatrix}=\begin{pmatrix}0\cr 0\end{pmatrix}, \end{align*} so $y=0$ and $x$ is free. Hence \begin{align*} E_1(B)=\{(x,0):x\in\mathbb{R}\}=\operatorname{span}\{(1,0)\}, \end{align*} and $\dim E_1(B)=1$. Since $\chi_B(X)=(X-1)^2$, the only possible eigenvalue of $B$ is $1$, and its eigenspace supplies only one independent eigenvector. Therefore no basis of $\mathbb{R}^2$ can consist entirely of eigenvectors of $B$, so $B$ is not diagonalizable. The two matrices have the same eigenvalue with the same algebraic multiplicity, but different eigenspace dimensions; eigenvalues alone do not determine diagonalisation. [/example] ## Beyond and Connected Topics Eigenvalues sit at a crossroads of algebra, geometry, analysis, and differential equations. The finite-dimensional story begins in [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra), where determinants, bases, linear maps, and diagonalisation are developed systematically. From there, the natural algebraic continuation is canonical form: Jordan normal form refines eigenvalue data by describing the nilpotent chains inside each generalized eigenspace, while rational canonical form gives a version that does not require the characteristic polynomial to split over the field. In Lie theory, eigenvalues appear through adjoint actions, root space decompositions, and representations. The pages [Lie Algebras I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations) and [Lie Algebras II: Structure and Classification](/page/Lie%20Algebras%20II%3A%20Structure%20and%20Classification) develop settings where eigenspace decompositions become structural decompositions of algebras rather than just tools for matrices. In differential equations, eigenvalues govern linear systems and stability near equilibria. The page [Cambridge IA Differential Equations](/page/Cambridge%20IA%20Differential%20Equations) is the natural continuation for seeing eigenvalues control modes, exponential growth, oscillation, and phase portraits. In functional analysis, eigenvalues become part of spectral theory: for operators on infinite-dimensional spaces, the spectrum may contain scalars that are not eigenvalues, and compact self-adjoint operators recover a version of finite-dimensional diagonalisation with analytic subtleties. ## References Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Lie Algebras I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations). Androma, [Lie Algebras II: Structure and Classification](/page/Lie%20Algebras%20II%3A%20Structure%20and%20Classification). Androma, [Cambridge IA Differential Equations](/page/Cambridge%20IA%20Differential%20Equations). Sheldon Axler, *Linear Algebra Done Right* (2015). Kenneth Hoffman and Ray Kunze, *Linear Algebra* (1971). Peter D. Lax, *Linear Algebra and Its Applications* (2007).