A [linear map](/page/Linear%20Map) can mix coordinates so thoroughly that no direction seems to survive. A shear tilts most lines, a rotation moves every real line in the plane unless the angle is special, and a projection collapses whole directions to zero. Eigenvectors mark the exceptional directions that do survive: along them, the map does nothing more complicated than multiply by a scalar. The point of the theory is not merely to solve the equation $Av=\lambda v$. The equation asks whether a transformation has intrinsic one-dimensional pieces. If it does, the transformation can often be split into simpler modes; if it does not, the obstruction tells us something structural about the field, the characteristic polynomial, or the presence of shearing. [example: A Rotation with No Real Eigenvector] Let $V=\mathbb{R}^2$, and let $T:V\to V$ be rotation by angle $\pi/2$, so \begin{align*} T(x,y)=(-y,x). \end{align*} We show that there is no nonzero real vector $v\in \mathbb{R}^2$ and no real scalar $\lambda\in\mathbb{R}$ satisfying $T(v)=\lambda v$. Suppose, for contradiction, that $v=(x,y)$ is a real eigenvector with real eigenvalue $\lambda$. Then $v\ne 0$ and \begin{align*} T(x,y)=\lambda(x,y). \end{align*} Using the formula for $T$ and scalar multiplication in $\mathbb{R}^2$, this equation is \begin{align*} (-y,x)=(\lambda x,\lambda y). \end{align*} Equality of ordered pairs gives \begin{align*} -y=\lambda x. \end{align*} It also gives \begin{align*} x=\lambda y. \end{align*} From $-y=\lambda x$, multiplying both sides by $-1$ gives \begin{align*} y=-\lambda x. \end{align*} Substituting this value of $y$ into $x=\lambda y$ gives \begin{align*} x=\lambda(-\lambda x). \end{align*} Multiplying the scalars on the right gives \begin{align*} x=-\lambda^2 x. \end{align*} Adding $\lambda^2 x$ to both sides gives \begin{align*} (1+\lambda^2)x=0. \end{align*} Because $\lambda\in\mathbb{R}$, we have $\lambda^2\ge 0$, so $1+\lambda^2>0$. Hence $x=0$. Then $y=-\lambda x$ gives \begin{align*} y=-\lambda\cdot 0=0. \end{align*} Thus $v=(0,0)$, contradicting the requirement that an eigenvector be nonzero. Therefore rotation by $\pi/2$ has no real eigenvector. [/example] This failure is the right place to begin. Eigenvectors are not guaranteed over every field, and even when they exist there may not be enough of them to describe the whole space. The chapter develops the definitions, the polynomial test for finding eigenvalues, the geometry of eigenspaces, and the diagonalization questions that make eigenvectors useful. ## Definition ### Eigenvectors and Eigenvalues The simplest possible invariant behaviour is that a nonzero vector stays on its own line. The map may stretch it, shrink it, reverse it, or send it to zero, but the output still belongs to the same one-dimensional subspace. The scalar recording that action is the eigenvalue. [definition: Eigenvector] Let $V$ be a [vector space](/page/Vector%20Space) over a field $k$, and let $T:V\to V$ be a linear map. A vector $v\in V$ is an eigenvector of $T$ if $v\ne 0$ and there exists $\lambda\in k$ such that \begin{align*} T(v)=\lambda v. \end{align*} [/definition] The exclusion of $0$ is essential because $T(0)=\lambda 0$ for every $\lambda\in k$. Including $0$ would make every scalar look relevant to every linear map, while the theory is meant to detect directions. Once the direction has been identified, the scalar by which the map acts on that direction becomes a separate object of study. This lets us ask which scalars occur before choosing particular vectors in their directions. [definition: Eigenvalue] Let $V$ be a vector space over a field $k$, and let $T:V\to V$ be a linear map. A scalar $\lambda\in k$ is an eigenvalue of $T$ if there exists $v\in V$ with $v\ne 0$ such that \begin{align*} T(v)=\lambda v. \end{align*} [/definition] For a fixed eigenvalue, computations should keep all matching vectors together instead of treating them one by one. We therefore need a named subspace that stores every solution of the same eigenvalue equation and makes the kernel method available. [definition: Eigenspace] Let $V$ be a vector space over a field $k$, let $T:V\to V$ be a linear map, and let $\lambda\in k$. The eigenspace of $T$ with eigenvalue $\lambda$ is \begin{align*} E_\lambda(T):=\ker(T-\lambda I), \end{align*} where $I:V\to V$ is the identity map. [/definition] The nonzero elements of $E_\lambda(T)$ are precisely the eigenvectors with eigenvalue $\lambda$. Writing the condition as a kernel also tells us how to compute: subtract $\lambda I$ and solve a homogeneous linear system. [example: Computing an Eigenspace] Let $T:\mathbb{R}^2\to\mathbb{R}^2$ be given by \begin{align*} T(x,y)=(3x+y,2y). \end{align*} We compute the eigenspaces for $\lambda=3$ and $\lambda=2$ by solving $T(x,y)=\lambda(x,y)$. For $\lambda=3$, the equation is \begin{align*} (3x+y,2y)=3(x,y). \end{align*} Scalar multiplication in $\mathbb{R}^2$ gives \begin{align*} 3(x,y)=(3x,3y). \end{align*} Thus \begin{align*} (3x+y,2y)=(3x,3y). \end{align*} Equality of ordered pairs gives \begin{align*} 3x+y=3x. \end{align*} Subtracting $3x$ from both sides gives \begin{align*} y=0. \end{align*} The second coordinate equation is \begin{align*} 2y=3y. \end{align*} Subtracting $2y$ from both sides again gives \begin{align*} 0=y. \end{align*} So the solutions are exactly the vectors of the form $(x,0)$ with $x\in\mathbb{R}$. Since \begin{align*} (x,0)=x(1,0), \end{align*} we get \begin{align*} E_3(T)=\operatorname{span}\{(1,0)\}. \end{align*} For $\lambda=2$, the equation is \begin{align*} (3x+y,2y)=2(x,y). \end{align*} Scalar multiplication gives \begin{align*} 2(x,y)=(2x,2y). \end{align*} Thus \begin{align*} (3x+y,2y)=(2x,2y). \end{align*} Equality of ordered pairs gives \begin{align*} 3x+y=2x. \end{align*} Subtracting $2x$ from both sides gives \begin{align*} x+y=0. \end{align*} The second coordinate equation is \begin{align*} 2y=2y, \end{align*} which imposes no further restriction. From $x+y=0$ we have $x=-y$, so every solution has the form \begin{align*} (x,y)=(-y,y). \end{align*} Factoring out $y$ gives \begin{align*} (-y,y)=y(-1,1). \end{align*} Therefore \begin{align*} E_2(T)=\operatorname{span}\{(-1,1)\}. \end{align*} The example shows the standard workflow: choose a candidate scalar, write out $T(x,y)=\lambda(x,y)$ coordinate by coordinate, and solve the resulting homogeneous linear system. [/example] ### Matrix Language Most computations are performed after choosing a basis. The abstract vector equation then becomes a matrix equation, and the same definition can be stated for coordinate columns. This is a computational translation, not a different concept. [definition: Matrix Eigenvector] Let $A\in k^{n\times n}$. A nonzero vector $v\in k^n$ is an eigenvector of $A$ if there exists $\lambda\in k$ such that \begin{align*} Av=\lambda v. \end{align*} [/definition] Changing basis changes the column used to represent a vector, but it preserves the underlying statement that a line is carried into itself. Eigenvectors belong to the linear map; matrices are the coordinate language used to compute them. For a matrix $A\in k^{n\times n}$ and a scalar $\lambda\in k$, we write \begin{align*} E_\lambda(A):=\ker(A-\lambda I). \end{align*} This is the matrix eigenspace notation corresponding to $E_\lambda(T)$ for the represented linear map. ## Detecting Eigenvectors The definition gives a direct test when the scalar is known, but it does not explain how to find candidate scalars. The equation $Av=\lambda v$ is equivalent to $(A-\lambda I)v=0$, so a nonzero solution exists exactly when $A-\lambda I$ has a nontrivial kernel. In finite dimensions, the determinant detects that failure of invertibility. This motivates a polynomial whose roots are the only possible eigenvalues. The polynomial converts the geometric search for invariant lines into an algebraic root-finding problem over the base field. [definition: Characteristic Polynomial] Let $A\in k^{n\times n}$. The characteristic polynomial of $A$ is \begin{align*} p_A(t):=\det(tI-A)\in k[t]. \end{align*} [/definition] Some texts use $\det(A-tI)$ instead. The two polynomials differ by the factor $(-1)^n$, so they have the same roots and therefore give the same eigenvalues. The determinant criterion is useful because it exactly matches the kernel condition. What must be ruled out is a mismatch between the algebra and the geometry: a root of $p_A(t)$ that produces no eigenvector, or an eigenvector whose eigenvalue is invisible to $p_A(t)$. The detection result below removes both possibilities by tying roots of the characteristic polynomial exactly to nontrivial kernels of $A-\lambda I$. [quotetheorem:918] This theorem is the first major bridge in the subject. The geometry asks for invariant lines; the algebra asks for roots of $p_A(t)$. Whether those roots lie in the chosen field becomes a mathematical issue rather than a technicality. [example: Field Dependence of Eigenvectors] Let $T:\mathbb{R}^2\to\mathbb{R}^2$ be given by $T(x,y)=(-y,x)$. In the standard basis, the matrix of $T$ has first column $(0,1)$ and second column $(-1,0)$, so its entries are $a_{11}=0$, $a_{12}=-1$, $a_{21}=1$, and $a_{22}=0$. Therefore $tI-A$ has entries $t$, $1$, $-1$, and $t$ in those positions. Using the determinant formula for a $2\times 2$ matrix, \begin{align*} p_A(t)=\det(tI-A)=t\cdot t-1\cdot(-1)=t^2+1. \end{align*} If $\lambda\in\mathbb{R}$, then \begin{align*} p_A(\lambda)=\lambda^2+1. \end{align*} Since $\lambda^2\ge 0$ for real $\lambda$, we have $\lambda^2+1>0$, so $p_A(\lambda)\ne 0$. By *[Eigenvalues as Roots of the Characteristic Polynomial](/theorems/918)*, $A$ has no real eigenvalue, and hence $T$ has no real eigenvector. Now consider the same coordinate formula on $\mathbb{C}^2$. Since $i^2=-1$, \begin{align*} p_A(i)=i^2+1=-1+1=0. \end{align*} For $v=(1,-i)$, applying $T$ gives \begin{align*} T(1,-i)=(-(-i),1)=(i,1). \end{align*} Multiplying $v$ by $i$ gives \begin{align*} iv=i(1,-i)=(i,-i^2)=(i,1). \end{align*} Hence \begin{align*} T(v)=iv. \end{align*} Because $v=(1,-i)\ne(0,0)$, this is a complex eigenvector with eigenvalue $i$. The same coordinate formula therefore has no eigenvectors over $\mathbb{R}$ but has eigenvectors after the field is enlarged to $\mathbb{C}$. [/example] The example leaves a general existence question. If eigenvalues can fail to exist only because the characteristic polynomial has no root in the chosen field, then fields with enough roots should force eigenvectors to appear. Over an [algebraically closed field](/page/Algebraically%20Closed%20Field), every positive-degree polynomial has a root, so the obstruction seen over $\mathbb{R}$ disappears for finite-dimensional operators. [quotetheorem:9880] This is why complex vector spaces are especially natural in spectral questions. Over $\mathbb{C}$, finite-dimensional operators always have at least one eigenline; over $\mathbb{R}$, rotations can avoid such lines. ## Eigenspaces and Invariant Lines An eigenvector determines more than a vector: it determines a line. If $T(v)=\lambda v$, then every scalar multiple of $v$ is carried into the same line. This observation leads to the broader notion of an invariant subspace, which is the right language for decomposing a transformation. [definition: Invariant Subspace] Let $V$ be a vector space over a field $k$, and let $T:V\to V$ be a linear map. A subspace $W\subset V$ is invariant under $T$ if \begin{align*} T(W)\subset W. \end{align*} [/definition] A one-dimensional invariant subspace should be the coordinate-free version of an eigendirection, but this needs a check. The possible gap is that invariance is a statement about a whole line, while the eigenvector equation is a statement about a chosen nonzero vector on that line. In one dimension, the action of $T$ on the line must be scalar multiplication, which is exactly the missing link. [quotetheorem:9881] The theorem clarifies what an eigenvector represents. The vector itself can be rescaled, but the invariant line is intrinsic. The eigenvalue records the action of $T$ on that line. [example: Projection onto a Line] Let $P:\mathbb{R}^2\to\mathbb{R}^2$ be the projection \begin{align*} P(x,y)=(x,0). \end{align*} We compute how $P$ acts on the two coordinate axes and then check that a vector off those axes need not be an eigenvector. First take a nonzero vector on the $x$-axis, say $(x,0)$ with $x\ne 0$. Substituting $y=0$ into the formula for $P$ gives \begin{align*} P(x,0)=(x,0). \end{align*} Since \begin{align*} 1(x,0)=(x,0), \end{align*} we have \begin{align*} P(x,0)=1(x,0). \end{align*} Thus every nonzero vector on the $x$-axis is an eigenvector with eigenvalue $1$. Now take a nonzero vector on the $y$-axis, say $(0,y)$ with $y\ne 0$. Substituting $x=0$ into the formula for $P$ gives \begin{align*} P(0,y)=(0,0). \end{align*} Since \begin{align*} 0(0,y)=(0,0), \end{align*} we have \begin{align*} P(0,y)=0(0,y). \end{align*} Thus every nonzero vector on the $y$-axis is an eigenvector with eigenvalue $0$. The vector $(1,1)$ is not an eigenvector. If it were an eigenvector, there would be some $\lambda\in\mathbb{R}$ such that \begin{align*} P(1,1)=\lambda(1,1). \end{align*} The left side is \begin{align*} P(1,1)=(1,0), \end{align*} and the right side is \begin{align*} \lambda(1,1)=(\lambda,\lambda). \end{align*} So we would have \begin{align*} (1,0)=(\lambda,\lambda). \end{align*} Equality of ordered pairs gives $\lambda=1$ from the first coordinate and $\lambda=0$ from the second coordinate, which is impossible. Hence $(1,1)$ is not an eigenvector. The projection preserves the coordinate axes as eigendirections, but it does not preserve every line through the origin. [/example] The projection example shows that an eigenvalue can be zero, and this case has a precise algebraic meaning. It asks whether the map kills a nonzero vector, so the right criterion should be stated in terms of the kernel. [quotetheorem:9882] For square matrices, this says that $0$ is an eigenvalue exactly when the matrix is singular. Eigenvectors therefore record both preferred directions and directions lost by the transformation. ## Bases of Eigenvectors ### Diagonalization One eigenvector gives one invariant line. The most useful case is when the whole vector space is made from invariant lines. Then the transformation has no hidden mixing between basis directions, and its matrix becomes diagonal. [definition: Diagonalizable Linear Map] Let $V$ be a finite-dimensional vector space over a field $k$, and let $T:V\to V$ be a linear map. The map $T$ is diagonalizable over $k$ if there exists a basis $\mathcal{B}=(v_1,\ldots,v_n)$ of $V$ and scalars $\lambda_1,\ldots,\lambda_n\in k$ such that \begin{align*} T(v_i)=\lambda_i v_i \end{align*} for each $i\in\{1,\ldots,n\}$. [/definition] A diagonal matrix is therefore not just a convenient shape. It is the coordinate expression of a map whose basis vectors are eigenvectors. The remaining issue is equivalence: diagonal entries describe independent scalar actions only when the chosen basis consists of eigenvectors, and an eigenbasis should in turn force the representing matrix to have no off-diagonal mixing. [quotetheorem:8325] The theorem shifts diagonalization from matrix manipulation to geometry. To diagonalize is to find enough invariant lines to span the whole space. [example: Diagonalizing a Two-Dimensional Map] Let $T:\mathbb{R}^2\to\mathbb{R}^2$ be given by \begin{align*} T(x,y)=(3x+y,2y). \end{align*} We verify that $v_1=(1,0)$ and $v_2=(-1,1)$ are eigenvectors, then use them as a basis to compute the representing matrix of $T$. For $v_1=(1,0)$, substituting $x=1$ and $y=0$ into the formula for $T$ gives \begin{align*} T(1,0)=(3\cdot 1+0,2\cdot 0). \end{align*} Thus \begin{align*} T(1,0)=(3,0). \end{align*} Since \begin{align*} 3(1,0)=(3,0), \end{align*} we have \begin{align*} T(v_1)=3v_1. \end{align*} So $v_1$ is an eigenvector with eigenvalue $3$. For $v_2=(-1,1)$, substituting $x=-1$ and $y=1$ gives \begin{align*} T(-1,1)=(3(-1)+1,2\cdot 1). \end{align*} The first coordinate is $3(-1)+1=-3+1=-2$, and the second coordinate is $2\cdot 1=2$, so \begin{align*} T(-1,1)=(-2,2). \end{align*} Since \begin{align*} 2(-1,1)=(-2,2), \end{align*} we have \begin{align*} T(v_2)=2v_2. \end{align*} So $v_2$ is an eigenvector with eigenvalue $2$. The vectors $v_1$ and $v_2$ are linearly independent. Suppose \begin{align*} a(1,0)+b(-1,1)=(0,0). \end{align*} Expanding the left side gives \begin{align*} (a,0)+(-b,b)=(a-b,b). \end{align*} Thus \begin{align*} (a-b,b)=(0,0). \end{align*} Equality of ordered pairs gives $b=0$ from the second coordinate. Substituting $b=0$ into $a-b=0$ gives $a=0$. Hence the only linear relation between $v_1$ and $v_2$ has coefficients $a=b=0$, so $v_1$ and $v_2$ form a basis of $\mathbb{R}^2$. In the basis $\mathcal{B}=(v_1,v_2)$, the first basis vector is sent to $3v_1+0v_2$, and the second basis vector is sent to $0v_1+2v_2$. Therefore the coordinate columns of the representing matrix are \begin{align*} [T(v_1)]_{\mathcal{B}}=(3,0) \end{align*} and \begin{align*} [T(v_2)]_{\mathcal{B}}=(0,2). \end{align*} So the matrix of $T$ in the basis $\mathcal{B}$ is diagonal: \begin{align*} [T]_{\mathcal{B}}=\begin{pmatrix}3&0\cr 0&2\end{pmatrix}. \end{align*} The [change of basis](/page/Change%20Of%20Basis) has separated the two independent eigendirections: one coordinate is multiplied by $3$, and the other is multiplied by $2$. [/example] ### Independence of Eigenvectors A common route to an eigenbasis is to find eigenvectors with different eigenvalues. Distinct eigenvalues prevent accidental linear dependence, because each eigenvector responds differently when $T$ is applied. [quotetheorem:8488] The independence theorem suggests a clean route to diagonalization: find as many distinct eigenvalues as the dimension. The obstruction to an eigenbasis is possible linear dependence among candidate eigenvectors; distinct eigenvalues remove that obstruction automatically. When there are $n$ such eigenvalues in an $n$-dimensional space, the resulting independent eigenvectors already have the right number to form a basis. That distinct-eigenvalue test is sufficient, not necessary. The sharper test compares, for each eigenvalue $\lambda$, the eigenspace $M_\lambda$ with the multiplicity of $\lambda$ in the characteristic polynomial. The number $g_\lambda=\dim M_\lambda$ is the geometric multiplicity, while $a_\lambda$ denotes the algebraic multiplicity. An operator is diagonalisable when there is a basis of eigenvectors, equivalently when its matrix is diagonal in a suitable basis. Thus diagonalization requires enough eigendirections in total: after the characteristic polynomial splits, each eigenvalue must contribute as many independent eigenvectors as its algebraic multiplicity demands. Repeated eigenvalues can still produce enough eigenvectors, but repetition opens the door to a new kind of failure. ## Multiplicity and Failure of Diagonalization A repeated root of the characteristic polynomial raises a question that roots alone cannot answer. Does the repeated eigenvalue supply several independent eigendirections, or does it supply only one direction together with hidden shearing? To separate these possibilities, we compare two multiplicities. The first multiplicity belongs to the characteristic polynomial. It counts how many times the scalar appears as a root, before any eigenspace has been computed. [definition: Algebraic Multiplicity] Let $A\in k^{n\times n}$, and let $\lambda\in k$ be an eigenvalue of $A$. The algebraic multiplicity of $\lambda$ is its multiplicity as a root of the characteristic polynomial $p_A(t)$. [/definition] Algebraic multiplicity alone does not tell us how many independent eigenvectors exist. We need a second number, measured inside the eigenspace, because diagonalization depends on independent directions rather than on repeated polynomial roots. [definition: Geometric Multiplicity] Let $A\in k^{n\times n}$, and let $\lambda\in k$ be an eigenvalue of $A$. The geometric multiplicity of $\lambda$ is \begin{align*} \dim E_\lambda(A). \end{align*} [/definition] The comparison between these two numbers identifies the obstruction to diagonalization. Algebraic multiplicity is a promise made by the characteristic polynomial; geometric multiplicity measures how many independent eigenvectors are delivered. [quotetheorem:919] The upper bound is where diagonalization can fail. A repeated eigenvalue may have fewer independent eigenvectors than its algebraic multiplicity allows. [example: A Repeated Eigenvalue with Too Few Eigenvectors] Let $T:\mathbb{R}^2\to\mathbb{R}^2$ be given by \begin{align*} T(x,y)=(x+y,y). \end{align*} In the standard basis, $T(1,0)=(1,0)$ and $T(0,1)=(1,1)$, so the matrix of $T$ is \begin{align*} A=\begin{pmatrix}1&1\cr 0&1\end{pmatrix}. \end{align*} Thus \begin{align*} tI-A=\begin{pmatrix}t&0\cr 0&t\end{pmatrix}-\begin{pmatrix}1&1\cr 0&1\end{pmatrix}=\begin{pmatrix}t-1&-1\cr 0&t-1\end{pmatrix}. \end{align*} Using the determinant formula for a $2\times 2$ matrix, \begin{align*} p_A(t)=\det(tI-A)=(t-1)(t-1)-(-1)\cdot 0=(t-1)^2. \end{align*} Therefore $\lambda=1$ is a root of $p_A(t)$ with algebraic multiplicity $2$. Now compute the eigenspace for $\lambda=1$. The equation $T(x,y)=1(x,y)$ is \begin{align*} (x+y,y)=(x,y). \end{align*} Equality of ordered pairs gives \begin{align*} x+y=x. \end{align*} Subtracting $x$ from both sides gives \begin{align*} y=0. \end{align*} The second coordinate equation is \begin{align*} y=y, \end{align*} so it imposes no additional restriction. Hence every solution has the form $(x,0)$ with $x\in\mathbb{R}$, and since \begin{align*} (x,0)=x(1,0), \end{align*} we get \begin{align*} E_1(T)=\operatorname{span}\{(1,0)\}. \end{align*} Thus the geometric multiplicity is \begin{align*} \dim E_1(T)=1. \end{align*} The only eigendirection is the $x$-axis, so this repeated eigenvalue supplies one independent eigenvector rather than two; consequently there is no basis of $\mathbb{R}^2$ consisting of eigenvectors, and $T$ is not diagonalizable. [/example] This is the simplest Jordan-type failure. The map preserves the eigenline, but it also shears parallel to that line, so a second independent eigendirection does not appear. The full finite-dimensional criterion should therefore require every eigenvalue to supply the maximum possible number of independent eigenvectors. [quotetheorem:8491] This criterion gives a complete diagnostic. First check that the characteristic polynomial splits over the chosen field; then compare each eigenspace dimension with the corresponding algebraic multiplicity. ## Symmetry and Orthogonality The shearing failure above is impossible for symmetric matrices. Symmetry ties the matrix to the Euclidean [inner product](/page/Inner%20Product), and that compatibility forces eigenspaces for different eigenvalues to sit at right angles. [definition: Symmetric Matrix] A matrix $A\in\mathbb{R}^{n\times n}$ is symmetric if \begin{align*} A^\top=A. \end{align*} [/definition] Symmetry can be read as a rule for moving the matrix from one side of the dot product to the other. This rule makes two different eigenvalues incompatible with a nonzero inner product between their eigenvectors, which is the next structural payoff. [quotetheorem:3280] Orthogonality of existing eigenspaces is only part of the story; we still need enough eigenvectors to span the whole space. The strongest finite-dimensional theorem is often stated over $\mathbb{C}^n$ for Hermitian matrices. Here $A^\dagger$ means the conjugate transpose of $A$, a matrix is Hermitian when $A^\dagger=A$, and a matrix $P$ is unitary when $P^\dagger P=I$. For real matrices, conjugate transpose reduces to ordinary transpose, so the real symmetric case is the real form of this Hermitian statement. [quotetheorem:925] This theorem explains the prominence of eigenvectors in geometry and applied mathematics. Symmetric matrices arise from quadratic forms, covariance matrices, and [self-adjoint operators](/page/Self-Adjoint%20Operators), and their eigenvectors give preferred orthogonal directions. [example: Principal Axes of a Quadratic Form] Let $A:\mathbb{R}^2\to\mathbb{R}^2$ be defined by \begin{align*} A(x,y)=(2x+y,x+2y). \end{align*} We first verify the two eigendirections. For $v_1=(1,1)$, \begin{align*} A(1,1)=(2\cdot 1+1,1+2\cdot 1)=(3,3). \end{align*} Since \begin{align*} 3(1,1)=(3,3), \end{align*} we have $A(v_1)=3v_1$. For $v_2=(1,-1)$, \begin{align*} A(1,-1)=(2\cdot 1+(-1),1+2(-1))=(1,-1). \end{align*} Since \begin{align*} 1(1,-1)=(1,-1), \end{align*} we have $A(v_2)=v_2$. The vectors are perpendicular because \begin{align*} v_1\cdot v_2=(1,1)\cdot(1,-1)=1\cdot 1+1\cdot(-1)=0. \end{align*} Their lengths are \begin{align*} \|v_1\|=\sqrt{1^2+1^2}=\sqrt{2} \end{align*} and \begin{align*} \|v_2\|=\sqrt{1^2+(-1)^2}=\sqrt{2}. \end{align*} Thus the corresponding orthonormal eigenvectors are \begin{align*} u_1=\frac{1}{\sqrt{2}}(1,1) \end{align*} and \begin{align*} u_2=\frac{1}{\sqrt{2}}(1,-1). \end{align*} For $x=(x_1,x_2)$, the quadratic form is \begin{align*} x\cdot Ax=(x_1,x_2)\cdot(2x_1+x_2,x_1+2x_2). \end{align*} Expanding the dot product gives \begin{align*} x\cdot Ax=x_1(2x_1+x_2)+x_2(x_1+2x_2). \end{align*} Distributing each term gives \begin{align*} x\cdot Ax=2x_1^2+x_1x_2+x_1x_2+2x_2^2. \end{align*} Combining the two middle terms gives \begin{align*} x\cdot Ax=2x_1^2+2x_1x_2+2x_2^2. \end{align*} Now write $x$ in the orthonormal eigenbasis as \begin{align*} x=y_1u_1+y_2u_2. \end{align*} By linearity of $A$ and the eigenvector equations $A(u_1)=3u_1$ and $A(u_2)=u_2$, \begin{align*} Ax=y_1A(u_1)+y_2A(u_2)=3y_1u_1+y_2u_2. \end{align*} Therefore \begin{align*} x\cdot Ax=(y_1u_1+y_2u_2)\cdot(3y_1u_1+y_2u_2). \end{align*} Expanding by bilinearity of the dot product gives \begin{align*} x\cdot Ax=3y_1^2(u_1\cdot u_1)+y_1y_2(u_1\cdot u_2)+3y_1y_2(u_2\cdot u_1)+y_2^2(u_2\cdot u_2). \end{align*} Since $u_1$ and $u_2$ are orthonormal, $u_1\cdot u_1=1$, $u_2\cdot u_2=1$, and $u_1\cdot u_2=u_2\cdot u_1=0$. Substituting these values gives \begin{align*} x\cdot Ax=3y_1^2\cdot 1+y_1y_2\cdot 0+3y_1y_2\cdot 0+y_2^2\cdot 1. \end{align*} Hence \begin{align*} x\cdot Ax=3y_1^2+y_2^2. \end{align*} In the original coordinates the level set $x\cdot Ax=1$ is described by $2x_1^2+2x_1x_2+2x_2^2=1$, while in the orthonormal eigenbasis it is described by $3y_1^2+y_2^2=1$; the eigenvectors give the two perpendicular principal axes of this ellipse. [/example] ## Dynamical and Algebraic Meaning ### Iteration and Modes Eigenvectors matter in applications because many processes repeatedly apply the same linear map. On an eigenvector, repeated application is reduced to powers of a scalar. Growth, decay, and oscillation can then be read from eigenvalues. [quotetheorem:9883] If a vector decomposes into eigenvectors, each component evolves independently. This is the algebraic core of solving linear recurrences and linear differential systems. [example: A Linear Recurrence from Eigenvectors] Let $A:\mathbb{R}^2\to\mathbb{R}^2$ be given by \begin{align*} A(x,y)=(2x,y/2). \end{align*} The coordinate directions are eigenvector directions for $A$. Indeed, \begin{align*} A(1,0)=(2\cdot 1,0/2)=(2,0). \end{align*} Since \begin{align*} 2(1,0)=(2,0), \end{align*} we have $A(1,0)=2(1,0)$. Similarly, \begin{align*} A(0,1)=(2\cdot 0,1/2)=(0,1/2). \end{align*} Since \begin{align*} \frac{1}{2}(0,1)=(0,1/2), \end{align*} we have $A(0,1)=\frac{1}{2}(0,1)$. For the recurrence $x_{m+1}=A(x_m)$ with $x_0=(3,4)$, first decompose the initial vector along these two eigendirections: \begin{align*} 3(1,0)+4(0,1)=(3,0)+(0,4)=(3,4). \end{align*} We claim that for every $m\ge 0$, \begin{align*} x_m=3\cdot 2^m(1,0)+4\cdot 2^{-m}(0,1). \end{align*} At $m=0$, this gives \begin{align*} 3\cdot 2^0(1,0)+4\cdot 2^{-0}(0,1)=3(1,0)+4(0,1)=(3,4)=x_0. \end{align*} Now suppose the formula holds for some $m\ge 0$. Then \begin{align*} x_m=3\cdot 2^m(1,0)+4\cdot 2^{-m}(0,1)=(3\cdot 2^m,4\cdot 2^{-m}). \end{align*} Applying the recurrence and the formula for $A$ gives \begin{align*} x_{m+1}=A(x_m)=A(3\cdot 2^m,4\cdot 2^{-m}). \end{align*} Substituting into $A(x,y)=(2x,y/2)$ gives \begin{align*} A(3\cdot 2^m,4\cdot 2^{-m})=(2(3\cdot 2^m),(4\cdot 2^{-m})/2). \end{align*} The first coordinate is \begin{align*} 2(3\cdot 2^m)=3\cdot 2^{m+1}. \end{align*} The second coordinate is \begin{align*} (4\cdot 2^{-m})/2=4\cdot 2^{-m}\cdot 2^{-1}=4\cdot 2^{-(m+1)}. \end{align*} Therefore \begin{align*} x_{m+1}=(3\cdot 2^{m+1},4\cdot 2^{-(m+1)}). \end{align*} Writing this again in the eigenvector directions gives \begin{align*} x_{m+1}=3\cdot 2^{m+1}(1,0)+4\cdot 2^{-(m+1)}(0,1). \end{align*} Thus the formula holds for all $m\ge 0$. The component along $(1,0)$ is multiplied by $2$ at each step and grows like $2^m$, while the component along $(0,1)$ is multiplied by $1/2$ at each step and decays like $2^{-m}$. [/example] ### Representation-Theoretic Direction Eigenvectors also organise algebraic actions. When a group, algebra, or [Lie algebra](/page/Lie%20Algebra) acts on a vector space, finding vectors on which selected operators act by scalars is often the first step toward decomposing the representation. In Lie theory, eigenvectors generalise into weight vectors and eigenspaces generalise into weight spaces. The same equation $T(v)=\lambda v$ becomes part of a larger system in which several commuting or nearly commuting operators are studied together. ## Beyond and Connected Topics The natural continuation is diagonalization, where eigenvectors are assembled into bases and matrices are simplified by change of basis. The main question becomes whether the space splits into enough invariant lines. The [characteristic polynomial](/page/Characteristic%20Polynomial) gives the algebraic mechanism for detecting eigenvalues. It connects eigenvectors to determinants, polynomial roots, and the choice of field. The [minimal polynomial](/page/Minimal%20Polynomial) refines this information. It detects diagonalizability and records the nilpotent pieces that remain when eigenvectors alone do not span the space. For symmetric and self-adjoint operators, the spectral theorem explains why orthogonal eigenbases exist. This is the version most visible in quadratic forms, numerical linear algebra, and [Hilbert space](/page/Hilbert%20Space) theory. In Lie theory and representation theory, eigenvectors become weight vectors. The pages [Lie Algebras I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations), [Lie Algebras II: Structure and Classification](/page/Lie%20Algebras%20II%3A%20Structure%20and%20Classification), and [Lie Algebras III: Representation Theory](/page/Lie%20Algebras%20III%3A%20Representation%20Theory) develop this idea when many operators interact. ## References Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Characteristic Polynomial](/page/Characteristic%20Polynomial). Androma, [Minimal Polynomial](/page/Minimal%20Polynomial). 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, [Lie Algebras III: Representation Theory](/page/Lie%20Algebras%20III%3A%20Representation%20Theory). Sheldon Axler, *Linear Algebra Done Right* (2015). Kenneth Hoffman and Ray Kunze, *Linear Algebra* (1971). Roger A. Horn and Charles R. Johnson, *Matrix Analysis* (2013).