# Eigenvalue and Eigenvector A [linear map](/page/Linear%20Map) $T : V \to V$ on a finite-dimensional vector space $V$ over a field $F$ acts on every vector simultaneously, and understanding $T$ in full generality can be daunting. However, a remarkable simplification occurs whenever we can find vectors whose image under $T$ points in the same direction as the original -- vectors that $T$ merely scales. These are the **eigenvectors** of $T$, and the corresponding scale factors are the **eigenvalues**. The theory of eigenvalues and eigenvectors is one of the central pillars of linear algebra, with applications ranging from systems of differential equations to quantum mechanics, from Google's PageRank algorithm to the stability analysis of dynamical systems. This article develops the theory from its geometric motivation through the characteristic polynomial and multiplicity theory, culminating in the diagonalisation criterion and the spectral theorem for Hermitian matrices. ## Motivation [motivation] ### The diagonal dream Suppose we are given a linear map $T : V \to V$ on a finite-dimensional vector space $V$ over a field $F$, and suppose we can find a basis $\mathcal{B} = (v_1, v_2, \ldots, v_n)$ of $V$ such that $T$ acts on each basis vector by simply scaling it: \begin{align*} T(v_1) &= \lambda_1 v_1, \\ T(v_2) &= \lambda_2 v_2, \\ &\;\;\vdots \\ T(v_n) &= \lambda_n v_n, \end{align*} for scalars $\lambda_1, \lambda_2, \ldots, \lambda_n \in F$. In this situation the matrix of $T$ with respect to $\mathcal{B}$ is the diagonal matrix $D = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)$. Diagonal matrices are the simplest matrices to work with: computing $D^k$ reduces to raising each diagonal entry to the $k$-th power, solving the system $Dx = b$ reduces to $n$ independent one-variable equations, and the matrix exponential $e^{tD}$ is simply $\operatorname{diag}(e^{\lambda_1 t}, \ldots, e^{\lambda_n t})$. This makes diagonal representations indispensable in applications. For a system of linear ordinary differential equations $x'(t) = Ax(t)$ where $A$ is a real $n \times n$ matrix, if $A$ is diagonalisable with $A = PDP^{-1}$, then the substitution $y = P^{-1}x$ decouples the system into $n$ independent equations $y_i'(t) = \lambda_i y_i(t)$, each with the elementary solution $y_i(t) = y_i(0) e^{\lambda_i t}$. ### What goes wrong Not every linear map admits such a convenient basis. Consider the matrix \begin{align*} A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \end{align*} acting on $\mathbb{R}^2$. The only eigenvalue is $\lambda = 0$, but the eigenspace is the span of $(1, 0)^T$, which is one-dimensional. There is no basis of $\mathbb{R}^2$ consisting of eigenvectors of $A$, so $A$ cannot be diagonalised. Such matrices are called **defective**. The central question of this article is: given a linear map $T$, how do we determine whether a basis of eigenvectors exists, and if it does, how do we find it? ### Eigenvectors as the building blocks Geometrically, an eigenvector of $T$ determines an **invariant direction** -- a one-dimensional subspace $\operatorname{span}(v)$ that $T$ maps into itself. The eigenvalue $\lambda$ records whether $T$ stretches ($|\lambda| > 1$), compresses ($|\lambda| < 1$), reflects ($\lambda < 0$), or annihilates ($\lambda = 0$) vectors along that direction. Understanding all the invariant directions of $T$, together with the dimensions of the spaces they generate, is the key to deciding whether $T$ can be diagonalised. [/motivation] ## Formal Definition Let $V$ be a finite-dimensional vector space over a field $F$, and let $T : V \to V$ be a linear map. [definition:Eigenvalue And Eigenvector] A scalar $\lambda \in F$ is called an **eigenvalue** of $T$ if there exists a nonzero vector $v \in V$ such that \begin{align*} T(v) = \lambda v. \end{align*} Any nonzero vector $v \in V$ satisfying $T(v) = \lambda v$ is called an **eigenvector** of $T$ corresponding to the eigenvalue $\lambda$. When $T$ is represented by a matrix $A \in M_{n \times n}(F)$ with respect to some basis, we equivalently say that $\lambda$ is an eigenvalue of $A$ and $v \in F^n \setminus \{\mathbf{0}\}$ is an eigenvector of $A$ if $Av = \lambda v$. [/definition] The requirement that $v \neq \mathbf{0}$ is essential: since $T(\mathbf{0}) = \lambda \cdot \mathbf{0}$ for every $\lambda \in F$, allowing the zero vector would make every scalar an eigenvalue and render the definition useless. [definition:Eigenspace] Let $\lambda \in F$ be an eigenvalue of $T : V \to V$. The **eigenspace** of $T$ corresponding to $\lambda$ is the [set](/page/Set) \begin{align*} E_\lambda = \ker(T - \lambda I) = \{ v \in V : T(v) = \lambda v \}. \end{align*} This is a subspace of $V$, and its nonzero elements are precisely the eigenvectors corresponding to $\lambda$. [/definition] The eigenspace $E_\lambda$ is indeed a subspace because it is the kernel of the linear map $T - \lambda I : V \to V$. Note that $\lambda$ is an eigenvalue of $T$ if and only if $E_\lambda \neq \{\mathbf{0}\}$, which happens if and only if $T - \lambda I$ is not injective, which in turn happens if and only if $T - \lambda I$ is singular. ## Examples [example:TwoByTwoEigenvalues] **Finding eigenvalues of a $2 \times 2$ matrix.** Let \begin{align*} A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} \in M_{2 \times 2}(\mathbb{R}). \end{align*} To find the eigenvalues we seek $\lambda \in \mathbb{R}$ such that $A - \lambda I$ is singular, i.e., $\det(A - \lambda I) = 0$. We compute \begin{align*} A - \lambda I &= \begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{pmatrix}, \\ \det(A - \lambda I) &= (4 - \lambda)(3 - \lambda) - 2 \\ &= \lambda^2 - 7\lambda + 12 - 2 \\ &= \lambda^2 - 7\lambda + 10 \\ &= (\lambda - 2)(\lambda - 5). \end{align*} So $A$ has eigenvalues $\lambda_1 = 2$ and $\lambda_2 = 5$. For $\lambda_1 = 2$: we solve $(A - 2I)v = \mathbf{0}$: \begin{align*} \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \end{align*} which gives $2v_1 + v_2 = 0$, so $E_2 = \operatorname{span}\!\left(\begin{pmatrix} 1 \\ -2 \end{pmatrix}\right)$. For $\lambda_2 = 5$: we solve $(A - 5I)v = \mathbf{0}$: \begin{align*} \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \end{align*} which gives $-v_1 + v_2 = 0$, so $E_5 = \operatorname{span}\!\left(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\right)$. [/example] [example:ThreeByThreeEigenvalues] **A $3 \times 3$ matrix with a repeated eigenvalue.** Let \begin{align*} B = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \in M_{3 \times 3}(\mathbb{R}). \end{align*} Since $B$ is upper-triangular, its eigenvalues are the diagonal entries: $\lambda_1 = 2$ (repeated) and $\lambda_2 = 3$. More precisely, the characteristic polynomial is \begin{align*} \chi_B(\lambda) = (2 - \lambda)^2(3 - \lambda). \end{align*} For $\lambda_1 = 2$: \begin{align*} B - 2I = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}. \end{align*} The kernel has dimension $1$ (only the first coordinate is free), giving $E_2 = \operatorname{span}\!\left(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\right)$. Even though $\lambda_1 = 2$ appears as a root of multiplicity $2$ in the characteristic polynomial, the eigenspace is only one-dimensional. This discrepancy between algebraic and geometric multiplicity will be the focus of a later section. For $\lambda_2 = 3$: $E_3 = \operatorname{span}\!\left(\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\right)$, which is one-dimensional. [/example] ## The Characteristic Polynomial The examples above suggest a systematic approach: $\lambda$ is an eigenvalue of $A$ if and only if $\det(A - \lambda I) = 0$. The expression $\det(A - \lambda I)$, viewed as a function of $\lambda$, is a polynomial, and its roots are precisely the eigenvalues. [definition:Characteristic Polynomial] Let $A \in M_{n \times n}(F)$. The **characteristic polynomial** of $A$ is the polynomial $\chi_A \in F[\lambda]$ defined by \begin{align*} \chi_A(\lambda) = \det(A - \lambda I). \end{align*} This is a monic polynomial of degree $n$ in $\lambda$ (up to sign convention; some authors define $\chi_A(\lambda) = \det(\lambda I - A)$, which gives the same roots but a different leading coefficient). [/definition] The characteristic polynomial is well-defined as an invariant of the linear map $T$ rather than of any particular matrix representation: if $B = P^{-1}AP$ for some invertible $P$, then \begin{align*} \det(B - \lambda I) = \det(P^{-1}AP - \lambda P^{-1}P) = \det(P^{-1}(A - \lambda I)P) = \det(A - \lambda I), \end{align*} so similar matrices share the same characteristic polynomial. We may therefore speak of the characteristic polynomial $\chi_T$ of a linear map $T$ without ambiguity. The following theorem formalises the connection between eigenvalues and the characteristic polynomial. [quotetheorem:918] This result reduces the problem of finding eigenvalues to the algebraic problem of finding roots of a polynomial. Over an algebraically closed field such as $\mathbb{C}$, the characteristic polynomial always splits into linear factors, so every linear map on a finite-dimensional complex vector space has at least one eigenvalue. Over $\mathbb{R}$, the characteristic polynomial may have no real roots (consider a rotation by $\pi/2$ in $\mathbb{R}^2$, whose characteristic polynomial is $\lambda^2 + 1$), in which case the map has no real eigenvalues. ## Algebraic and Geometric Multiplicity When the characteristic polynomial has repeated roots, the eigenvalue theory becomes more subtle. Two different notions of "multiplicity" capture different aspects of a repeated eigenvalue. [definition:Algebraic Multiplicity] Let $\lambda_0$ be an eigenvalue of $A \in M_{n \times n}(F)$. The **algebraic multiplicity** of $\lambda_0$, denoted $a(\lambda_0)$, is the multiplicity of $\lambda_0$ as a root of the characteristic polynomial $\chi_A$. That is, $a(\lambda_0)$ is the largest integer $k$ such that $(\lambda - \lambda_0)^k$ divides $\chi_A(\lambda)$ in $F[\lambda]$. [/definition] [definition:Geometric Multiplicity] Let $\lambda_0$ be an eigenvalue of $T : V \to V$. The **geometric multiplicity** of $\lambda_0$, denoted $g(\lambda_0)$, is the dimension of the eigenspace $E_{\lambda_0}$: \begin{align*} g(\lambda_0) = \dim E_{\lambda_0} = \dim \ker(T - \lambda_0 I). \end{align*} [/definition] The algebraic multiplicity counts how many times $\lambda_0$ appears as a root, while the geometric multiplicity counts how many linearly independent eigenvectors correspond to $\lambda_0$. These two quantities are related by a fundamental inequality. [quotetheorem:919] The matrix $B$ from the earlier example illustrates this inequality sharply: the eigenvalue $\lambda = 2$ has algebraic multiplicity $a(2) = 2$ but geometric multiplicity $g(2) = 1$. The gap between these two quantities is the obstruction to diagonalisation, as we shall see. Another key structural property is that eigenvectors corresponding to distinct eigenvalues never interfere with each other. [quotetheorem:920] This result is worth pausing over. It says that if $\lambda_1, \lambda_2, \ldots, \lambda_k$ are distinct eigenvalues of $T$ and we pick any nonzero eigenvector $v_i \in E_{\lambda_i}$ for each $i$, then the collection $\{v_1, v_2, \ldots, v_k\}$ is linearly independent. An important corollary is that if a linear map on an $n$-dimensional space has $n$ distinct eigenvalues, then it is automatically diagonalisable -- the $n$ eigenvectors form a basis. However, having fewer than $n$ distinct eigenvalues does not preclude diagonalisability; it simply means we need to examine each eigenspace more carefully. ## Diagonalisation We now arrive at the central question: when can a linear map be represented by a diagonal matrix? [definition:Diagonalisable] A linear map $T : V \to V$ on a finite-dimensional vector space $V$ over $F$ is **diagonalisable** if there exists a basis of $V$ consisting entirely of eigenvectors of $T$. Equivalently, a matrix $A \in M_{n \times n}(F)$ is diagonalisable if there exists an invertible matrix $P \in M_{n \times n}(F)$ such that $P^{-1}AP$ is diagonal. [/definition] The columns of $P$ are eigenvectors of $A$, and the diagonal entries of $P^{-1}AP$ are the corresponding eigenvalues. The following theorem gives a precise necessary and sufficient condition. [quotetheorem:921] This theorem tells us that diagonalisability is controlled entirely by two conditions: first, the characteristic polynomial must split into linear factors over $F$ (so that all eigenvalues lie in $F$), and second, for each eigenvalue, the geometric multiplicity must equal the algebraic multiplicity. The second condition is where things can fail, as the matrix $B$ from our earlier example demonstrates. [example:DiagonalisationWorked] **Diagonalising a matrix.** Returning to the matrix from our first example, \begin{align*} A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}, \end{align*} we found eigenvalues $\lambda_1 = 2$ and $\lambda_2 = 5$ with corresponding eigenvectors $v_1 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ and $v_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. Since we have two distinct eigenvalues in a $2$-dimensional space, the eigenvectors form a basis, and $A$ is diagonalisable. Setting \begin{align*} P = \begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}, \end{align*} we compute $P^{-1} = \frac{1}{3}\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}$, and one verifies that \begin{align*} P^{-1}AP = \frac{1}{3}\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}. \end{align*} As a payoff, computing $A^{100}$ is now straightforward: $A^{100} = P \begin{pmatrix} 2^{100} & 0 \\ 0 & 5^{100} \end{pmatrix} P^{-1}$. [/example] [example:DefectiveMatrix] **A defective (non-diagonalisable) matrix.** The matrix \begin{align*} B = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \end{align*} has characteristic polynomial $\chi_B(\lambda) = (2 - \lambda)^2(3 - \lambda)$, so the eigenvalues are $\lambda_1 = 2$ with $a(2) = 2$ and $\lambda_2 = 3$ with $a(3) = 1$. We computed earlier that $g(2) = 1$. Since $g(2) < a(2)$, the diagonalisation criterion fails: $B$ is not diagonalisable. The eigenspace $E_2$ is "too small" to supply the two linearly independent eigenvectors needed to fill out a basis. Such a matrix is called **defective**. Understanding defective matrices requires the theory of [Jordan normal form](/theorems/864), which is developed in a separate article. [/example] ## Trace, Determinant, and Eigenvalues The characteristic polynomial encodes two familiar matrix invariants as elementary symmetric [functions](/page/Function) of the eigenvalues. [quotetheorem:922] To see why these relations hold, write the characteristic polynomial of $A \in M_{n \times n}(F)$ in factored form (over the algebraic closure of $F$): \begin{align*} \chi_A(\lambda) = (-1)^n(\lambda - \lambda_1)(\lambda - \lambda_2) \cdots (\lambda - \lambda_n), \end{align*} where $\lambda_1, \lambda_2, \ldots, \lambda_n$ are the eigenvalues counted with algebraic multiplicity. Expanding and comparing with $\chi_A(\lambda) = \det(A - \lambda I)$, the coefficient of $(-\lambda)^{n-1}$ gives $\lambda_1 + \lambda_2 + \cdots + \lambda_n = \operatorname{tr}(A)$, and setting $\lambda = 0$ gives $\chi_A(0) = \det(A) = (-1)^n(-\lambda_1)(-\lambda_2)\cdots(-\lambda_n) = \lambda_1 \lambda_2 \cdots \lambda_n$. These relations have important consequences. Since the trace and determinant can be expressed purely in terms of eigenvalues, and since similar matrices share the same eigenvalues (they share the same characteristic polynomial), the trace and determinant are **similarity invariants**: if $B = P^{-1}AP$, then $\operatorname{tr}(B) = \operatorname{tr}(A)$ and $\det(B) = \det(A)$. This also provides a quick check: a matrix is singular (non-invertible) if and only if at least one of its eigenvalues is zero. ## The Cayley--Hamilton Theorem One of the most striking results in linear algebra asserts that every matrix satisfies its own characteristic polynomial. [quotetheorem:865] The statement is deceptively simple but requires care in interpretation. For a matrix $A \in M_{n \times n}(F)$ with characteristic polynomial $\chi_A(\lambda) = \lambda^n + c_{n-1}\lambda^{n-1} + \cdots + c_1\lambda + c_0$, the Cayley--Hamilton theorem asserts that \begin{align*} A^n + c_{n-1}A^{n-1} + \cdots + c_1 A + c_0 I = O, \end{align*} where $O$ denotes the $n \times n$ zero matrix. Note that we substitute the matrix $A$ for the scalar variable $\lambda$ and replace the constant term $c_0$ by $c_0 I$ -- we are evaluating the polynomial in the ring $M_{n \times n}(F)$, not in $F$. [example:CayleyHamiltonApplication] **Using Cayley--Hamilton to compute a matrix inverse.** Let $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$. The characteristic polynomial is \begin{align*} \chi_A(\lambda) = \lambda^2 - 5\lambda - 2. \end{align*} By the Cayley--Hamilton theorem, $A^2 - 5A - 2I = O$, so $A^2 = 5A + 2I$. Rearranging: \begin{align*} A^2 - 5A &= 2I, \\ A(A - 5I) &= 2I, \\ A^{-1} &= \frac{1}{2}(A - 5I) = \frac{1}{2}\begin{pmatrix} -4 & 2 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}. \end{align*} This technique expresses $A^{-1}$ as a polynomial in $A$, which can be useful when $A$ has a special structure (e.g., sparse or banded) that makes matrix powers cheap to compute. [/example] The Cayley--Hamilton theorem also connects to the theory of the **minimal polynomial** $m_A(\lambda)$, the monic polynomial of smallest degree such that $m_A(A) = O$. The Cayley--Hamilton theorem guarantees that $m_A$ divides $\chi_A$, so the degree of $m_A$ is at most $n$. The minimal polynomial provides a finer invariant than the characteristic polynomial: two similar matrices must share the same minimal polynomial, and the eigenvalues of $A$ are precisely the roots of $m_A$. ## The Spectral Theorem The theory developed so far applies to arbitrary matrices over arbitrary fields, and we have seen that diagonalisability can fail. A remarkable feature of **Hermitian** (or real symmetric) matrices is that they are always diagonalisable, and moreover they admit an orthonormal basis of eigenvectors. [quotetheorem:924] The first property -- that all eigenvalues are real -- can be understood as follows. Let $A \in M_{n \times n}(\mathbb{C})$ be Hermitian ($A = A^*$, where $A^* = \overline{A}^T$ denotes the conjugate transpose), and suppose $Av = \lambda v$ for some nonzero $v \in \mathbb{C}^n$. Then \begin{align*} \lambda \langle v, v \rangle = \langle v, \lambda v \rangle = \langle v, Av \rangle = \langle A^*v, v \rangle = \langle Av, v \rangle = \langle \lambda v, v \rangle = \overline{\lambda} \langle v, v \rangle, \end{align*} where $\langle \cdot, \cdot \rangle$ denotes the standard Hermitian inner product on $\mathbb{C}^n$. Since $v \neq \mathbf{0}$, we have $\langle v, v \rangle > 0$, and dividing gives $\lambda = \overline{\lambda}$, so $\lambda \in \mathbb{R}$. The second property -- orthogonality of eigenspaces -- follows from a similar computation. If $Av = \lambda v$ and $Aw = \mu w$ with $\lambda \neq \mu$, then \begin{align*} \lambda \langle w, v \rangle = \langle w, Av \rangle = \langle Aw, v \rangle = \mu \langle w, v \rangle, \end{align*} using the fact that $\mu$ is real (so $\overline{\mu} = \mu$). Since $\lambda \neq \mu$, we conclude $\langle w, v \rangle = 0$. The full force of the spectral theorem asserts that diagonalisation can be achieved by a unitary change of basis. [quotetheorem:925] That is, for every Hermitian matrix $A \in M_{n \times n}(\mathbb{C})$, there exists a unitary matrix $U$ (satisfying $U^*U = I$) such that \begin{align*} U^* A U = \operatorname{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n), \end{align*} where $\lambda_1, \ldots, \lambda_n \in \mathbb{R}$ are the eigenvalues. The columns of $U$ form an orthonormal basis of eigenvectors. For real symmetric matrices ($A = A^T$ with $A \in M_{n \times n}(\mathbb{R})$), $U$ can be taken to be a real orthogonal matrix ($U^T U = I$). The spectral theorem has profound applications. In the theory of **quadratic forms**, a real symmetric matrix $A$ defines a quadratic form $Q(x) = x^T A x$, and diagonalising $A$ via an orthogonal change of coordinates reduces $Q$ to the sum $\sum_{i=1}^n \lambda_i y_i^2$. The signs of the eigenvalues then determine whether $Q$ is positive definite, negative definite, or indefinite -- a classification that governs the nature of critical points in multivariable calculus and optimisation theory. In quantum mechanics, observables are modelled by Hermitian operators on a [Hilbert space](/page/Hilbert%20Space), and the spectral theorem guarantees that their eigenvalues (the possible measurement outcomes) are real numbers, while the eigenvectors (the observable's eigenstates) form a complete orthonormal system. ## Problems [problem] Let $A = \begin{pmatrix} 3 & -1 \\ 1 & 1 \end{pmatrix}$. Find the eigenvalues and eigenspaces of $A$. Determine whether $A$ is diagonalisable and, if so, find an invertible matrix $P$ such that $P^{-1}AP$ is diagonal. **Difficulty: 1** [/problem] [solution] The characteristic polynomial is \begin{align*} \chi_A(\lambda) &= \det\begin{pmatrix} 3 - \lambda & -1 \\ 1 & 1 - \lambda \end{pmatrix} = (3 - \lambda)(1 - \lambda) + 1 = \lambda^2 - 4\lambda + 4 = (\lambda - 2)^2. \end{align*} So $\lambda = 2$ is the only eigenvalue, with algebraic multiplicity $a(2) = 2$. For the eigenspace: $A - 2I = \begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}$, which has rank $1$. Therefore $g(2) = \dim\ker(A - 2I) = 2 - 1 = 1$, with $E_2 = \operatorname{span}\!\left(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\right)$. Since $g(2) = 1 < 2 = a(2)$, the diagonalisation criterion fails. The matrix $A$ is not diagonalisable. [/solution] [problem] Let $A \in M_{n \times n}(F)$ be a matrix satisfying $A^2 = A$ (such a matrix is called **idempotent**). Prove that the only possible eigenvalues of $A$ are $0$ and $1$. **Difficulty: 1** [/problem] [solution] Let $\lambda$ be an eigenvalue of $A$ with eigenvector $v \neq \mathbf{0}$. Then $Av = \lambda v$. Applying $A$ to both sides: \begin{align*} A^2 v = A(\lambda v) = \lambda Av = \lambda^2 v. \end{align*} Since $A^2 = A$, we also have $A^2 v = Av = \lambda v$. Combining: \begin{align*} \lambda^2 v = \lambda v \implies (\lambda^2 - \lambda)v = \mathbf{0}. \end{align*} Since $v \neq \mathbf{0}$, we must have $\lambda^2 - \lambda = 0$, i.e., $\lambda(\lambda - 1) = 0$. Therefore $\lambda = 0$ or $\lambda = 1$. [/solution] [problem] Let $A \in M_{n \times n}(\mathbb{R})$ be a real symmetric matrix. Use the spectral theorem to prove that $A$ is positive definite (i.e., $x^T A x > 0$ for all $x \neq \mathbf{0}$) if and only if all eigenvalues of $A$ are strictly positive. **Difficulty: 2** [/problem] [solution] By the spectral theorem, there exists an orthogonal matrix $P$ such that $P^T A P = D = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)$ where $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $A$. For any $x \in \mathbb{R}^n$, set $y = P^T x$. Since $P$ is orthogonal, $y = \mathbf{0}$ if and only if $x = \mathbf{0}$. Then \begin{align*} x^T A x = x^T P D P^T x = (P^T x)^T D (P^T x) = y^T D y = \sum_{i=1}^n \lambda_i y_i^2. \end{align*} ($\Rightarrow$) Suppose $A$ is positive definite. For each $j$, take $x = Pe_j$ (so $y = e_j$). Then $x^T A x = \lambda_j > 0$. ($\Leftarrow$) Suppose all $\lambda_i > 0$. For any $x \neq \mathbf{0}$, the vector $y = P^T x \neq \mathbf{0}$, so at least one $y_i \neq 0$, and every term $\lambda_i y_i^2 \geq 0$ with at least one strictly positive. Hence $x^T A x = \sum_{i=1}^n \lambda_i y_i^2 > 0$. [/solution] ## References - Axler, S. *Linear Algebra Done Right*, 3rd edition, Springer, 2015. - Halmos, P. R. *Finite-Dimensional Vector Spaces*, Springer, 1987. - Strang, G. *Introduction to Linear Algebra*, 6th edition, Wellesley-Cambridge Press, 2023.