# Eigenvalues & the Spectral Theorem > Measurement outcomes *are* eigenvalues; the states you collapse into *are* eigenvectors. The > spectral theorem — that observables can be diagonalized in an orthonormal eigenbasis — is the > mathematical content of the measurement postulate, the reason phase estimation works, and the tool > that lets us define functions of operators (like $e^{-iHt}$). This is the technical heart of > Course 0.1. ## Learning Objectives After this lesson you will be able to: 1. Compute eigenvalues/eigenvectors via the characteristic polynomial and diagonalize a matrix. 2. Define **normal** operators and state the **spectral theorem** for them. 3. Write an operator in its **spectral decomposition** $A = \sum_k \lambda_k P_k$. 4. Explain **degeneracy** and project onto eigenspaces. 5. Prove that operators are **simultaneously diagonalizable iff they commute**. --- ## Intuition An **eigenvector** of $A$ is a special direction that $A$ merely stretches (by the **eigenvalue**), without rotating it: $A|v\rangle = \lambda|v\rangle$. If you can find a full basis of eigenvectors, the operator becomes trivial to understand — in that basis it's just "multiply coordinate $k$ by $\lambda_k$," a diagonal matrix. The spectral theorem says the operators of physics (Hermitian observables, unitary gates — all *normal*) always admit such an **orthonormal** eigenbasis. That's why we can always think of an observable as "a set of outcomes $\lambda_k$ each attached to a perpendicular measurement direction." --- ## Theory ### Eigenvalues and eigenvectors For an operator $A$ on $\mathbb{C}^n$, a nonzero $|v\rangle$ is an **eigenvector** with **eigenvalue** $\lambda\in\mathbb{C}$ if $$ A|v\rangle = \lambda|v\rangle \;\Longleftrightarrow\; (A - \lambda I)|v\rangle = 0 . $$ A nonzero solution exists iff $A-\lambda I$ is singular, i.e. iff $\lambda$ is a root of the **characteristic polynomial** $$ p_A(\lambda) = \det(A - \lambda I) = 0 . $$ Over $\mathbb{C}$, $p_A$ has degree $n$ and (by the fundamental theorem of algebra) exactly $n$ roots counted with multiplicity — *every* complex matrix has at least one eigenvalue. The **eigenspace** for $\lambda$ is $\ker(A-\lambda I)$; its dimension is the **geometric multiplicity**, and $\lambda$ is **degenerate** if that exceeds $1$. ### Diagonalization $A$ is **diagonalizable** if there is a basis of eigenvectors. Stacking them as columns of $V$, $$ A = V \Lambda V^{-1}, \qquad \Lambda = \operatorname{diag}(\lambda_1,\dots,\lambda_n). $$ From this, $\operatorname{Tr}(A) = \sum_k \lambda_k$ and $\det(A) = \prod_k \lambda_k$ (eigenvalues are the trace/determinant's "atoms"). Not every matrix is diagonalizable (e.g. $\begin{psmallmatrix}0&1\\0&0\end{psmallmatrix}$ is *defective*), but the operators we care about always are — because they are normal. ### Normal operators and the spectral theorem $A$ is **normal** if it commutes with its adjoint: $$ A^\dagger A = A A^\dagger . $$ Hermitian ($A^\dagger=A$) and unitary ($A^\dagger = A^{-1}$) operators are both normal. The payoff: > **Spectral theorem (finite-dimensional).** $A$ is normal **iff** it is **unitarily > diagonalizable**: there exists an orthonormal eigenbasis $\{|v_k\rangle\}$ and eigenvalues > $\lambda_k$ with > $$ A = \sum_k \lambda_k\,|v_k\rangle\langle v_k| = U\Lambda U^\dagger, \quad U = \sum_k |v_k\rangle\langle e_k|\ \text{unitary}. $$ **Proof of the key direction (normal ⇒ orthonormal eigenbasis), by induction on $n$.** Every operator on $\mathbb{C}^n$ has an eigenvalue $\lambda_1$ with a unit eigenvector $|v_1\rangle$ (root of $p_A$). Consider $W = \{|w\rangle : \langle v_1|w\rangle = 0\}$, the orthogonal complement of $|v_1\rangle$. **Claim: $W$ is invariant under $A$.** First, normality gives that $|v_1\rangle$ is also an eigenvector of $A^\dagger$ with eigenvalue $\overline{\lambda_1}$ (standard lemma: for normal $A$, $\|(A-\lambda I)x\| = \|(A^\dagger - \overline\lambda I)x\|$, so they share eigenvectors). Then for $|w\rangle\in W$: $\langle v_1|A w\rangle = \langle A^\dagger v_1|w\rangle = \lambda_1\langle v_1|w\rangle = 0$, so $A|w\rangle\in W$. The restriction $A|_W$ is normal on the $(n-1)$-dimensional $W$; by induction it has an orthonormal eigenbasis, which together with $|v_1\rangle$ gives one for all of $\mathbb{C}^n$. ∎ ### Spectral decomposition and eigenprojectors Grouping repeated eigenvalues, write the **distinct** eigenvalues $\lambda_k$ with orthogonal projectors $P_k$ onto their eigenspaces: $$ A = \sum_k \lambda_k P_k, \qquad P_k = \sum_{i:\,\lambda_i = \lambda_k}|v_i\rangle\langle v_i|, \qquad P_k P_l = \delta_{kl}P_k,\quad \sum_k P_k = I . $$ This is the form the measurement postulate uses (Term 1.3): outcome $\lambda_k$ occurs with probability $\langle\psi|P_k|\psi\rangle$ and collapses the state to $P_k|\psi\rangle/\|P_k|\psi\rangle\|$. ### Functions of operators The spectral decomposition lets us apply any function $f:\mathbb{C}\to\mathbb{C}$ to a normal operator: $$ f(A) := \sum_k f(\lambda_k)\,P_k . $$ In particular $e^{-iHt} = \sum_k e^{-i\lambda_k t}P_k$ — the engine of time evolution (Term 1.6) and of building gates from Hamiltonians. We develop this fully in [0.1.6](06-special-operators.md). ### Simultaneous diagonalization > **Theorem.** Two normal operators $A,B$ are simultaneously diagonalizable (share an orthonormal > eigenbasis) **iff** $[A,B] = 0$. **Proof (⇐, the useful direction).** If they share an eigenbasis $\{|v_k\rangle\}$ with $A|v_k\rangle = a_k|v_k\rangle$, $B|v_k\rangle = b_k|v_k\rangle$, then $AB|v_k\rangle = a_k b_k|v_k\rangle = BA|v_k\rangle$ on a basis, so $AB = BA$. Conversely if $[A,B]=0$, $B$ preserves each eigenspace of $A$ (if $A|v\rangle = a|v\rangle$ then $A(B|v\rangle) = BA|v\rangle = aB|v\rangle$, so $B|v\rangle$ is in the same eigenspace); diagonalizing $B$ within each eigenspace of $A$ yields a common eigenbasis. ∎ Physically: **commuting observables are jointly measurable** with definite simultaneous values — the formal counterpart of the commutator discussion in [0.1.4](04-linear-operators-matrices.md). --- ## Worked Examples ### Example 1 — Diagonalizing Pauli-$X$ Characteristic polynomial of $X = \begin{psmallmatrix}0&1\\1&0\end{psmallmatrix}$: $\det(X-\lambda I) = \lambda^2 - 1 = 0 \Rightarrow \lambda = \pm 1$. For $\lambda = +1$: $(X-I)|v\rangle = 0 \Rightarrow$ components equal $\Rightarrow |v_+\rangle = \tfrac1{\sqrt2}(1,1)^T = |+\rangle$. For $\lambda = -1$: $|v_-\rangle = \tfrac1{\sqrt2}(1,-1)^T = |-\rangle$. These are orthonormal (as the spectral theorem guarantees for the Hermitian $X$). Spectral decomposition: $$ X = (+1)|+\rangle\langle+| + (-1)|-\rangle\langle-| . $$ Check trace $= 1 - 1 = 0$ ✓ and det $= (1)(-1) = -1$ ✓, matching [0.1.4](04-linear-operators-matrices.md). ### Example 2 — A degenerate operator and its eigenprojector Let $A = \begin{psmallmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 5\end{psmallmatrix}$. Eigenvalue $\lambda = 2$ is **doubly degenerate** with eigenspace $\operatorname{span}\{|0\rangle,|1\rangle\}$, and $\lambda = 5$ with eigenvector $|2\rangle$. The eigenprojectors are $$ P_2 = |0\rangle\langle0| + |1\rangle\langle1| = \operatorname{diag}(1,1,0),\quad P_5 = |2\rangle\langle2| = \operatorname{diag}(0,0,1), $$ with $A = 2P_2 + 5P_5$, $P_2 + P_5 = I$, $P_2 P_5 = 0$. Note the eigenvectors *within* the degenerate space are not unique — any orthonormal pair spanning it works — but the **projector** $P_2$ is unique. --- ## Hands-on (Python) ```python import numpy as np X = np.array([[0, 1], [1, 0]], dtype=complex) # For Hermitian/normal matrices, use eigh: returns REAL, sorted eigenvalues and an # ORTHONORMAL eigenbasis (columns of vecs). Reserve np.linalg.eig for general matrices. vals, vecs = np.linalg.eigh(X) print(vals) # [-1. 1.] print(np.round(vecs, 3)) # columns are |->, |+> (orthonormal) # Verify diagonalization A = V Λ V†: Lam = np.diag(vals) print(np.allclose(vecs @ Lam @ vecs.conj().T, X)) # True # Spectral decomposition A = Σ λ_k |v_k>Solution Already diagonal: eigenvalues $+1$ (eigenvector $|0\rangle$) and $-1$ ($|1\rangle$). Spectral form $Z = (+1)|0\rangle\langle0| + (-1)|1\rangle\langle1|$. **E2 (easy).** Show that the eigenvalues of any **projector** $P = P^\dagger = P^2$ are $0$ or $1$.
Solution If $P|v\rangle = \lambda|v\rangle$ then $P^2|v\rangle = \lambda^2|v\rangle$; but $P^2 = P$ gives $\lambda^2 = \lambda$, so $\lambda\in\{0,1\}$. (Eigenvalue $1$: vectors in the subspace; $0$: its orthogonal complement.)
**E3 (medium).** Prove that a Hermitian operator has **real** eigenvalues and that eigenvectors for distinct eigenvalues are orthogonal.
Solution Let $A=A^\dagger$, $A|v\rangle = \lambda|v\rangle$ with $\langle v|v\rangle = 1$. Then $\lambda = \langle v|A|v\rangle = \overline{\langle v|A^\dagger|v\rangle} = \overline{\langle v|A|v\rangle} = \overline\lambda$, so $\lambda\in\mathbb{R}$. For $A|v\rangle = a|v\rangle$, $A|w\rangle = b|w\rangle$ with $a\neq b$: $a\langle w|v\rangle = \langle w|A|v\rangle = \langle Aw|v\rangle = b\langle w|v\rangle$ (using $b$ real). So $(a-b)\langle w|v\rangle = 0 \Rightarrow \langle w|v\rangle = 0$. ∎
**E4 (medium).** Diagonalize the Hadamard $H = \tfrac1{\sqrt2}\begin{psmallmatrix}1&1\\1&-1\end{psmallmatrix}$. What are its eigenvalues?
Solution $\det(H-\lambda I) = (\tfrac1{\sqrt2}-\lambda)(-\tfrac1{\sqrt2}-\lambda) - \tfrac12 = \lambda^2 - 1 = 0$, so $\lambda = \pm1$ (consistent with $H^2 = I$). $H$ is Hermitian and unitary, an involution; its eigenvectors are the two fixed/anti-fixed directions of the reflection it represents (numerically, $\approx(0.924, 0.383)^T$ for $+1$ and $(-0.383, 0.924)^T$ for $-1$).
**E5 (hard).** Suppose normal operators $A,B$ satisfy $[A,B]=0$ and $A$ has *non-degenerate* spectrum. Prove that every eigenvector of $A$ is also an eigenvector of $B$, hence $B = f(A)$ for some function $f$.
Solution If $A|v\rangle = a|v\rangle$ with $a$ non-degenerate, then $A(B|v\rangle) = BA|v\rangle = aB|v\rangle$, so $B|v\rangle$ lies in the (1-dimensional) eigenspace of $A$ for $a$; thus $B|v\rangle = b|v\rangle$ for some scalar $b$ — $|v\rangle$ is an eigenvector of $B$. Defining $f(a) := b$ on the spectrum of $A$ gives $B = \sum_a f(a)|v_a\rangle\langle v_a| = f(A)$. ∎ (This is why, for a system with a non-degenerate Hamiltonian, every conserved quantity is a function of the energy.)
--- ## Checkpoint 1. What equation defines an eigenvalue, and how does the characteristic polynomial find them? 2. Define a *normal* operator and state the spectral theorem. 3. Write a general spectral decomposition and give the eigenprojector properties. 4. How do you define $f(A)$ for a normal $A$? Give the example used for time evolution. 5. State the commute ⟺ simultaneously-diagonalizable theorem and its physical meaning.
Answers 1. $A|v\rangle = \lambda|v\rangle$, i.e. $\det(A-\lambda I)=0$; the roots of that degree-$n$ polynomial are the eigenvalues. 2. $A^\dagger A = A A^\dagger$; then $A$ has an orthonormal eigenbasis and $A = \sum_k\lambda_k|v_k\rangle\langle v_k| = U\Lambda U^\dagger$. 3. $A = \sum_k\lambda_k P_k$ with $P_k P_l = \delta_{kl}P_k$, $\sum_k P_k = I$, $P_k$ Hermitian. 4. $f(A) = \sum_k f(\lambda_k)P_k$; e.g. $e^{-iHt} = \sum_k e^{-i\lambda_k t}P_k$. 5. Normal $A,B$ share an orthonormal eigenbasis iff $[A,B]=0$; physically, commuting observables are simultaneously measurable with definite joint values.
--- ## Further Reading - **[Axl]** Axler, *Linear Algebra Done Right*, Ch. 5, 7 — eigenvalues, the spectral theorem (a clean, determinant-free treatment). - **[NC]** Nielsen & Chuang, §2.1.7–2.1.8 — eigendecomposition and the spectral decomposition. - **[HJ]** Horn & Johnson, *Matrix Analysis*, Ch. 1, 2 — normal matrices, Schur/spectral theorems. --- ← Prev: [Linear Operators & Matrices](04-linear-operators-matrices.md) · Up: [Term 0](../README.md) · Next: [Special Operators](06-special-operators.md) →