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Derivation

Eigenvalues as Roots of the Characteristic Polynomial

D-080 Home PU-104 Threads symmetry Depends on Determinant as the Unique Alternating Multilinear Form
Statement

Let \(A\) be a linear operator on a finite-dimensional vector space \(V\), represented in some basis by an \(n\times n\) matrix. A scalar \(\lambda\) is an eigenvalue of \(A\) — meaning \(A\mathbf{v}=\lambda\mathbf{v}\) for some nonzero \(\mathbf{v}\) — if and only if \(\lambda\) is a root of the characteristic polynomial \(p_A(\lambda)=\det(A-\lambda I)\). Moreover \(p_A\) is invariant under change of basis, so the eigenvalues are properties of the operator itself, not of the matrix that happens to represent it.

Why it matters

This result is the bridge from the geometric idea of an eigenvector — a direction the operator merely rescales — to a concrete computational recipe: convert the eigenvalue problem into finding the roots of a single scalar polynomial. Every diagonalisation, every normal-mode frequency of a coupled system, and every quantised energy level of an \(n\)-state quantum system is ultimately extracted this way.

The basis-independence is the deeper payoff. It certifies that "the spectrum of \(A\)" is a well-posed notion: a physicist may compute in whatever basis is convenient and still report the same eigenvalues an experiment would return. The invariant coefficients of \(p_A\) — most notably the trace and determinant — are themselves frame-independent observables, tying this algebraic fact directly to the symmetry thread's message that physics lives in invariants.

Assumptions
The space is finite-dimensional, \(\dim V=n<\infty\).Without finite dimension the determinant is undefined and \(p_A\) does not exist; the spectrum can then contain continuous or residual parts that are not eigenvalues at all, so the polynomial characterisation collapses.
The scalars form a field (typically \(\mathbb{R}\) or \(\mathbb{C}\)).Over a mere ring, \(\det(A-\lambda I)=0\) need not force a nonzero kernel, so the equivalence "singular \(\Leftrightarrow\) has an eigenvector" fails and the whole argument unravels.
A basis is fixed so that \(A\) is a genuine matrix.The map \(\lambda\mapsto\det(A-\lambda I)\) needs a matrix. Eigenvalues still exist abstractly without a basis, but the polynomial recipe requires the finite matrix form.
Roots are counted in an algebraically closed field for the count \(n\) to hold.Over \(\mathbb{R}\) a real matrix may have fewer than \(n\) real eigenvalues, because complex-conjugate root pairs stay hidden; the statement "exactly \(n\) roots with multiplicity" is a fact about \(\mathbb{C}\), guaranteed by the fundamental theorem of algebra.
Derivation
1
\[ A\mathbf{v}=\lambda\mathbf{v},\qquad \mathbf{v}\neq\mathbf{0} \]
Definition of an eigenvalue–eigenvector pair: a nonzero vector that \(A\) merely rescales. A
2
\[ A\mathbf{v}-\lambda\mathbf{v}=\mathbf{0}\;\Longrightarrow\;(A-\lambda I)\,\mathbf{v}=\mathbf{0} \]
Move both terms to one side and factor, writing \(\lambda\mathbf{v}=\lambda I\mathbf{v}\); the identity \(I\) is essential so both terms are operators acting on \(\mathbf{v}\). A
3
\[ \lambda\ \text{is an eigenvalue}\iff \ker(A-\lambda I)\neq\{\mathbf{0}\} \]
By Step 2 an eigenvector is exactly a nonzero element of \(\ker(A-\lambda I)\); demanding \(\mathbf{v}\neq\mathbf{0}\) is demanding a nontrivial kernel. A
4
\[ \ker(A-\lambda I)\neq\{\mathbf{0}\}\iff (A-\lambda I)\ \text{is not invertible} \]
Rank–nullity on a finite-dimensional space: an operator is injective iff surjective iff invertible, so a nontrivial kernel is precisely failure of invertibility. B
5
\[ (A-\lambda I)\ \text{not invertible}\iff \det(A-\lambda I)=0 \]
By the determinant's multilinear characterisation — the unique alternating multilinear form on the columns with \(\det I=1\) — the determinant vanishes exactly when the columns are linearly dependent, i.e. when the operator is singular. (Uses determinant-multilinear-characterization.) C
6
\[ p_A(\lambda)\;\equiv\;\det(A-\lambda I)\quad\Longrightarrow\quad \lambda\in\operatorname{spec}(A)\iff p_A(\lambda)=0 \]
Define the characteristic polynomial and chain Steps 3–5. This is the eigenvalue-as-root equivalence. A
7
\[ p_A(\lambda)=(-1)^n\lambda^n+(-1)^{n-1}(\operatorname{tr}A)\,\lambda^{n-1}+\cdots+\det A \]
Expand the determinant by the multilinear (Leibniz) rule. Each matrix entry is at most linear in \(\lambda\) — only the diagonal carries \(-\lambda\) — so \(p_A\) has degree exactly \(n\), with leading term from the product of the diagonal factors \((a_{ii}-\lambda)\) and constant term \(p_A(0)=\det A\). C
8
\[ \det(S^{-1}AS-\lambda I)=\det\!\big(S^{-1}(A-\lambda I)S\big)=\det(A-\lambda I) \]
Under a change of basis \(A\mapsto S^{-1}AS\), and \(\lambda I=S^{-1}(\lambda I)S\). Multiplicativity of the determinant gives \(\det(S^{-1})\det(A-\lambda I)\det S\), and \(\det(S^{-1})\det S=1\). Hence \(p_A\) — and every one of its coefficients and roots — is basis-independent. C
Result
\[ \boxed{\;\lambda\in\operatorname{spec}(A)\iff p_A(\lambda)=\det(A-\lambda I)=0,\qquad p_A\ \text{invariant under}\ A\mapsto S^{-1}AS\;} \]

Reading. The eigenvalues of \(A\) are exactly the roots of one degree-\(n\) polynomial built from \(A\), and that polynomial is identical in every basis. The spectrum is therefore an intrinsic fingerprint of the operator: similar matrices share the same characteristic polynomial, hence the same eigenvalues with the same algebraic multiplicities. The invariant coefficients \(\operatorname{tr}A=\sum_i\lambda_i\) and \(\det A=\prod_i\lambda_i\) are frame-independent scalars.

Units check. The subtraction \(a_{ii}-\lambda\) is only meaningful if \(\lambda\) carries the same units as the entries of \(A\), so \(p_A\) is dimensionally homogeneous of total degree \(n\) in those units. For a Hamiltonian matrix (entries in joules) the roots come out in joules — energies; for a linear rate matrix (entries in \(\mathrm{s^{-1}}\)) the eigenvalues are inverse-time growth/decay rates governing \(e^{\lambda t}\).

Limiting cases
  • Diagonal \(A=\operatorname{diag}(d_1,\dots,d_n)\): \(p_A(\lambda)=\prod_i(d_i-\lambda)\); eigenvalues are the diagonal entries, read off directly.
  • Triangular \(A\): the determinant is the product of the diagonal, so \(p_A(\lambda)=\prod_i(a_{ii}-\lambda)\) regardless of off-diagonal content — the spectrum sits on the diagonal.
  • \(n=1\): \(p_A(\lambda)=a_{11}-\lambda\), the single root \(\lambda=a_{11}\); a scalar is its own eigenvalue.
  • Identity \(A=I\): \(p_A(\lambda)=(1-\lambda)^n\), one eigenvalue \(\lambda=1\) of algebraic multiplicity \(n\).
Breaks when
  • Infinite dimension. On an infinite-dimensional Hilbert space there is no determinant and no characteristic polynomial. The spectrum can include continuous and residual parts that are not eigenvalues (e.g. the position operator has no normalisable eigenvectors), so the polynomial recipe simply does not apply.
  • Real field, complex spectrum. A real \(90^\circ\) rotation \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) has \(p_A(\lambda)=\lambda^2+1\) with no real roots. The equivalence still holds over \(\mathbb{C}\), but "there exists a real eigenvalue" fails — there is no invariant real direction.
  • Algebraic vs geometric multiplicity. The polynomial locates eigenvalues and their algebraic multiplicities, but a root of multiplicity \(m\) need not supply \(m\) independent eigenvectors. Defective (non-diagonalisable) matrices such as a Jordan block violate any attempt to count eigenvectors from root multiplicity alone.
Failure modes
  • Dropping the \(I\). Writing \(\det(A-\lambda)=0\) is meaningless — one cannot subtract a scalar from a matrix. The scalar must be promoted to \(\lambda I\).
  • Sign and degree slips. Forgetting that each diagonal factor contributes \(-\lambda\) leads to a wrong leading coefficient; it is \((-1)^n\), and \(p_A(0)=\det A\), not \(1\).
  • Confusing algebraic and geometric multiplicity. Assuming a double root guarantees two independent eigenvectors; this fails for defective matrices.
  • Believing the eigenvector is basis-independent. Only the eigenvalue is invariant under \(A\mapsto S^{-1}AS\); the eigenvector components transform as \(\mathbf{v}\mapsto S^{-1}\mathbf{v}\).
  • Setting individual entries to zero. The condition is the single scalar equation \(\det(A-\lambda I)=0\), not the vanishing of separate matrix entries.
Discussion

The derivation collapses an existential statement — "there exists a nonzero \(\mathbf{v}\) with \(A\mathbf{v}=\lambda\mathbf{v}\)" — into a purely algebraic one, "\(p_A(\lambda)=0\)." The bridge is the determinant acting as a singularity detector: as the unique alternating multilinear normalised form on the columns, it vanishes exactly when those columns are linearly dependent. This is why the multilinear characterisation is the load-bearing prior result — it is precisely what makes "singular" and "\(\det=0\)" one and the same statement.

Basis-independence promotes the polynomial from a mere computational device to a genuine invariant. Because \(p_A(\lambda)=\det(A-\lambda I)\) is unchanged under similarity, so are all of its coefficients. Two of them are invariants physicists meet constantly: the trace \(\operatorname{tr}A=\sum_i\lambda_i\) and the determinant \(\det A=\prod_i\lambda_i\). These symmetric functions of the eigenvalues are coordinate-free scalars — observable content that survives any change of frame, exactly the symmetry thread's central claim.

At the highest level of rigour the characteristic polynomial sits inside a richer web of invariants governed by the Cayley–Hamilton theorem, \(p_A(A)=0\): the operator satisfies its own characteristic equation. This links the spectrum to the minimal polynomial, whose roots are the same eigenvalues but with possibly smaller multiplicities; the gap between the two encodes the Jordan structure — precisely the defective behaviour the raw root count cannot see. The characteristic polynomial thus pins down the eigenvalues completely, but the operator only up to the finer data of its generalised eigenspaces.

Common misconceptions. The characteristic polynomial gives eigenvalues, not eigenvectors: the vectors are found afterwards by solving \((A-\lambda I)\mathbf{v}=\mathbf{0}\) for each root. And a shared characteristic polynomial does not imply similarity — \(\begin{pmatrix}1&1\\0&1\end{pmatrix}\) and \(\begin{pmatrix}1&0\\0&1\end{pmatrix}\) share \(p(\lambda)=(1-\lambda)^2\) yet are not similar, since only the identity is similar to the identity.

Worked examples
1
\[ A=\begin{pmatrix}2&1\\1&2\end{pmatrix},\qquad p_A(\lambda)=\det(A-\lambda I) \]
Symmetric coupling matrix — e.g. two identical oscillators with equal coupling. Set up the characteristic determinant. A
2
\[ p_A(\lambda)=(2-\lambda)(2-\lambda)-(1)(1)=\lambda^2-4\lambda+3 \]
Expand the \(2\times2\) determinant. Consistency checks: \(p_A(0)=3=\det A\) and the \(\lambda\)-coefficient is \(-\operatorname{tr}A=-4\). A
3
\[ \lambda^2-4\lambda+3=(\lambda-1)(\lambda-3)=0\;\Rightarrow\;\lambda=1,\ 3 \]
Factor the quadratic. Roots are real and distinct because \(A\) is real symmetric. A
\[ \boxed{\;\lambda_1=1,\quad\lambda_2=3\;}\]

Reading. The two normal modes have effective stiffnesses 1 and 3 in the matrix's units. The sum \(1+3=4=\operatorname{tr}A\) and product \(1\cdot3=3=\det A\) confirm the invariants.

Units check. If the entries are in \(\mathrm{N/m}\), both eigenvalues are in \(\mathrm{N/m}\) — effective spring constants of the decoupled modes.

1
\[ A=\begin{pmatrix}0&-\omega\\ \omega&0\end{pmatrix},\qquad p_A(\lambda)=\det(A-\lambda I) \]
Generator of rotation / harmonic phase flow, with \(\omega>0\) an angular frequency in \(\mathrm{s^{-1}}\). B
2
\[ p_A(\lambda)=(-\lambda)(-\lambda)-(-\omega)(\omega)=\lambda^2+\omega^2 \]
Expand: diagonal product \(\lambda^2\) plus \(\omega^2\) from the off-diagonal. Here \(\operatorname{tr}A=0\) so there is no \(\lambda^1\) term, and \(\det A=\omega^2\). B
3
\[ \lambda^2+\omega^2=0\;\Rightarrow\;\lambda=\pm i\omega \]
Over \(\mathbb{R}\) there are no roots — the flow rotates every vector, leaving no invariant real direction; over \(\mathbb{C}\) the roots form a conjugate pair. C
\[ \boxed{\;\lambda_\pm=\pm i\omega\;}\]

Reading. The purely imaginary eigenvalues signal undamped oscillation: solutions of \(\dot{\mathbf{x}}=A\mathbf{x}\) evolve as \(e^{\pm i\omega t}\), sinusoids of angular frequency \(\omega\). The absence of a real root illustrates the real-field break.

Units check. Entries are in \(\mathrm{s^{-1}}\); the eigenvalues \(\pm i\omega\) are in \(\mathrm{s^{-1}}\), correctly the rate in \(e^{\lambda t}\).

Problems
  1. Compute the characteristic polynomial and eigenvalues of \(A=\begin{pmatrix}3&0\\0&5\end{pmatrix}\), and verify \(\operatorname{tr}A\) and \(\det A\) against the roots.
    Solution \(p_A(\lambda)=(3-\lambda)(5-\lambda)=\lambda^2-8\lambda+15\); roots \(\lambda=3,5\). Check: \(\operatorname{tr}A=8=3+5\), \(\det A=15=3\cdot5\). Diagonal case — the eigenvalues are the diagonal entries.
  2. Find the eigenvalues of the triangular matrix \(A=\begin{pmatrix}2&7\\0&4\end{pmatrix}\). Does the off-diagonal \(7\) affect them?
    Solution \(p_A(\lambda)=(2-\lambda)(4-\lambda)-0\cdot7=(2-\lambda)(4-\lambda)\); roots \(\lambda=2,4\). The \(7\) does not affect the eigenvalues — for triangular matrices the spectrum is the diagonal; the off-diagonal only changes the eigenvectors.
  3. Show explicitly that \(A=\begin{pmatrix}2&1\\1&2\end{pmatrix}\) and \(B=S^{-1}AS\) with \(S=\begin{pmatrix}1&1\\0&1\end{pmatrix}\) have the same characteristic polynomial.
    Solution \(S^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}\). Then \(AS=\begin{pmatrix}2&3\\1&3\end{pmatrix}\) and \(B=S^{-1}AS=\begin{pmatrix}1&0\\1&3\end{pmatrix}\). So \(p_B(\lambda)=(1-\lambda)(3-\lambda)-0\cdot1=\lambda^2-4\lambda+3\), identical to \(p_A\); roots \(1,3\), confirming basis-independence.
  4. A \(3\times3\) matrix has \(p_A(\lambda)=-\lambda^3+6\lambda^2-11\lambda+6\). Find its eigenvalues, trace, and determinant without knowing \(A\).
    Solution Factor: \(-\lambda^3+6\lambda^2-11\lambda+6=-(\lambda-1)(\lambda-2)(\lambda-3)\); roots \(\lambda=1,2,3\). Trace \(=\) sum of roots \(=6\); determinant \(=p_A(0)=6=1\cdot2\cdot3\). Both consistent with the coefficient relations.
  5. The matrix \(A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\) has a repeated eigenvalue. Find it, determine its geometric multiplicity, and explain why the polynomial alone did not reveal that.
    Solution \(p_A(\lambda)=(1-\lambda)^2\), so \(\lambda=1\) with algebraic multiplicity 2. Solving \((A-I)\mathbf{v}=\mathbf{0}\) gives \(\begin{pmatrix}0&1\\0&0\end{pmatrix}\mathbf{v}=\mathbf{0}\Rightarrow v_2=0\), so eigenvectors are multiples of \((1,0)^T\) — geometric multiplicity 1. The matrix is defective (a Jordan block). The characteristic polynomial reports algebraic multiplicity but cannot count independent eigenvectors, which requires solving the kernel.