Target Audience: Anyone comfortable finding eigenvalues who wants the
structure theory that explains when the usual recipe stops working.
Prerequisites:Vector Spaces and Linear Maps
and
Determinants. Rank, kernel and the determinant are used constantly, and the characteristic
polynomial is a determinant.
How to Read This Page
This page has one question at its centre:
when can a matrix be diagonalised, and what happens when it cannot? Sections 1 to
4 answer the first half. Sections 5 to 8 answer the second, and end at the Jordan
canonical form — the honest replacement for diagonalisation when no basis of
eigenvectors exists. Sections 9 and 10 then find the class of matrices where the problem
never arises at all.
The
explorer in Section 8
carries the weight. Given a matrix it finds the eigenvalues, measures every generalised
eigenspace, reads the Jordan block sizes off the resulting kernel sequence, and then
checks its own answer — that the block sizes total the algebraic
multiplicity, that their count is the geometric multiplicity, and that the assembled
Jordan form has the same characteristic polynomial and kernel sequence as the matrix it
came from.
All of it is exact rational arithmetic. Eigenvalue work is where floating point does the
most damage: a defective matrix and a diagonalisable one differ by a rank test, and a rank
test against accumulated rounding error is a coin toss.
1. Eigenvalues and Eigenvectors
Definition
A scalar \(\lambda\) is an eigenvalue of \(A\) when there is a
non-zero vector \(\mathbf{v}\) with \[ A\mathbf{v} = \lambda\mathbf{v}.
\] Such a \(\mathbf{v}\) is an eigenvector for \(\lambda\).
The condition \(\mathbf{v} \neq \mathbf{0}\) is part of the definition, not a technical
aside. Without it every scalar would be an eigenvalue, since \(A\mathbf{0} =
\lambda\mathbf{0}\) holds for all \(\lambda\).
What an eigenvector is, geometrically
A direction the map does not turn. On that line \(A\) acts purely by scaling, with
factor \(\lambda\) — stretching for \(\lambda \gt 1\), shrinking for \(0 \lt
\lambda \lt 1\), and flipping for \(\lambda \lt 0\). Everything else the matrix does to
space is rotation and shear; the eigenvectors are where it does neither.
Collecting all the vectors for one \(\lambda\) gives the eigenspace \[
E_\lambda = \{\mathbf{v} : A\mathbf{v} = \lambda\mathbf{v}\} = \ker(A - \lambda I), \]
which is a subspace. So an eigenvector is never unique: any non-zero multiple works, and
so does any combination of eigenvectors for the same eigenvalue.
Rewriting the definition as a kernel
\(A\mathbf{v} = \lambda\mathbf{v}\) with \(\mathbf{v} \neq \mathbf{0}\) is the same as
\((A - \lambda I)\mathbf{v} = \mathbf{0}\) with \(\mathbf{v} \neq \mathbf{0}\), which is
the same as \(\ker(A - \lambda I) \neq \{\mathbf{0}\}\). That single rewriting turns an
eigenvalue question into a rank question, and it is the hinge of the next section.
On the site
Eigenvalues and the Characteristic Polynomial
proves the equivalence above, the existence of eigenvalues over \(\mathbb{C}\), the
independence of eigenvectors for distinct eigenvalues, and a diagonalisability
criterion.
2. The Characteristic Polynomial
A kernel is non-trivial exactly when the determinant vanishes. So the eigenvalues of \(A\)
are precisely the roots of
\[ p(x) = \det(xI - A), \]
the characteristic polynomial, which is monic of degree \(n\) by
construction.
Worked: a matrix that hides its eigenvalues
The \(4 \times 4\) matrix with rows \((0,1,0,0)\), \((0,0,0,0)\), \((0,-3,3,1)\),
\((0,1,-1,5)\) has diagonal entries \(0, 0, 3, 5\) — and eigenvalues \(0\) and \(4\).
Its characteristic polynomial is \(x^4 - 8x^3 + 16x^2 = x^2(x - 4)^2\). Only a
triangular matrix gives its eigenvalues away on the diagonal; in general the diagonal
tells you nothing, and the polynomial has to be computed.
Do eigenvalues always exist?
Over \(\mathbb{C}\), yes. Every non-constant complex polynomial has a root, so \(p(x)\)
has one and every complex square matrix has at least one eigenvalue.
Over \(\mathbb{R}\), no. Rotation by \(90^\circ\) has \(p(x) = x^2 + 1\), no real root,
and no fixed direction — which is geometrically obvious the moment it is stated.
The Fundamental Theorem of Algebra
is what guarantees the complex case, and it is the reason the theory in Sections 6 to 8
is stated over \(\mathbb{C}\).
Two free checks
\(\operatorname{tr} A = \sum \lambda_i\) and \(\det A = \prod \lambda_i\), both counted
with multiplicity. They cost nothing and catch most arithmetic slips: if the eigenvalues
do not add to the trace, something is wrong before any eigenvector is computed.
3. Two Multiplicities
An eigenvalue can be counted two different ways, and the whole of the rest of this page
lives in the gap between them.
Definition
What it counts
Algebraic
Multiplicity of \(\lambda\) as a root of \(p(x)\)
How many times the polynomial says \(\lambda\) is there
Geometric
\(\dim\ker(A - \lambda I)\)
How many independent eigenvectors \(\lambda\) actually supplies
The lower bound holds because an eigenvalue has at least one eigenvector by definition.
When the upper bound is strict for some eigenvalue, the matrix is called
defective.
The standard example
\(\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}\) has \(p(x) = (x - 2)^2\), so
\(\lambda = 2\) has algebraic multiplicity \(2\). But \(A - 2I = \begin{pmatrix} 0 &
1 \\ 0 & 0 \end{pmatrix}\) has rank \(1\), so its kernel is one-dimensional and the
geometric multiplicity is \(1\). One eigenvector, two dimensions to fill: this matrix is
defective, and it is the smallest example there is.
A repeated eigenvalue is not the problem
It is easy to absorb the rule as “repeated eigenvalues cause trouble”. They
do not. The identity matrix repeats one eigenvalue \(n\) times and is already diagonal;
so is \(\operatorname{diag}(2, 2, 5)\). What causes trouble is a
shortage of eigenvectors, and the only way to know is to compute the kernel.
4. Diagonalisability
\(A\) is diagonalisable when \(A = PDP^{-1}\) for some diagonal \(D\).
Equivalently — and more usefully — when \(A\) has \(n\) independent eigenvectors: they
become the columns of \(P\), and the eigenvalues sit along \(D\) in the matching order.
The test
\(A\) is diagonalisable exactly when geometric multiplicity equals algebraic
multiplicity for every eigenvalue. Checking one is not enough.
Distinct eigenvalues: sufficient, not necessary
\(n\) distinct eigenvalues force diagonalisability, because eigenvectors for distinct
eigenvalues are independent. The argument is worth remembering: if a non-trivial
combination vanished, applying \(A - \lambda_j I\) would annihilate the \(j\)th term
while scaling the others by non-zero factors, giving a shorter non-trivial relation —
and repeating that leaves a non-zero multiple of a single eigenvector equal to
\(\mathbf{0}\).
Why anyone wants this
\(A = PDP^{-1}\) gives \(A^{k} = PD^{k}P^{-1}\), and \(D^{k}\) merely raises each
diagonal entry to the \(k\)th power. The middle terms telescope.
So iteration, matrix exponentials and systems of linear differential equations all
decouple into \(n\) independent one-dimensional problems in an eigenbasis. The system
\(\dfrac{d\mathbf{x}}{dt} = A\mathbf{x}\) becomes \(\dfrac{dy_i}{dt} = \lambda_i y_i\),
each solving as \(y_i = c_i e^{\lambda_i t}\).
5. The Minimal Polynomial
The characteristic polynomial has a blind spot. Compare \(\operatorname{diag}(2, 2)\) with
\(\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}\): both have \(p(x) = (x - 2)^2\),
yet the first is diagonal and the second is defective. Something finer is needed.
Definition
\(m_A(x)\) is the monic polynomial of least degree with \(m_A(A) = 0\).
It exists because the powers \(I, A, A^2, \ldots\) cannot remain independent forever in a
space of dimension \(n^2\), so some combination of them vanishes.
Divisibility, and what it gives
\(m_A\) divides every polynomial that annihilates \(A\). The proof is one
division: if \(f(A) = 0\), write \(f = qm_A + r\) with \(\deg r \lt \deg m_A\); then
\(r(A) = 0\) too, and minimality forces \(r = 0\).
By Cayley–Hamilton — every matrix satisfies its own
characteristic polynomial, \(p_A(A) = 0\) — it follows that \(m_A\) divides \(p_A\).
They have the same roots, namely the eigenvalues; what differs is the exponents.
The Minimal Polynomial
proves existence, divisibility, uniqueness, and that its roots are exactly the
eigenvalues.
The detector
\(A\) is diagonalisable exactly when \(m_A\) has no repeated root.
The blind spot, resolved
For \(\operatorname{diag}(2,2)\), \(m_A(x) = x - 2\) — no repeated root, diagonalisable.
For the Jordan block, \(m_A(x) = (x - 2)^2\) — repeated root, defective. The
characteristic polynomial could not separate them; the minimal polynomial does it
immediately.
Do not “prove” Cayley–Hamilton by substitution
Substituting \(A\) for \(x\) in \(\det(xI - A)\) to get \(\det(A - A) = 0\) is tempting
and meaningless: it puts a matrix where a scalar belongs, and the determinant of a
matrix of matrices is not defined. The theorem is true, but it needs a real proof.
6. The Primary Decomposition
If a matrix is short of eigenvectors, the fix is to widen what counts as one.
Generalised eigenvectors
\(\mathbf{v}\) is a generalised eigenvector for \(\lambda\) when \((A - \lambda
I)^{k}\mathbf{v} = \mathbf{0}\) for some \(k \geq 1\). The
generalised eigenspace is \(G_\lambda = \ker(A - \lambda I)^{n}\).
The point of the widening is a dimension count. The ordinary eigenspace has dimension
equal to the geometric multiplicity, which can fall short. The generalised
eigenspace always has dimension equal to the algebraic multiplicity — so
the pieces now add up.
one summand per distinct eigenvalue, each invariant under \(A\). That is the
primary decomposition, proved on
The Primary Decomposition Theorem.
The kernel sequence
Set \(d_k = \dim\ker(A - \lambda I)^{k}\). The sequence rises and then plateaus, and
every number in it means something:
\(d_1\) is the geometric multiplicity;
the value it plateaus at is the algebraic multiplicity;
the index where it first plateaus is the size of the largest Jordan block.
The explorer prints this sequence for every eigenvalue, because the entire block
structure of Section 8 is read out of it.
Worked: two shapes with the same dimensions
A \(4 \times 4\) with a single eigenvalue \(3\) of algebraic multiplicity \(4\) can have
kernel dimensions \(0, 2, 3, 4\) — geometric multiplicity \(2\), largest block \(3\) —
or \(0, 4\), meaning four independent eigenvectors and a diagonal matrix. Same
characteristic polynomial, entirely different matrices.
7. Nilpotent Operators and Jordan Chains
On \(G_\lambda\) the operator \(A - \lambda I\) is nilpotent: some power
of it is zero. So once nilpotent operators are understood, the general case follows by
adding \(\lambda\) back along the diagonal. The whole difficulty of the subject is
concentrated here.
Nilpotent
\(N^{k} = 0\) for some \(k \geq 1\). Its only eigenvalue is \(0\), so its characteristic
polynomial is \(x^{n}\), and the least such \(k\) is the size of its largest block.
Jordan chain
A list \(\mathbf{v},\; N\mathbf{v},\; N^{2}\mathbf{v},\; \ldots\) continued until it
reaches \(\mathbf{0}\). The last non-zero vector in a chain is an ordinary eigenvector;
the ones above it are the generalised ones feeding into it.
The structure theorem says the space breaks up completely into such chains: there is a
basis consisting of Jordan chains, and each chain contributes exactly one block.
The Structure Theorem for Nilpotent Operators
proves it by recursing on the image and lifting generators back.
Worked: a nilpotent of size four
The matrix sending \(e_2 \mapsto e_1\), \(e_3 \mapsto e_2\) and killing \(e_1, e_4\) has
kernel dimensions \(0, 2, 3, 4\). Reading them: geometric multiplicity \(2\), so two
chains; the sequence plateaus at index \(3\), so the longest chain has length \(3\).
Blocks \(3\) and \(1\), totalling \(4\). And indeed \(N^3 = 0\) while \(N^2 \neq 0\).
Reading the block sizes
The number of blocks of size exactly \(j\) is \[ 2d_j - d_{j - 1} - d_{j + 1}. \] Two
consequences are worth carrying: the number of blocks is the geometric
multiplicity, and the total of their sizes is the algebraic multiplicity. The
explorer checks both on whatever it computes.
8. The Jordan Canonical Form
Jordan block
A square matrix with a single \(\lambda\) along the diagonal and \(1\)s on the
superdiagonal. A block of size \(1\) is an ordinary diagonal entry, so a diagonalisable
matrix is exactly the case where every block has size \(1\).
The theorem
Every complex square matrix is similar to a block-diagonal matrix of Jordan blocks, and
that form is unique up to the order of the blocks.
Putting Sections 6 and 7 together gives the proof: split \(\mathbb{C}^n\) by eigenvalue
using the primary decomposition, apply the nilpotent structure theorem to \(A - \lambda
I\) on each piece, and read each chain as a block.
The Jordan Canonical Form
assembles it and proves uniqueness.
The characteristic polynomial is not enough
\(\operatorname{diag}(2,2)\) and one block \(J_2(2)\) share a characteristic polynomial
and are not similar. What separates them is the kernel sequence — or equivalently the
minimal polynomial. Any claim about Jordan structure that rests on \(p(x)\) alone is
unsupported.
Interactive: The Jordan Form Explorer
Choose a matrix. The explorer computes the characteristic and minimal polynomials, finds
each eigenvalue with both multiplicities, measures the kernel sequence, reads the Jordan
block sizes off it, assembles the Jordan form — and then verifies its own answer. Every
value is an exact fraction.
A
Eigenvalues
The Jordan form J, assembled from the block sizes
Checks the explorer runs on its own answer
The block structure is computed from kernel dimensions, never tabulated. The same
routine was tested by building matrices from known blocks, hiding the structure by a
change of basis, and requiring it to be recovered — 85 cases.
9. Orthogonality and Gram–Schmidt
Sections 1 to 8 used only the vector space structure. Adding an inner product brings
length and angle back, and with them a much better-behaved class of matrices.
Orthonormal basis
A basis that is pairwise orthogonal with every vector of norm \(1\). Coordinates
relative to one are simply inner products, \(\mathbf{v} = \sum \langle \mathbf{v},
\mathbf{b}_i \rangle \mathbf{b}_i\), with no system to solve.
Gram–Schmidt
Given an independent list, subtract from each vector its projection onto the span of
those already processed; normalise at the end. The result is an orthonormal basis for
the same span, so every finite-dimensional inner product space has one, and any
orthonormal set extends to a basis.
Orthogonality, Orthonormal Bases, and the Gram–Schmidt Process
proves correctness by induction.
Orthogonal matrices
\(Q^{T}Q = I\), so \(Q^{-1} = Q^{T}\) — no work at all to undo. Such maps preserve norms
and inner products, which makes them exactly the rigid motions fixing the origin:
rotations when \(\det Q = 1\), reflections when \(\det Q = -1\).
“Orthogonal matrix” means orthonormal columns
The name is a historical misfortune. Columns that are merely pairwise orthogonal are not
enough — they must also have norm \(1\), or \(Q^{T}Q\) is diagonal rather than the
identity.
10. The Spectral Theorem
For real symmetric matrices
If \(S^{T} = S\) then all eigenvalues of \(S\) are real, eigenvectors for distinct
eigenvalues are orthogonal, and there is an orthonormal basis of
eigenvectors: \[ S = QDQ^{T} \] with \(Q\) orthogonal and \(D\) real diagonal.
The Spectral Theorem for Symmetric Matrices
proves it by induction on \(n\), using three lemmas: the eigenvalues are real, distinct
eigenvalues give orthogonal eigenvectors, and the orthogonal complement of an eigenvector
is invariant.
Why this closes the page
A symmetric matrix is never defective. The theorem hands over a full
basis of eigenvectors, so geometric multiplicity always equals algebraic multiplicity,
and the entire apparatus of Sections 5 to 8 is unnecessary. That is the real reason
symmetric matrices are the ones everybody prefers to work with.
Worked: orthogonality across, but not within
The symmetric preset has eigenvalues \(1\) (twice) and \(4\). Eigenvectors \((-1,1,0)\)
and \((-1,0,1)\) span the \(\lambda = 1\) eigenspace; \((1,1,1)\) spans the \(\lambda =
4\) one.
Both of the first two are orthogonal to \((1,1,1)\) — that is the theorem. But they are
not orthogonal to each other: their inner product is \(1\). Orthogonality is
guaranteed between eigenspaces, never automatically within one;
Gram–Schmidt inside the eigenspace is what finishes the job.
11. Summary
Eigenvalues are the roots of \(\det(xI - A)\), because an eigenvector
is a non-zero kernel element of \(A - \lambda I\).
Two multiplicities, and \(1 \leq\) geometric \(\leq\) algebraic.
Diagonalisable means equality at every eigenvalue. A repeated eigenvalue is not the
problem; a shortage of eigenvectors is.
The minimal polynomial sees what the characteristic one cannot —
diagonalisable exactly when \(m_A\) has no repeated root.
Generalised eigenspaces make the dimensions add up, giving the primary
decomposition, one summand per eigenvalue.
On each summand \(A - \lambda I\) is nilpotent, and nilpotent operators
break into Jordan chains.
Hence the Jordan canonical form, unique up to block order, with the
block sizes read off the kernel sequence by \(2d_j - d_{j - 1} - d_{j + 1}\).
Symmetric matrices escape all of it: the Spectral Theorem gives an
orthonormal eigenbasis, so they are never defective.
Unit 11, complete
This closes the linear algebra unit. Vector spaces supplied the language, determinants
supplied the single number that detects collapse, and this page used that number to find
the directions a map leaves alone — and to classify every map that has too few of them.
12. Quick Reference
Result
Statement
Proof on this site
Eigenvalues and \(p(x)\)
\(\lambda\) is an eigenvalue \(\iff \det(A - \lambda I) = 0\)