MAT247H1/Unit 9
Unit 9 · Algebra II

Primary Decomposition & Jordan Form

MAT240H1 left a gap: some operators refuse to diagonalize. This unit closes it. Over $\mathbb{C}$ every operator is block-diagonal with blocks that are as close to diagonal as possible, and the list of block sizes is a complete invariant. Nothing is left unclassified.

By the end you can
Why it matters & where it comes from
Why it exists

Similarity classes were left open in MAT240H1: two matrices could share a characteristic polynomial and still be different operators. The Jordan form is a normal form — one representative per class — so "are these similar?" becomes a finite check.

In the real world

Solving $x'=Ax$ when $A$ is defective produces terms like $te^{\lambda t}$, and the power of $t$ is the Jordan block size minus one. Resonance in a driven oscillator is a repeated eigenvalue with a nontrivial Jordan block. The matrix exponential is computed block by block.

Story & open questions

Camille Jordan published the form in 1870, though Weierstrass had equivalent results earlier. The Jordan form is famously numerically unstable — arbitrarily small perturbations turn a Jordan block into a diagonalizable matrix — which is why numerical work uses the Schur form of Unit 6 instead.

Primary decomposition

Theorem Primary decomposition

Let $T$ be an operator on a finite-dimensional $V$ over $F$, and suppose its minimal polynomial factors as $m(t)=p_1(t)^{a_1}\cdots p_k(t)^{a_k}$ with the $p_i$ distinct irreducible polynomials. Put $W_i=N\!\left(p_i(T)^{a_i}\right)$. Then

$V=W_1\oplus\cdots\oplus W_k$, each $W_i$ is $T$-invariant, and the minimal polynomial of $T|_{W_i}$ is $p_i^{a_i}$.
Proof Coprimality does the work
Set $q_i(t)=m(t)/p_i(t)^{a_i}$. The $q_i$ have no common irreducible factor, so $\gcd(q_1,\dots,q_k)=1$ and there are polynomials $h_i$ with $\sum_ih_iq_i=1$ (MAT240H1, Unit 9).
Let $E_i=h_i(T)q_i(T)$. Then $\sum_iE_i=I$. Also $p_i(T)^{a_i}E_i=h_i(T)m(T)=0$, so $R(E_i)\subseteq W_i$.
Hence every $v=\sum_iE_iv$ lies in $\sum_iW_i$. For directness, if $v\in W_i\cap\sum_{j\ne i}W_j$ then $v$ is killed by both $p_i^{a_i}$ and $q_i$, which are coprime, so $v=0$. ∎
This is the Chinese remainder theorem. Splitting a space according to coprime factors of an annihilating polynomial is the same argument that splits $\mathbb{Z}/n$ into prime-power pieces. The Jordan form is what you get when the irreducible factors are all linear.

Generalized eigenvectors

Definition

For an eigenvalue $\lambda$, the generalized eigenspace is $K_\lambda=\{v\in V:(T-\lambda I)^mv=0\text{ for some }m\ge1\}$. Its nonzero elements are generalized eigenvectors. Always $E_\lambda\subseteq K_\lambda$.

Theorem Dimensions

Over a field where the characteristic polynomial splits, $\dim K_\lambda$ equals the algebraic multiplicity of $\lambda$, and $V=\bigoplus_\lambda K_\lambda$. Moreover the chain $N(T-\lambda I)\subseteq N((T-\lambda I)^2)\subseteq\cdots$ strictly increases until it stabilises, at which point it equals $K_\lambda$.

This is primary decomposition with $p_i(t)=t-\lambda_i$: over $\mathbb{C}$ every irreducible polynomial is linear, so the decomposition is into generalized eigenspaces.

Example A chain that does not stop at step one

$A=\left(\begin{smallmatrix}2&1&0\\0&2&1\\0&0&2\end{smallmatrix}\right)$.

$N(A-2I)$ is spanned by $e_1$: dimension $1$, so the geometric multiplicity is $1$ while the algebraic multiplicity is $3$.
$(A-2I)^2$ has rank $1$, so its null space has dimension $2$, spanned by $e_1,e_2$. And $(A-2I)^3=0$, so $K_2=\mathbb{R}^3$.
The chain $1,2,3$ climbing one at a time is the signature of a single Jordan block of size $3$.

The Jordan canonical form

Definition

A Jordan block $J_k(\lambda)$ is the $k\times k$ matrix with $\lambda$ on the diagonal, $1$ on the superdiagonal, and $0$ elsewhere. A Jordan matrix is block diagonal with Jordan blocks.

Theorem Jordan canonical form

Let $T$ be an operator on a finite-dimensional complex vector space — or over any field where the characteristic polynomial splits. Then there is a basis in which $[T]$ is a Jordan matrix, and this matrix is unique up to the order of the blocks.

Consequently two matrices are similar if and only if they have the same Jordan form up to reordering.

Theorem Reading off the blocks

For a fixed eigenvalue $\lambda$, write $r_m=\dim N\!\left((T-\lambda I)^m\right)$. Then:

The number of Jordan blocks for $\lambda$ is $r_1$, the geometric multiplicity.
The total size of those blocks is the algebraic multiplicity of $\lambda$.
The largest block has size equal to the exponent of $(t-\lambda)$ in the minimal polynomial.
The number of blocks of size at least $m$ is $r_m-r_{m-1}$.
Example Enumerate the possibilities

Suppose $\dim V=4$, $f(t)=(t-5)^4$ and $m(t)=(t-5)^2$.

All blocks carry eigenvalue $5$; sizes sum to $4$; the largest is $2$.
Partitions of $4$ with largest part $2$: $2+2$ and $2+1+1$. So exactly two candidates.
They are told apart by $r_1$: two blocks in the first case, three in the second. Equivalently $\dim E_5$ is $2$ or $3$. This is why the two polynomials alone do not always determine the form.
Example Build the basis

For $A=\left(\begin{smallmatrix}3&1\\0&3\end{smallmatrix}\right)$, find a Jordan basis.

$E_3=\operatorname{span}\{e_1\}$ is one-dimensional, so there is a single block of size $2$ — $A$ is already in Jordan form.
A Jordan chain is $v_2$ with $(A-3I)v_2=v_1\ne0$ and $(A-3I)v_1=0$. Take $v_2=e_2$, giving $v_1=e_1$. The basis $\{v_1,v_2\}$ produces exactly $J_2(3)$.
Common pitfalls

• Assuming the characteristic and minimal polynomials determine the Jordan form. They do for $n\le3$ but not in general — the $4$-dimensional example above is the smallest counterexample.
• Confusing the number of blocks with the largest block size. The count is the geometric multiplicity; the size comes from the minimal polynomial.
• Applying the theorem when the characteristic polynomial does not split. Over $\mathbb{R}$ a rotation has no Jordan form; use the rational form of Unit 10.
• Putting the $1$s on the subdiagonal. Both conventions appear in textbooks; pick the one your course uses and stay with it.

Practice

Original problems written for this guide in the style of Friedberg, Insel & Spence exercises. Write a full solution before opening the answer.

Practice 1 · Read the form

An operator on a $3$-dimensional space has $f(t)=(t-2)^3$ and $m(t)=(t-2)^2$. Find its Jordan form.

Answer
Sizes sum to $3$ with largest $2$: only $2+1$. So the form is $J_2(2)\oplus J_1(2)$, with $\dim E_2=2$.
Practice 2 · Diagonalizable test

When is the Jordan form diagonal?

Answer
Exactly when every block has size $1$, equivalently when the minimal polynomial is a product of distinct linear factors.
Practice 3 · Compute the chain

For $A=\left(\begin{smallmatrix}1&1&0\\0&1&0\\0&0&1\end{smallmatrix} \right)$, find the Jordan form.

Solution
$f(t)=(t-1)^3$. $N(A-I)$ is spanned by $e_1,e_3$, so $r_1=2$: two blocks. Sizes summing to $3$ with two blocks means $2+1$, so $J_2(1)\oplus J_1(1)$ and $m(t)=(t-1)^2$.
Practice 4 · Ambiguity

Give two non-similar $4\times4$ matrices with the same characteristic and minimal polynomials.

Solution
$J_2(0)\oplus J_2(0)$ and $J_2(0)\oplus J_1(0)\oplus J_1(0)$. Both have $f(t)=t^4$ and $m(t)=t^2$, but the first has two blocks and the second has three, so they are not similar.
Practice 5 · Primary decomposition

Decompose $V$ for an operator with $m(t)=(t-1)^2(t+3)$.

Solution
$V=N((T-I)^2)\oplus N(T+3I)$. The first summand may be bigger than $E_1$; the second equals $E_{-3}$ since that factor appears to the first power.
Practice 6 · Nilpotent

Show $T$ is nilpotent iff its only eigenvalue is $0$, and describe the Jordan form.

Solution
If $T^k=0$ then any eigenvalue satisfies $\lambda^k=0$, so $\lambda=0$. Conversely if the only eigenvalue is $0$ then $f(t)=t^n$ and Cayley–Hamilton gives $T^n=0$. The Jordan form is blocks $J_k(0)$ — zeros on the diagonal, ones above.
Practice 7 · Counting blocks

If $r_1=3$, $r_2=5$, $r_3=6$ for eigenvalue $\lambda$, list the block sizes.

Solution
Blocks of size $\ge1$: $r_1=3$. Size $\ge2$: $r_2-r_1=2$. Size $\ge3$: $r_3-r_2=1$. Size $\ge4$: $0$. So one block of size $3$, one of size $2$, one of size $1$ — total $6$, matching $r_3$.
Practice 8 · Powers

Compute $J_3(\lambda)^m$ for $\lambda\ne0$.

Solution
Write $J=\lambda I+N$ with $N$ nilpotent of index $3$. Since $\lambda I$ and $N$ commute, the binomial theorem gives $J^m=\lambda^mI+m\lambda^{m-1}N+\binom{m}{2}\lambda^{m-2}N^2$, and higher terms vanish. The polynomial-times-exponential growth in ODE solutions comes from exactly this.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT247H1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Use the current UofT MAT247H1 calendar topics as the coverage spine, and pair lecture notes with a Friedberg-Insel-Spence-style linear algebra text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

Minimal polynomial factors, primary decomposition splits $V$, each piece is eigenvalue-plus-nilpotent, and the nilpotent part has a chain basis. The Jordan form is those three steps assembled.

Study routine. Practice going both ways: from a matrix to its Jordan form, and from prescribed polynomials to the list of possible forms. Exams ask the second more often than the first.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Block count

What determines the number of Jordan blocks for $\lambda$?

Guide
The geometric multiplicity $\dim E_\lambda$: one block per independent eigenvector.
Extension 2 · Largest block

What determines the size of the largest block?

Guide
The exponent of $(t-\lambda)$ in the minimal polynomial.
Extension 3 · Field requirement

When does a Jordan form fail to exist?

Guide
When the characteristic polynomial does not split over the field — for instance a real rotation. Passing to $\mathbb{C}$, or using the rational canonical form, resolves it.
Extension 4 · Similarity test

How do you decide whether two matrices are similar?

Guide
Compute both Jordan forms; they are similar iff the forms agree up to reordering blocks. Matching characteristic and minimal polynomials is necessary but not sufficient.
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