Jordan canonical form

In one sentence

The Jordan form is a picture of block sizes, every feature of that picture is a rank you can compute, and the questions are almost all "how many structures are possible" or "which structure is this".

Why the exam asks it

Because it is the complete answer to a question the earlier chapters kept failing to settle — when are two matrices similar — and because counting the possible forms for given data is a self-contained puzzle that needs no computation at all, only bookkeeping.

The idea

Over a field where $\chi_A$ splits (always over $\mathbb{C}$), $A$ is similar to a block-diagonal matrix of Jordan blocks

$J_k(\lambda) = \begin{pmatrix}\lambda & 1 & & \\ & \lambda & \ddots & \\ & & \ddots & 1 \\ & & & \lambda\end{pmatrix},$

unique up to the order of the blocks. So two matrices are similar exactly when they have the same Jordan form, and that is the classification the characteristic and minimal polynomials failed to provide.

Reading the picture

Draw, for each eigenvalue $\lambda$, one column of boxes per Jordan block, the column height being the block size. Then:

feature of the picture	what it equals
total boxes for $\lambda$	algebraic multiplicity of $\lambda$
number of columns	geometric multiplicity, $= \dim\ker(A - \lambda I)$
height of the tallest column	multiplicity of $(x-\lambda)$ in $m_A$

Every one of these is computable. In particular the number of columns is $n - \operatorname{rank}(A - \lambda I)$, which is a single row reduction.

Take the three nilpotent structures on four boxes and compare them in pairs. All three have characteristic polynomial $x^4$, so that polynomial separates nothing at all.

Left and middle. $J_2\oplus J_2$ and $J_2 \oplus J_1 \oplus J_1$ have the same tallest column, so the same minimal polynomial $x^2$. They differ in the number of columns, $2$ against $3$, which appears as $\operatorname{rank}(N) = 2$ against $1$. So the minimal polynomial does not fix the number of blocks.

Left and right. $J_2 \oplus J_2$ and $J_3 \oplus J_1$ have the same number of columns, both $2$, hence the same $\operatorname{rank}(N) = 2$ and the same kernel dimension. They differ in the tallest column, $2$ against $3$, so their minimal polynomials are $x^2$ and $x^3$ — visible one rank further down, where $\operatorname{rank}(N^2)$ is $0$ against $1$. So the number of blocks does not fix the minimal polynomial.

The two invariants are independent, and each direction gets set as its own question. The second is the one behind options of the form "these two nilpotent maps have isomorphic kernels, so they have the same minimal polynomial" — false, with $J_2\oplus J_2$ against $J_3\oplus J_1$ as the witness. It also shows why one rank is never enough to identify a structure: you need $\operatorname{rank}(N)$ and $\operatorname{rank}(N^2)$, which is exactly what the block-counting formula below says.

Four boxes is the smallest size where any of this can happen, which is why $4\times4$ is where the question is always set.

The full rank formula

The number of Jordan blocks for $\lambda$ of size $\ge k$ is

$\operatorname{rank}(A-\lambda I)^{k-1} - \operatorname{rank}(A - \lambda I)^{k},$

so the number of blocks of size exactly $k$ is

$\operatorname{rank}(A-\lambda I)^{k-1} - 2\operatorname{rank}(A-\lambda I)^{k} + \operatorname{rank}(A-\lambda I)^{k+1}.$

You rarely need the second formula, but the first — computing the ranks of successive powers until they stop dropping — reconstructs the whole picture mechanically.

Counting possible forms

"How many matrices up to similarity have $\chi_A = (x-1)^n$?" is asking for the number of partitions of $n$, since each Jordan structure is a partition into block sizes. For $n = 4$ there are $5$: $4$, $3{+}1$, $2{+}2$, $2{+}1{+}1$, $1{+}1{+}1{+}1$.

Adding a constraint on $m_A$ fixes the largest part. With $\chi_A = (x-1)^4$ and $m_A = (x-1)^2$, the largest part is $2$, leaving $2{+}2$ and $2{+}1{+}1$ — exactly the two drawn above.

With several eigenvalues, multiply the counts: choose a partition independently for each.

Consequences worth carrying

• $A$ is diagonalisable $\iff$ every block has size $1$ $\iff$ $m_A$ has no repeated factor.
• $A$ is cyclic (one block per eigenvalue) $\iff$ $m_A = \chi_A$ $\iff$ every eigenvalue has geometric multiplicity $1$.
• $A$ and $A^T$ are always similar: transposing reverses each block, which is a relabelling.
• A nilpotent $n\times n$ matrix has $A^n = 0$, and the smallest such power is the largest block size.

Where intuition breaks

"Same characteristic and minimal polynomial, so similar." True up to $3\times3$, false from $4\times4$: $J_2\oplus J_2$ against $J_2\oplus J_1\oplus J_1$. Tempting precisely because it holds in every small case anyone checks by hand.

"Two nilpotent maps with kernels of the same dimension have the same minimal polynomial." $J_2\oplus J_2$ and $J_3\oplus J_1$ on four boxes: both have $\operatorname{rank}(N) = 2$, so both kernels are two-dimensional, and their minimal polynomials are $x^2$ and $x^3$. Tempting because the kernel is the first invariant anyone computes and it does fix the number of blocks. It says nothing about their sizes, and the sizes are what the minimal polynomial reads.

"Every matrix has a Jordan form." Only when $\chi_A$ splits over the field. Over $\mathbb{R}$ a rotation has none, because $x^2+1$ does not factor. Over $\mathbb{C}$ it is automatic, and the stem usually says $\mathbb{C}$ for exactly this reason.

"The Jordan form is unique." Up to the order of the blocks. Options sometimes claim a canonical ordering that is not part of the theorem.

"Number of Jordan blocks = number of distinct eigenvalues." It is the number of distinct eigenvalues only when each has geometric multiplicity $1$. In general the block count for one eigenvalue is its geometric multiplicity.

"$m_A$ has degree $n$, so $A$ is diagonalisable." The opposite: $m_A = \chi_A$ means one block per eigenvalue, which is diagonalisable only if all blocks have size $1$, that is only if the eigenvalues are distinct.

"$A^2 = 0$ implies $A$ has a single $2\times2$ block." It implies every block has size $\le 2$. In $4\times4$ that still leaves two structures.

The exam's angle

1. Translate everything into the box picture immediately. Algebraic multiplicity, geometric multiplicity, minimal polynomial: total, columns, tallest column.
2. For counting questions, count partitions, with the largest part fixed by $m_A$ and the total fixed by $\chi_A$. No matrices need be written down.
3. To identify a given matrix, compute $\operatorname{rank}(A-\lambda I)$ and then $\operatorname{rank}(A-\lambda I)^2$. Two numbers usually determine the structure completely.
4. Check the field before asserting a Jordan form exists.
5. Remember $A \sim A^T$ — it makes several options that look asymmetric collapse.

The night before

• Jordan form exists when $\chi_A$ splits; always over $\mathbb{C}$. Unique up to block order.
• Similar $\iff$ same Jordan form. This is the classification.
• Total boxes $=$ algebraic multiplicity. Columns $=$ geometric multiplicity $= n - \operatorname{rank}(A-\lambda I)$. Tallest column $=$ multiplicity in $m_A$.
• Blocks of size $\ge k$: $\operatorname{rank}(A-\lambda I)^{k-1} - \operatorname{rank}(A-\lambda I)^k$.
• Counting structures $=$ counting partitions; $m_A$ fixes the largest part.
• Block count and minimal polynomial are independent. On four boxes: $2{+}2$ and $2{+}1{+}1$ share $m_A$ and differ in block count; $2{+}2$ and $3{+}1$ share the block count and differ in $m_A$. One rank never identifies a structure.
• $\chi_A = (x-1)^4$, $m_A = (x-1)^2$: exactly two forms, $2{+}2$ and $2{+}1{+}1$.
• Diagonalisable $\iff$ all blocks size $1$. Cyclic $\iff$ $m_A = \chi_A$ $\iff$ one block per eigenvalue.
• $A$ and $A^T$ are always similar.