Jordan canonical form

This is a basic fact of Linear Algebra. When a matrix cannot be diagonalized, at least we can obtain a particular well-behaved form. Consider, for simplicity, an endomorphism $F$ in a $C$ -vector space, and suppose it has only one eigenvalue (if not, we can decompose the vector space in subspaces satisfying this property. Is not elemental).

We will deal with a particular case, for the sake of exposition. Consider a function $F : C^{4} \to C^{4}$ such that the characteristic polynomial is $P (λ) = (2 - λ)^{4}$ . Define

G = F - 2 I d .

We have $G^{4} = 0$ , by Cayley-Hamilton theorem. Also,

K e r (G) \subseteq K e r (G^{2}) \subseteq K e r (G^{3}) \dots

Observe that:

if $w \in K e r (G)$ then $F (w) = 2 w$ , and $w$ is an eigenvector.
if $w \in K e r (G^{2}) - K e r (G)$ , then the pair ${G (w), w}$ is independent and:
a) $F (G (w)) = 2 G (w)$
b) $F (w) = G (w) + 2 w$
so they can be part of a basis giving rise to a block $(\begin{matrix} 2 & 1 \\ 0 & 2 \end{matrix})$ .
if $w \in K e r (G^{3}) - K e r (G^{2})$ , then the set ${G^{2} (w), G (w), w}$ is independent and:
a) $F (G^{2} (w)) = 2 G^{2} (w)$
b) $F (G (w)) = G^{2} (w) + 2 G (w)$
c) $F (w) = G (w) + 2 w$
so they can be part of a basis giving rise to a block $(\begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{matrix})$ .
and so on.

Therefore, we have the algebraic multiplicity of the eigenvalue $λ = 2$ , which is 4 in this case, the dimension $d_{1}$ of the eigenspace $K e r (G)$ , and the sequence of dimensions $d_{k}$ of the generalized eigenspaces, defined as $K e r (G^{k})$ , $2 \leq k \leq 4$ . I.e., we have the sequence:

d_{1} \leq d_{2} \leq d_{3} \leq d_{4} = 4

We have also the following lemma:
Lemma. It is satisfied that $d_{k + 1} \leq d_{k} + d_{1}$ for every $k$ . $◼$
Proof
For each $k \geq 1$ consider the linear map

φ_{k} : \ker (G^{k + 1}) ⟶ \ker (G^{k}), φ_{k} (v) = G (v) .

Its kernel is

\ker (φ_{k}) = {v \in \ker (G^{k + 1}) : G (v) = 0} = \ker (G),

so $\dim \ker (φ_{k}) = d_{1}$ . Its image lies in $\ker (G^{k})$ , so $\dim im (φ_{k}) \leq \dim \ker (G^{k}) = d_{k} .$
But then,

\dim \ker (G^{k + 1}) = \dim \ker (φ_{k}) + \dim im (φ_{k}) \leq d_{1} + d_{k},

i.e.

d_{k + 1} \leq d_{k} + d_{1} .

$◼$

We have several cases:

4,4,4,4. Then we have enough eigenvectors, and the matrix is diagonalizable. We have

(\begin{matrix} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{matrix})

3,4,4,4. We can take $w \in K e r (G^{2}) - K e r (G)$ , so we have the pair ${G (w), w}$ , which can be extended with two independent vectors $v_{1}, v_{2} \in K e r (G)$ . We have in that basis:

(\begin{matrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{matrix})

2,4,4,4. We can take $w_{1}, w_{2} \in K e r (G^{2}) - K e r (G)$ , so we have the sets ${G (w_{1}), w_{1}}$ and ${G (w_{2}), w_{2}}$ , which constitute a basis in which the matrix is:

(\begin{matrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{matrix})

1,4,4,4; 1,1,4,4; 1,2,4,4; 1,1,1,4; 1,2,2,4. These cannot happen, because of the lemma above.
1,2,3,4. We can take $w \in K e r (G^{4}) - K e r (G^{3})$ , and the set ${G^{3} (w), G^{2} (w), G (w), w}$ is a basis with associated matrix

(\begin{matrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{matrix})

2,3,3,4. Imposible, since for $w \in K e r (G^{4}) - K e r (G^{3})$ , the set ${G^{3} (w), G^{2} (w), G (w), w}$ is independent, and only $G^{3} (w)$ can belong to $k e r (G)$ . So we would have 5 independent vectors.
2,3,4,4. We can take $w \in K e r (G^{3}) - K e r (G^{2})$ and the set ${G^{2} (w), G (w), w}$ is independent. We take $v \in k e r (G)$ independent and we obtain a basis with matrix

(\begin{matrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{matrix})

3,3,4,4. Analogous to case 6. We can take $w \in K e r (G^{3}) - K e r (G^{2})$ and the set ${G^{2} (w), G (w), w}$ is independent, but only $G^{2} (w)$ belongs to $k e r (G)$ . Then we would have five independent vectors.
3,3,3,4. Idem. For $w \in K e r (G^{4}) - K e r (G^{3})$ , the set ${G^{3} (w), G^{2} (w), G (w), w}$ is independent, and only $G^{3} (w)$ can belong to $k e r (G)$ . We would have 6 independent vectors.

Related: power of a Jordan block.

Old stuff

In general, we have that for an endomorphism $f : V \mapsto V$

I m (f) ≅ V / K e r (f)

and so $d i m (V) = d i m (K e r (f)) + d i m (I m (f))$ .

But we don't always have that $V = K e r (f) \oplus I m (f)$ . Now, if $f \circ f = f$ , i.e., $f$ is idempotent, the previous formula holds: for every $v \in V$ we have that $v = v - f (v) + f (v) = (v - f (v), f (v))$ and conversely, for $(h, w) \in K e r (f) \oplus I m (f)$ we can take $h + w \in V$ such that $(h + w - f (h + w), f (h + w)) = (h, w)$
Pasted image 20211017184359.png
This is the same as being a projection.

More generally, we have
Proposition
If $f$ is such that $I m (f) = I m (f^{2})$ , we have

V = K e r (f) \oplus I m (f)

Proof
First, let's observe that $f |_{I m (f)}$ is an isomorphism. This is so because it is surjective and dimensions are equals.
We can construct an isomorphism $V \to K e r (f) \times I m (f)$ at the following way:

v \mapsto (v - f (x), f (x))

where $x$ is such that $f^{2} (x) = f (v)$ ( $x$ exists because $I m (f) = I m (f^{2})$ ).

It is easy to prove injectivity and surjectivity. Only rest to show that is well defined. Let us suppose that, for $v \in V$ we get $x \neq y$ such that $f^{2} (x) = f^{2} (y) = f (v)$ . Then, since the restriction of $f$ is an isomorphism, $f (x) = f (y)$ .
$◼$
In the same way, we can prove the following proposition
Proposition
For any endomorphism $f : V \to V$ there exists $m \geq 1$ such that

K e r (f^{m}) \oplus I m (f^{m})

Proof
Observe that $I m (f) \supseteq I m (f^{2}) \supseteq I m (f^{3}) \supseteq \dots$
And therefore, there exist $m$ such that $I m (f^{m}) = I m (f^{m + 1}) = \dots = I m (f^{2 m}) = \dots$ .
At this point, we only have to apply the previous proposition to the transformation $h = f^{m}$ .
$◼$

Now, we are ready to show the Jordan canonical form. Let $V$ be a complex vector space, and $f \in e n d (V)$ . If $f$ is decomposable, we take a decomposition $V = ⨁_{i} U_{i}$ , with $f (U_{i}) \subseteq U_{i}$ (invariant subspace) and proceed with every $f_{i} = f |_{U_{i}}$ . So assume $f$ is indecomposable.
Consider $λ \in C$ such that $f (v) = λ v$ . It exists, because we can choose a basis and construct the polynomial equation $d e t (A - λ I) = 0$ , and $C$ is algebraically closed. So, for certain $λ$ , $K e r (f - λ i d)$ has, at least, dimension 1. Moreover, observe that $λ$ is the only (possibly multiple) solution to the previous polynomial equation

We define $h = f - λ i d$ , and apply the previous proposition, so we get

V = k e r (h^{m}) \oplus i m (h^{m})

for $m \geq 1$ .
Since $f$ is indecomposable and $v \in k e r (h^{m})$ (since $f (v) - λ v = 0$ ) we have that $V = k e r (h^{m})$ , and therefore $h^{m} : V \mapsto V$ is the null transformation and $h$ is \emph{nilpotent}. Let $k$ be the nilpotency index, and $w \in V$ such that $h^{k - 1} (w) \neq 0$ but $h^{k} (w) = 0$ .

We can show that ${w, h (w), h^{2} (w), \dots, h^{k} (w)}$ is a basis for $V$ . First, observe that they are independent because if

\sum_{i} a_{i} h^{i} (w) = 0

we can apply $h$ several times to show that every $a_{i} = 0$ .

And second, we can show that they span $V$ . In fact, let $E = s p a n {w, h (w), h^{2} (w), \dots, h^{k} (w)}$ . Observe that:

$f (w) \in E$ , since $h (w) = f (w) - λ w$ .
$f^{2} (w) \in E$ , since $h^{2} (w) = f^{2} (w) - 2 λ f (w) + λ^{2} w$ . Therefore, $f (h (w)) \in E$ , because $f (h (w)) = f^{2} (w) - λ f (w)$ and so $h^{2} (w) = f (h (w)) - λ h (w)$ .
$f^{3} (w) \in E$ and therefore $f (h^{2} (w)) \in E$ . In fact, $f (h^{2} (w)) = h (f (h (w))) = h (h^{2} (w) + λ h (w)) = h^{3} (w) + λ h^{2} (w)$
In general, $f (h^{i} (w)) = h (f (h^{i - 1} (w))) = h (h^{i} + λ h^{i - 1})) = h^{i + 1} (w) + λ h^{i} (w)$ .

Since $f (E) \subseteq E$ and $f$ is, by hypothesis, indecomposable, we conclude that $V = E$ .
It only rests to show the form of the matrix of $f$ respect to the basis ${w, h (w), h^{2} (w), \dots, h^{k} (w)}$ . But because of the facts of the above enumeration we conclude that

f \equiv (\begin{array}{cccc} λ & 0 & 0 & \dots \\ 1 & λ & 0 & \dots \\ 0 & 1 & λ & \dots \\ \dots & \dots & \dots & \dots \end{array})

In conclusion: For any endomorphism $f$ of $V$ we can find a partition $U_{i}$ of $f$ -stables vector subspaces of $V$ .
In each of one, $f_{i}$ has the form of the matrix above (in a particular basis). This is so because, either $f_{i}$ is a escalar matrix ( $f_{i} = λ I d$ ) or $f_{i} - λ I d$ is nilpotent, and for nilpotent operators we have the special basis ${h^{k} (w), h^{k - 1} (w), \dots, w}$ .

By the way: it there exists other decomposition in $f$ -stables subspaces, but not satisfying that matrix form. Even the canonical Jordan form is not unique.