Simple proof of Jordan normal form

Question

A lot of proofs in linear algebra use the fact that any square matrix can be written in Jordan normal form. Unfortunately I can't see why this is the case, I didn't get what Wikipedia said and I just can't find a proof that I understand. Can you guys please tell me what is the simplest proof of it in your opinion ?

Answer 1

Since the above answers and comments seem to claim that there's no short elementary proof, here's one (due to Mark Wildon?), using nothing but rank-nullity theorem and the fundamental theorem of algebra.

We first prove the result for nilpotent matrices, i.e., such that $A^k=0$ for some $k$. In that case, $0=A^ku=\lambda^ku$ for an eigenvector $u$, so we will have all $\lambda_i=0$. We proceed by induction in the dimension of the underlying space $\mathcal{V}$, by applying the induction hypothesis to $A$ restricted to $\mathrm{Im}\, A$, noting that $\dim \mathrm{Im}\, A<\dim \mathcal{V}$, since otherwise $A$ would be invertible and so would be $A^k$. The induction hypothesis says that $\mathrm{Im}\, A$ has a basis of the form $$w_1,Aw_1,\dots, A^{b_1-1}w_1,w_2,Aw_2,\dots,A^{b_2-1}w_2,\dots,w_k,A w_k,\dots,A^{b_k-1}w_k,$$

and $A^{b_i}w_i=0$. (Here $k$ is the number of Jordan blocks for $A$ restricted to $\mathrm{Im}\, A$ and $b_i$ is the size of $i$-the Jordan block. I'm also changing the notation to the lower Jordan form, which amounts to a permutation of basis elements.). Now, we can find vectors $v_1,\dots,v_k$ such that $w_i=Av_i.$ The vectors $A^{b_1-1}w_1,\dots,A^{b_k-1}w_k$ are linearly independent and belong to $\mathrm{Ker}\, A$, so we can complement them to the basis $A^{b_1-1}w_1,\dots,A^{b_k-1}w_k,u_1,\dots,u_m$ of $\mathrm{Ker}\, A$. The claim is now that $$v_1,Av_1,\dots, A^{b_1}v_1,\dots, v_k,Av_k,\dots, A^{b_k}v_k,u_1,\dots,u_m$$ is a basis of $\mathcal{V}.$ Indeed, it has $k+\dim\mathrm{Im}\,A+m=k+\dim\mathrm{Im}\,A+\dim\mathrm{Ker}\,A-k=\dim \mathcal{V}$ elements. They are also linearly independent, for if a linear combination of these vectors equals zero, applying $A$ kills $A^{b_i}v_i$ and $u_i$ and sends other vectors to the respective basis vectors of $\mathrm{Im}\,A.$ So, the coefficients at these other vectors must all vanish, but then so do the coefficients at $A^{b_i}v_i$ and $u_i$ since those form a basis of $\mathrm{Ker}\,A.$ It is clear that $A$ is in the Jordan form w.r.t. to the above basis, and the induction step is complete.

The general case is derived as follows: let $\lambda$ be an eigenvalue of $A$, denote $A_\lambda=A-\lambda\mathbb{I}$, then $\{0\}\subsetneq \mathrm{Ker}\,A_\lambda\subseteq\mathrm{Ker}\,A_\lambda^2\subset \dots.$ As there can only be finitely many non-strict inclusions, there exists $k$ such that $\mathrm{Ker}\,A_\lambda^k=\mathrm{Ker}\,A_\lambda^{k+1}=\dots.$ If $v\in \mathrm{Ker}\,A_\lambda^k\cap \mathrm{Im}\,A_\lambda^k,$ then $v=A_\lambda^kw$ for some $w\in \mathcal{V}$ and $A_\lambda^{2k}w=A_\lambda^kv=0,$ i.e. $w\in \mathrm{Ker}\,A_\lambda^{2k}=\mathrm{Ker}\,A_\lambda^k,$ so $v=A_\lambda^kw=0.$ That is, $\mathrm{Ker}\,A_\lambda^k\cap \mathrm{Im}\,A_\lambda^k=\{0\}.$ Then, $$ \dim (\mathrm{Ker}\,A_\lambda^k+\mathrm{Im}\,A_\lambda^k)=\dim\mathrm{Ker}\,A_\lambda^k+\dim \mathrm{Im}\,A_\lambda^k-\dim(\mathrm{Ker}\,A_\lambda^k\cap \mathrm{Im}\,A_\lambda^k)=\dim\mathcal{V}. $$ We conclude that $\mathcal{V}=\mathrm{Ker}\,A_\lambda^k\oplus\mathrm{Im}\,A_\lambda^k.$ Also, because $A_\lambda A=AA_\lambda$ and hence $A^k_\lambda A=AA^k_\lambda,$ both $\mathrm{Ker}\,A_\lambda^k$ and $\mathrm{Im}\,A_\lambda^k$ are invariant subspaces for $A$.

We are ready to complete the proof by induction on $\dim\mathcal{V}.$ Since $\dim\mathrm{Ker}\,A_\lambda^k\geq 1,$ we have $\dim\mathrm{Im}\,A_\lambda^k<\dim\mathcal{V},$ hence by induction hypothesis, there exists a Jordan basis for the restriction of $A$ to $\mathrm{Im}\,A_\lambda^k$. The restriction of $A_\lambda$ to $\mathrm{Ker}\,A_\lambda^k$ is nilpotent, by definition, so by the above nilpotent case, there exists a Jordan basis for this restriction. But then the matrix of the restriction of $A=A_\lambda+\lambda\mathbb{I}$ to $\mathrm{Ker}\,A_\lambda^k$ in this basis clearly also has the required Jordan form. So, we have Jordan bases for $A$ restricted to both $\mathrm{Im}\,A_\lambda^k$ and $\mathrm{Ker}\,A_\lambda^k$, taking them together gives a Jordan basis for $A$ on the whole $\mathcal{V}.$

For completeness, we derive the existence of an eigenvalue from fundamental theorem of algebra: the dimension of the space of all matrices is $(\dim\mathcal{V})^2$, hence if $n\geq (\dim\mathcal{V})^2$, there exists coefficients $a_0,\dots a_n,$ not all zero, such that $a_nA^n+\dots+a_1A+a_0\mathbb{I}=0.$ By fundamental theorem of algebra, we can factorize $a_nx^n+\dots+a_1x+a_0=(x-\lambda_1)\dots (x-\lambda_k),$ then also $a_nA^n+\dots+a_1A+a_0\mathbb{I}=(A-\lambda_1\mathbb{I}) \dots (A-\lambda_k\mathbb{I}).$ If we had $\mathrm{Ker}\,(A-\lambda_i\mathbb{I})=\{0\}$ for all $i$, then each $(A-\lambda_i\mathbb{I})$ would be bijective, and them so would be the product, which is impossible.

Answer 2

If $A$ is $n \times n$, let $f$ be the corresponding endomorphism of $V = \mathbf{C}^n$.

Then $V$ can be given a $\mathbf{C}[X]$-module structure by defining $P(X)\cdot v = P(f)(v)$.

A system of representatives for the irreducible elements of $\mathbf{C}[X]$ is given by $X - \lambda, \ \lambda \in \mathbf{C}$. By the structure theorem for fintely generated modules over a PID, $V$ is isomorphic to a finite direct sum of modules of the form $\mathbf{C}[X]/(X-\lambda)^k$. (A summand $\mathbf{C}[X]$ cannot occur because $V$ is finite-dimensional over $\mathbf{C}$.)

This expression of $V$ as a direct sum is also a direct sum of $\mathbf{C}$-vector subspaces, which will be the subspaces on which the Jordan blocks act. Moreover, since each summand is closed under multiplication by $X$, it is stable under $f$.

The only thing left to check is that $f$ acts as a Jordan block on each summand. Without loss of generality, we may assume that the summand is $\mathbf{C}[X]/(X-\lambda)^k$ and that $f$ is multiplication by $X$.

Then it is easy to check that the matrix of $f$ in the basis $((X-\lambda)^{k-1},(X - \lambda)^{k-2},\dots,X - \lambda, 1)$ is the $k \times k$ Jordan block with diagonal element $\lambda$.

Of course, all of this is close to trivial with the structure theorem, but every other proof I've seen of the existence of the Jordan decomposition amounts to rewriting a proof of the structure theorem in vector space language. So the most transparent thing is just to read a proof of the existence part of the structure theorem.

Answer 3

Ref: “Algebra” by Artin.

Let ${ V }$ be a finite dimensional ${ \mathbb{C} - }$vector space. Let ${ T : V \to V }$ be a linear operator.

The goal is to pick a basis ${ \mathscr{B} }$ of ${ V }$ such that the matrix of ${ T }$ with respect to ${ \mathscr{B} }$ is simple to work with.

Note that we can upper triangularize ${ T , }$ that is there exists a basis ${ \mathscr{B} }$ of ${ V }$ with respect to which the matrix of ${ T }$ is upper triangular.

Can we upper triangularize ${ T }$ to a more specific form of upper triangular matrix?

We can answer this question naturally by studying generalized eigenvectors.

Def [Generalized eigenvectors]
An eigenvector with eigenvalue ${ \lambda }$ is a nonzero vector ${ v \neq 0 }$ such that ${ (T - \lambda I) v = 0 . }$
A generalized eigenvector with eigenvalue ${ \lambda }$ is a nonzero vector ${ v \neq 0 }$ such that ${ (T - \lambda I) ^k v = 0 }$ for some ${ k > 0 . }$

Def [Exponent of a generalized eigenvector]
Let ${ v }$ be a generalized eigenvector with eigenvalue ${ \lambda . }$ The smallest ${ d > 0 }$ such that ${ (T - \lambda I) ^d v = 0 }$ is called the exponent of ${ v . }$

Let ${ x }$ be a generalized eigenvector with eigenvalue ${ \lambda }$ and exponent ${ d . }$ What are the properties of the nonzero powers ${ (x, (T - \lambda I) x, \ldots, (T - \lambda I) ^{d-1} x) }$?

Obs: Let ${ x }$ be a generalized eigenvector with eigenvalue ${ \lambda }$ and exponent ${ d . }$ Let

$${ u _j := (T - \lambda I) ^j x \quad \text{for } j = 0, 1, \ldots, d - 1 . }$$

Consider any nontrivial linear combination

$${ c _j u _j + \ldots + c _{d-1} u _{d-1} , \quad j \leq d - 1 \text{ and } c _j \neq 0 . }$$

Then the linear combination is a generalized eigenvector with eigenvalue ${ \lambda }$ and exponent ${ d - j . }$
Pf: Note that

$${ (T - \lambda I) ^{d - j - 1} (c _j u _j + \ldots + c _{d-1} u _{d-1} ) = c _j u _{d - 1} \neq 0 }$$

and

$${ (T - \lambda I) ^{d-j} (c _j u _j + \ldots + c _{d-1} u _{d-1} ) = 0 . }$$

Hence the linear combination is a generalized eigenvector with eigenvalue ${ \lambda }$ and exponent ${ d - j , }$ as needed. ${ \blacksquare }$

Obs: Let ${ x }$ be a generalized eigenvector with eigenvalue ${ \lambda }$ and exponent ${ d . }$ Let

$${ u _j := (T - \lambda I) ^j x \quad \text{for } j = 0, 1, \ldots, d - 1 . }$$

Note that the list ${ \mathscr{B} := (u _0, \ldots, u _{d-1}) }$ is linearly independent.
What is the action of ${ T }$ on ${ \mathscr{B} }$?
The action of ${ T }$ on ${ \mathscr{B} }$ is

$${ {\begin{align} &\, (T - \lambda I) u _0 = u _1, \\ &\, (T - \lambda I) u _1 = u _2, \\ &\, \quad \vdots \\ &\, (T - \lambda I) u _{d-1} = 0 \end{align}} }$$

that is

$${ {\begin{align} &\, u _0 \rightsquigarrow u _1 + \lambda u _0 \\ &\, u _1 \rightsquigarrow u _2 + \lambda u _1 \\ &\, \quad \vdots \\ &\, u _{d-1} \rightsquigarrow \lambda u _{d-1}. \end{align}} }$$

Hence the action of ${ T }$ on ${ \mathscr{B} }$ is

$${ T(u _0, \ldots, u _{d-1}) = (u _0, \ldots, u _{d-1}) M }$$

where ${ M }$ is the ${ d \times d }$ matrix in the sequence

$${ \mathcal{J} _{\lambda} = {\begin{pmatrix} \lambda \end{pmatrix}} , {\begin{pmatrix} \lambda &\, \\ 1 &\lambda \end{pmatrix}} , {\begin{pmatrix} \lambda &\, &\, \\ 1 &\lambda &\, \\ \, &1 &\lambda \end{pmatrix}} , \ldots }$$

We call ${ \mathcal{J} _{\lambda} }$ a sequence of Jordan blocks. ${ \blacksquare }$

Thm [Jordan decomposition]
Let ${ V }$ be a finite dimensional ${ \mathbb{C} - }$vector space. Let ${ T : V \to V }$ be a linear operator.
There is a basis ${ \mathscr{B} }$ of ${ V }$ such that the matrix of ${ T }$ with respect to ${ \mathscr{B} }$ is of the form

$${ {\begin{pmatrix} J _1 &\, &\, &\, \\ \, &J _2 &\, &\, \\ \, &\, &\ddots &\, \\ \, &\, &\, &J _{\ell} \end{pmatrix}} }$$

where each ${ J _i \in \mathcal{J} _{\lambda _i} }$ for some ${ \lambda _i . }$

Pf: Consider the stronger proposition:

Prop: Let ${ V }$ be a finite dimensional ${ \mathbb{C} - }$vector space. Let ${ T : V \to V }$ be a linear operator. Then there exist generalized eigenvectors ${ v _1, \ldots, v _r }$ with eigenvalues ${ \lambda _1, \ldots , \lambda _r }$ respectively, such that

$${ V = \text{span}(\mathscr{B}(v _1, \lambda _1)) \oplus \ldots \oplus \text{span}(\mathscr{B}(v _r, \lambda _r)) }$$

where ${ \mathscr{B}(v _i, \lambda _i) := (v _i, (T - \lambda _i I) v _i, \ldots, (T - \lambda _i I) ^{\text{exp}(v _i) - 1} v _i) . }$

It suffices to show the stronger proposition.

We can proceed by induction on ${ n = \dim(V) . }$ (The base case ${ n = 1 }$ is clear).

Pick an eigenvalue ${ \lambda }$ of ${ T . }$ Consider the powers

$${ (T - \lambda I) ^1, (T - \lambda I) ^2, \ldots }$$

We have chains of subspaces

$${ {\begin{align} &\, \text{ker}((T - \lambda I) ^1) \subseteq \text{ker}((T - \lambda I) ^2) \subseteq \ldots \\ &\, \text{im}((T - \lambda I) ^1) \supseteq \text{im}((T - \lambda I) ^2 ) \supseteq \ldots \end{align}} }$$

By finite dimensionality the chains become eventually constant, that is

$${ {\begin{align} &\, \mathcal{K} := \text{ker}((T - \lambda I) ^m) = \text{ker}((T - \lambda I) ^{m+1} ) = \ldots \\ &\, \mathcal{I} := \text{im}((T - \lambda I) ^m ) = \text{im}((T - \lambda I) ^{m+1} ) = \ldots \end{align}} }$$

for some ${ m . }$

Suppose we show that ${ \mathcal{K}, \mathcal{I} }$ are ${ (T - \lambda I) }$ invariant subspaces such that ${ V = \mathcal{K} \oplus \mathcal{I} . }$ Then it suffices to find Jordan decompositions of ${ (T - \lambda I) \big \vert _{\mathcal{K}} }$ on ${ \mathcal{K} }$ and ${ (T - \lambda I) \big \vert _{\mathcal{I}} }$ on ${ \mathcal{I} . }$

Note that ${ \mathcal{K}, \mathcal{I} }$ are ${ (T - \lambda I) }$ invariant subspaces. For example ${ (T - \lambda I) \text{ker}((T - \lambda I) ^{m+1}) \subseteq \text{ker}((T - \lambda I) ^{m}) , }$ hence ${ (T - \lambda I) \mathcal{K} \subseteq \mathcal{K} . }$ Similarly ${ (T - \lambda I) \text{im}((T - \lambda I) ^m) \subseteq \text{im}((T - \lambda I) ^{m+1}), }$ hence ${ (T - \lambda I) \mathcal{I} \subseteq \mathcal{I} .}$

Note that ${ V = \mathcal{K} \oplus \mathcal{I} . }$ For example ${ \dim(\mathcal{K}) + \dim(\mathcal{I}) }$ ${ = n }$ and ${ \dim(\mathcal{K} + \mathcal{I}) = \dim(\mathcal{K}) + \dim(\mathcal{I}) - \dim(\mathcal{K} \cap \mathcal{I}) , }$ hence it suffices to show ${ \mathcal{K} \cap \mathcal{I} = \lbrace 0 \rbrace . }$ Let ${ z \in \mathcal{K} \cap \mathcal{I} . }$ Hence ${ (T - \lambda I) ^m z = 0 }$ and ${ z = (T - \lambda I) ^m v }$ for some ${ v . }$ Hence ${ (T - \lambda I) ^{2m} v = 0 , }$ that is ${ v \in \ker((T - \lambda I) ^{2m} ) = \mathcal{K} . }$ Hence ${ (T - \lambda I) ^m v = 0 , }$ that is ${ z = 0 , }$ as needed.

Since ${ \lambda }$ is an eigenvalue of ${ T , }$ the subspace ${ \mathcal{K} }$ is not ${ \lbrace 0 \rbrace . }$ Hence ${ \mathcal{I} }$ is a proper subspace of ${ V . }$
By induction hypothesis on ${ \mathcal{I} }$ and ${ (T - \lambda I) \big\vert _{\mathcal{I}} , }$ there exist generalized eigenvectors ${ v _1, \ldots, v _r }$ with eigenvalues ${ \lambda _1, \ldots, \lambda _r }$ respectively, such that

$${ \mathcal{I} = \text{span}(\mathscr{B}(v _1, \lambda _1)) \oplus \ldots \oplus \text{span}(\mathscr{B}(v _r, \lambda _r)) }$$

where ${ \mathscr{B}(v _i, \lambda _i) := (v _i, (T - \lambda I - \lambda _i I)v _i , \ldots, (T - \lambda I - \lambda _i I) ^{\text{exp} (v _i ) - 1 } v _i) . }$

By the version of the theorem for nilpotent linear operators (proved below), applied to ${ \mathcal{K} }$ and ${ (T - \lambda I) \big \vert _{\mathcal{K}}, }$ there exist generalized eigenvectors ${ w _1, \ldots, w _s }$ with eigenvalues ${ 0, \ldots, 0 }$ respectively, such that

$${ \mathcal{K} = \text{span}(\mathscr{B}(w _1, 0)) \oplus \ldots \oplus \text{span}(\mathscr{B}(w _s, 0)) }$$

where ${ \mathscr{B}(w _i, 0) := (w _i, (T - \lambda I - 0 I)w _i , \ldots, (T - \lambda I - 0 I) ^{\text{exp} (w _i ) - 1 } w _i) . }$
Hence we are done. ${ \blacksquare }$

Thm [Jordan decomposition for nilpotent linear operators]
Let ${ V }$ be a finite dimensional ${ \mathbb{C} - }$vector space. Let ${ T : V \to V }$ be a nilpotent linear operator. Then there exist generalized eigenvectors ${ v _1, \ldots, v _r }$ with eigenvalues ${ 0 , \ldots , 0 }$ respectively, such that

$${ V = \text{span}(\mathscr{B}(v _1, 0)) \oplus \ldots \oplus \text{span}(\mathscr{B}(v _r, 0)) }$$

where ${ \mathscr{B}(v _i, 0) := (v _i, (T - 0 I) v _i, \ldots, (T - 0 I) ^{\text{exp}(v _i) - 1} v _i) . }$

Pf: Since ${ T }$ is nilpotent, every nonzero vector is a generalized eigenvector with eigenvalue ${ 0 . }$

We can proceed by induction on ${ n = \dim(V) . }$ (The base case ${ n = 1 }$ is clear).

Since ${ T }$ is nilpotent, ${ \text{ker}(T) \neq \lbrace 0 \rbrace . }$ Hence ${ \dim (\text{im}(T)) = \dim (V) - \dim (\text{ker}(T)) }$ is strictly less than ${ \dim(V) , }$ that is ${ \text{im}(T) }$ is a proper subspace of ${ V . }$
By induction hypothesis on ${ \text{im}(T) }$ and ${ T \big \vert _{\text{im}(T)} , }$ there exist generalized eigenvectors ${ w _1, \ldots, w _r }$ with eigenvalues ${ 0 , \ldots , 0 }$ respectively, such that

$${ \text{im}(T) = \text{span}(\mathscr{B}(w _1, 0)) \oplus \ldots \oplus \text{span}(\mathscr{B}(w _r, 0)) }$$

where ${ \mathscr{B}(w _i, 0) = (w _i, (T - 0 I) w _i, \ldots, (T - 0 I) ^{\text{exp}(w _i) - 1} w _i) . }$

For each ${ i }$ pick an element ${ v _i }$ such that ${ T(v _i) = w _i . }$ Consider the lists ${ \mathscr{B}(v _i, 0) = ( v _i, w _i, (T - 0 I) w _i, \ldots, (T - 0 I) ^{\text{exp}(w _i) - 1} w _i ) .}$
Consider the sum ${ \text{span}\mathscr{B}(v _1, 0) + \ldots + \text{span}\mathscr{B}(v _r, 0) . }$ As a step towards proving the result, can we show this sum is a direct sum?
We will show this is a direct sum.
Let

$${ \tilde{v _1} + \ldots + \tilde{v _r} = 0, \quad \text{ each } \tilde{v _i} \in \text{span} \mathscr{B}(v _i, 0) . }$$

Hence

$${ T(\tilde{v _1}) + \ldots + T(\tilde{v _r}) = 0, \quad \text{ each } T(\tilde{v _i}) \in T(\text{span} \mathscr{B}(v _i, 0)) = \text{span} \mathscr{B}(w _i, 0). }$$

Since ${ \text{span} \mathscr{B}(w _i, 0) }$ are independent, hence each ${ T (\tilde{v _i}) = 0 , }$ that is each ${ \tilde{v _i} \in \text{span} \mathscr{B}(v _i, 0) \cap \text{ker}(T) .}$ Note that ${ \text{span} \mathscr{B}(v _i, 0) \cap \text{ker}(T) }$ ${ = \text{span}( (T - 0 I) ^{\text{exp}(w _i) - 1} w _i ) }$ ${ \subseteq \text{span} \mathscr{B}(w _i, 0) . }$ Hence each ${ \tilde{v _i} \in \text{span} \mathscr{B}(w _i, 0) , }$ that is

$${ \tilde{v _1} + \ldots + \tilde{v _r} = 0, \quad \text{ each } \tilde{v _i} \in \text{span} \mathscr{B}(w _i, 0). }$$

Since ${ \text{span} \mathscr{B}(w _i, 0) }$ are independent, each ${ \tilde{v _i} = 0 }$ as needed.

Hence consider the direct sum

$${ \text{span}\mathscr{B}(v _1, 0) \oplus \ldots \oplus \text{span}\mathscr{B}(v _r, 0) . }$$

Note that

$${ {\begin{aligned} &\, T(\text{span}\mathscr{B}(v _1, 0) \oplus \ldots \oplus \text{span}\mathscr{B}(v _r, 0)) \\ = &\, T(\text{span} \mathscr{B}(v _1, 0)) + \ldots + T(\text{span} \mathscr{B}(v _r, 0)) \\ = &\, \text{span} \mathscr{B}(w _1, 0) + \ldots + \text{span} \mathscr{B}(w _r, 0). \end{aligned}} }$$

Hence consider the linear operator

$${ T \big \vert _U, \quad U := \text{span}\mathscr{B}(v _1, 0) \oplus \ldots \oplus \text{span}\mathscr{B}(v _r, 0) .}$$

Can we express ${ V = U \oplus N }$ such that ${ N }$ is a ${ T }$ invariant subspace? Then it suffices to find a Jordan decomposition of ${ T \vert _N }$ on ${ N . }$

Note that ${ V = U + \text{ker}(T) . }$ Let ${ v \in V . }$ Since ${ T(U) = \text{im}(T) , }$ there is an element ${ u \in U }$ such that ${ T(u) = T(v) . }$ Hence ${ v = u + (v - u) }$ with ${ u \in U, (v - u) \in \text{ker}(T) , }$ as needed.

Hence we can extend the basis ${ (\mathscr{B}(v _1, 0), \ldots, \mathscr{B}(v _r, 0)) }$ of ${ U }$ to a basis ${ (\mathscr{B}(v _1, 0), \ldots, \mathscr{B}(v _r, 0); z _1, \ldots, z _{\ell} ) }$ of ${ V , }$ where each ${ z _i \in \text{ker}(T) . }$

Hence there are generalised eigenvectors ${ v _1, \ldots, v _r; z _1, \ldots, z _{\ell} }$ with eigenvalues ${ 0, \ldots, 0; 0, \ldots , 0 }$ such that

$${ V = \text{span} \mathscr{B} (v _1, 0) \oplus \ldots \oplus \text{span} \mathscr{B}(v _r, 0) \oplus \text{span} \mathscr{B}(z _1, 0) \oplus \ldots \oplus \text{span} \mathscr{B}(z _{\ell}, 0), }$$

as needed. Here the bases ${ \mathscr{B} (v _i, 0) = ( v _i, w _i, (T - 0 I) w _i, \ldots, (T - 0 I) ^{\text{exp}(w _i) - 1} w _i ) }$ and ${ \mathscr{B} (z _j , 0) = (z _j ) . }$ ${ \blacksquare }$