Constructing Jordan Bases from Kernel Filtrations

Learning goals and prerequisites

A Jordan block is easy to recognize once it is displayed. The substantial theorem is that enough compatible chains can be found to fill the whole space. This chapter proves that theorem, rather than assuming that arbitrary bases of successive kernels will fit together. It then proves the decomposition into generalized eigenspaces and the uniqueness of all block sizes.

We assume finite-dimensional vector spaces over a field $F$, basis extension, rank–nullity, polynomial division and Bézout identities, and the general Cayley–Hamilton theorem proved earlier. The quotient-space notation needed for the proof is recalled below. The split Jordan theorem requires the characteristic polynomial to split over $F$. This is automatic over $\mathbb C$, but not over $\mathbb R$. The zero-dimensional space has the empty Jordan basis; below, a nilpotency index is discussed only for a nonzero space.

1. Chains, conventions, and the compatibility problem

For an operator $T$ and a scalar $\lambda$, a Jordan chain of length $r$ is a list $v_1,\ldots,v_r$ satisfying

$$(T-\lambda I)v_1=0,\qquad (T-\lambda I)v_j=v_{j-1}\quad(2\le j\le r),\qquad v_1\ne0.$$

The bottom vector is $v_1$ and the top is $v_r$. The list is independent. Indeed, in a nonzero relation choose the largest index $j$ with nonzero coefficient and apply $(T-\lambda I)^{j-1}$. All lower-index terms vanish, leaving a nonzero coefficient times $v_1$, a contradiction.

In the ordered chain basis $v_1,\ldots,v_r$, the matrix is

$$J_r(\lambda)=\begin{pmatrix} \lambda&1&0&\cdots&0\\ 0&\lambda&1&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&0\\ 0&\cdots&0&\lambda&1\\ 0&\cdots&\cdots&0&\lambda \end{pmatrix}.$$

The ones are above the diagonal because the image of the $j$th basis vector contains the $(j-1)$st. Reversing the order changes this convention.

A chain bottom cannot always be lifted. For $N=J_3(0)\oplus[0]$, the vector $e_4$ lies in $\ker N$ but not in $NV$. There is no $w$ with $Nw=e_4$, even though another bottom $e_1$ can be lifted twice. Thus “choose a basis of the kernel and lift every vector” is not a valid construction. We must coordinate choices at every level.

2. Successive kernel quotients and an injective map

Let $N$ be nilpotent, with $N^s=0$. Set

$$K_j=\ker N^j\quad(0\le j\le s),\qquad K_0=\{0\},\qquad K_s=V.$$

Here $K_j/K_{j-1}$ consists of cosets $v+K_{j-1}$, with two representatives equivalent when their difference belongs to $K_{j-1}$. Addition and scalar multiplication act on representatives; changing representatives changes the result only by another element of the subspace, so these operations are well-defined. If a basis of $K_{j-1}$ is extended to one of $K_j$, the classes of the added vectors form a quotient basis: they span by the extended expansion, and a relation among their classes would contradict independence of that extended basis. Thus $\dim(K_j/K_{j-1})=\dim K_j-\dim K_{j-1}$.

For $1\le j<s$, define

$$\phi_j:K_{j+1}/K_j\longrightarrow K_j/K_{j-1},\qquad [v]\longmapsto[Nv].$$

This map is well-defined: $N^{j+1}v=0$ implies $Nv\in K_j$, and changing $v$ by an element of $K_j$ changes $Nv$ by an element of $K_{j-1}$. It is linear because $N$ and quotient operations are linear. It is injective: if $Nv\in K_{j-1}$, then $N^jv=N^{j-1}(Nv)=0$, so $v\in K_j$ and $[v]=0$.

Put $d_j=\dim K_j$ and $q_j=d_j-d_{j-1}$. The injective maps give

$$q_1\ge q_2\ge\cdots\ge q_s\ge0.$$

This restriction has been proved without assuming Jordan form. Increasing kernels alone do not impose it; their quotient maps do.

A basis lemma for a filtration

If a set $L_j\subseteq K_j$ represents a basis of $K_j/K_{j-1}$ for each $j$, then the union of all $L_j$ is a basis of $V$.

To prove spanning, take $v\in K_s$. Subtract a linear combination of $L_s$ with the same quotient class; the remainder lies in $K_{s-1}$. Repeat until the remainder lies in $K_0=0$. For independence, take a relation and let $j$ be its largest occupied level. Reducing modulo $K_{j-1}$ forces every coefficient on $L_j$ to vanish, since those classes form a quotient basis. Descending through the finite levels makes every coefficient vanish.

3. A complete nilpotent chain construction

Theorem 1. Every nilpotent operator on a finite-dimensional space has a basis made of chains.

Proof. At level $s$, choose representatives $L_s$ of a basis of $K_s/K_{s-1}$. Suppose representatives $L_{j+1}$ of a basis of $K_{j+1}/K_j$ have been chosen. Their images $N L_{j+1}$ represent the images of that quotient basis under the injective map $\phi_j$, so their classes are independent in $K_j/K_{j-1}$. Extend these classes to a basis of $K_j/K_{j-1}$. For each newly added class choose a representative in $K_j$, and define $L_j$ to consist of those new representatives together with the exact vectors $N L_{j+1}$.

Continue downward to $j=1$. The filtration lemma proves that the union of the $L_j$ is a basis of $V$. For each vector first introduced at level $r$, follow its successive images under $N$. At every lower level the image was included as an exact representative, not merely as a matching quotient class. A vector introduced at level $r$ has nonzero class modulo $K_{r-1}$, so $N^{r-1}v\ne0$, while $N^rv=0$. It therefore generates a chain of length exactly $r$.

Above level one, no two vectors in a given level have the same image chain position: injectivity makes the images of its chosen quotient basis independent at the next level. Consequently the basis union is partitioned into the chains starting at the newly introduced representatives. In bottom-to-top order each chain has the required relations. The resulting full basis is a nilpotent Jordan basis. $\square$

The construction introduces $q_j-q_{j+1}$ new chain tops at level $j$, where $q_{s+1}=0$. Thus the number of chains of length exactly $j$ is already visible during the construction.

Worked example 1: compatible levels in dimension five. Define $N$ on the standard basis of $\mathbb R^5$ by

$$Ne_1=0,\quad Ne_2=e_1,\quad Ne_3=e_2+e_4,\quad Ne_4=0,\quad Ne_5=e_4.$$

Its range is spanned by the independent vectors $e_1,e_2+e_4,e_4$, so $d_1=2$. Its squared range is spanned by $e_1$, so $d_2=4$. Since $N^3=0$, $d_3=5$. The quotient dimensions are $q_1=2,q_2=2,q_3=1$.

Choose $e_3$ at level three. Its image $e_2+e_4$ is the inherited representative at level two; append $e_5$ as the second quotient-basis representative there. At level one their images are $e_1,e_4$, a basis of $K_1$. The chains are

$$(e_1,e_2+e_4,e_3),\qquad(e_4,e_5).$$

Their five vectors are independent: the change-of-basis matrix is triangular with diagonal entries all one. Directly applying $N$ gives $J_3(0)\oplus J_2(0)$. The quotient construction has supplied actual vectors, not just the sizes $3,2$.

4. Separate eigenvalues by polynomial projections

Assume $\chi_T(t)=\prod_{i=1}^r p_i(t)$, where

$$p_i(t)=(t-\lambda_i)^{a_i},\qquad a_i\ge1,$$

and the $\lambda_i$ are distinct elements of $F$. Define $h_i=\chi_T/p_i$. Since $p_i$ and $h_i$ are relatively prime, polynomial Bézout gives $u_i h_i+v_i p_i=1$. Set $e_i=u_i h_i$.

The polynomial $e_i$ is congruent to one modulo $p_i$ and to zero modulo every $p_j$ with $j\ne i$. Therefore $\sum_i e_i-1$ is divisible by every $p_j$, and hence by their product. For completeness, if relatively prime $p,q$ both divide $f$, write $f=pg$. A Bézout identity $ap+bq=1$ gives $g=apg+bqg$, whose two terms are divisible by $q$ because $pg=f$ is. Thus $q\mid g$ and $pq\mid f$. Repeating proves the product-divisibility assertion.

Cayley–Hamilton now gives

$$\sum_i e_i(T)=I.$$

Also $p_i e_i=u_i\chi_T$, so $e_i(T)V$ lies in

$$G_i=\ker p_i(T)=\ker(T-\lambda_iI)^{a_i}.$$

Every vector is consequently a sum of vectors in the $G_i$.

This sum is direct. On $G_i$, the polynomial $e_i-1$ is a multiple of $p_i$, so $e_i(T)$ acts as the identity. On $G_j$ for $j\ne i$, $e_i$ is a multiple of $p_j$, so it acts as zero. Applying $e_i(T)$ to a relation $\sum_j w_j=0$ with $w_j\in G_j$ leaves only $w_i=0$. Finally each $G_i$ is $T$-invariant, since $p_i(T)Tw=T p_i(T)w=0$ whenever $w\in G_i$.

We have proved the primary decomposition

$$V=G_1\oplus\cdots\oplus G_r.$$

On $G_i$, the operator $N_i=T-\lambda_iI$ has $N_i^{a_i}=0$. The nilpotent chain theorem applies to it. Combining the resulting bases of all the direct summands proves the split Jordan existence theorem for $T$.

Worked example 2: explicit projections with a repeated root. Suppose a real operator has characteristic polynomial $(t-1)^2(t+1)$. Set

$$e_{-1}(t)=\frac{(t-1)^2}{4},\qquad e_1(t)=1-\frac{(t-1)^2}{4}.$$

At $t=-1$, the first polynomial equals one, while $e_1=(3-t)(t+1)/4$ is divisible by $t+1$. Modulo $(t-1)^2$, $e_1$ equals one and $e_{-1}$ equals zero. Thus $e_1(T)$ and $e_{-1}(T)$ project onto $\ker(T-I)^2$ and $\ker(T+I)$, respectively. This produces the invariant pieces before a chain basis is chosen inside the first one.

5. Uniqueness, nullities, and the field boundary

Fix an eigenvalue $\lambda$, and let

$$d_k(\lambda)=\dim\ker(T-\lambda I)^k,\qquad d_0(\lambda)=0.$$

On a $\lambda$-block of size $r$, the kernel contribution is $\min(k,r)$. On a block for a different eigenvalue $\mu$, write $T-\lambda I=(\mu-\lambda)I+R$ with $R^r=0$. It is invertible, with inverse

$$\frac1{\mu-\lambda}\sum_{j=0}^{r-1}\left(-\frac R{\mu-\lambda}\right)^j,$$

as multiplication makes the finite geometric sum telescope. Such a block contributes zero to the kernel. Therefore

$$d_k(\lambda)=\sum_r b_r\min(k,r),$$

where $b_r$ counts the size-$r$ blocks for $\lambda$. Subtracting consecutive expressions gives

$$d_k-d_{k-1}=\#\{\lambda\text{-blocks of size at least }k\},$$

and subtracting these increments gives

$$2d_k-d_{k-1}-d_{k+1}=\#\{\lambda\text{-blocks of size exactly }k\}.$$

All kernel dimensions are intrinsic to the operator, so every block size is uniquely determined up to order. Together with existence this proves Jordan canonical form, not merely a candidate shape.

The characteristic polynomial of a block is $(t-\lambda)^r$. Multiplying block determinants shows that the sum of its $\lambda$-block sizes is the algebraic multiplicity of $\lambda$. In particular the stabilized value of $d_k(\lambda)$ is that multiplicity, not generally the dimension of the entire space.

The minimal polynomial retains the largest size instead. On a size-$r$ block, divide an annihilating polynomial by $(t-\lambda)^r$ and express the remainder as $\sum_{j=0}^{r-1}b_j(t-\lambda)^j$. Applying its evaluation to the top chain vector gives $\sum_j b_jv_{r-j}=0$, so chain independence makes every remainder coefficient zero. Conversely $(T-\lambda I)^r$ is zero on that block. A polynomial kills a direct sum exactly when it kills every block, so it must be divisible by the largest required power of each distinct linear factor. Those powers are relatively prime, making their product the monic annihilator of least degree. Thus the exponent of $t-\lambda$ in the minimal polynomial is precisely the largest $\lambda$-block size.

Worked example 3: identify a complete block partition. A nilpotent operator on an eight-dimensional space has successive kernel dimensions $3,5,7,8$. With $d_0=0$ and $d_5=8$, the increments are $3,2,2,1,0$. There are three blocks in total, two of size at least two, two of size at least three, and one of size at least four. The block sizes are therefore $4,3,1$, summing to eight. The largest chain has length four.

Worked example 4: splitting cannot be replaced by a vacuous condition. The real matrix $R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ has characteristic polynomial $t^2+1$ and no real eigenvalues. It has no split Jordan form over $\mathbb R$, since the diagonal of such a form would supply real eigenvalues. Over $\mathbb C$, its distinct eigenvalues $i,-i$ give two size-one blocks. This is a change of ground field, not a contradiction.

6. Graded exercises

P01 — Foundation. Prove a chain $v_1,\ldots,v_r$ is independent, including the case $r=1$. Explain why a chain top in $K_r\setminus K_{r-1}$ has exactly length $r$.

P02 — Proof. Verify all parts of the induced map $K_{j+1}/K_j\to K_j/K_{j-1}$: codomain, independence of representative, linearity, and injectivity. Deduce the inequality between consecutive kernel increments.

P03 — Core. A nilpotent operator in dimension nine has kernel dimensions $3,6,8,9$ for its first four powers. Determine every Jordan block size and verify that the resulting blocks reproduce those dimensions.

P04 — Diagnosis. Can a nilpotent operator have $d_1=2,d_2=5$? Explain using the injective quotient map, without assuming Jordan form first.

P05 — Core. For a real operator with characteristic polynomial $(t-1)^2(t+1)$, use the two polynomials in Worked example 2 to prove directly that the projections sum to $I$, have product zero, and are idempotent.

P06 — Proof. Suppose $N^2=0$ on an $n$-dimensional space and $\operatorname{rank}N=r$. Determine the numbers of blocks of sizes two and one, and prove $2r\le n$.

P07 — Computation. Let $T=2I+N$ on $\mathbb R^4$, where $Ne_1=0$, $Ne_2=2e_1$, $Ne_3=e_1+e_2$, and $Ne_4=0$. Construct a Jordan basis and give its invertible change-of-basis matrix.

P08 — Field audit. Why does “every real eigenvalue has a full-sized eigenspace” not prove that a real matrix is diagonalizable over $\mathbb R$? Give a concrete example and identify the missing condition.

P09 — Proof. Prove that the minimal polynomial of $J_r(\lambda)$ is $(t-\lambda)^r$. Do not infer its degree merely from its characteristic polynomial.

P10 — Core and field audit. For $T=J_2(1)\oplus[1]\oplus J_2(-2)$ over a field $F$, compute $d_1(1),d_2(1),d_3(1)$ separately when $\operatorname{char}F\ne3$ and when $\operatorname{char}F=3$. Explain which generalized eigenspace is being counted in each case.

7. Full solutions

S01. For $r=1$, a nonzero one-vector list is independent by the definition of linear independence. For general $r$, suppose $\sum c_jv_j=0$ has a nonzero coefficient. Let $j$ be its largest such index. Applying $(T-\lambda I)^{j-1}$ kills every term with smaller index and leaves $c_jv_1=0$, impossible because $c_j\ne0$ and $v_1\ne0$. A top $v\in K_r\setminus K_{r-1}$ satisfies $N^rv=0$ and $N^{r-1}v\ne0$. Its bottom-to-top sequence $N^{r-1}v,\ldots,Nv,v$ has exactly $r$ nonzero entries, obeys the chain equations, and cannot be extended downward by an additional nonzero image because its next image is zero.

S02. If $v\in K_{j+1}$, then $N^j(Nv)=N^{j+1}v=0$, so $Nv\in K_j$. If $v-w\in K_j$, then $N^{j-1}(Nv-Nw)=N^j(v-w)=0$, making $Nv-Nw\in K_{j-1}$. Thus quotient classes do not depend on the representative. Linearity follows from $N(\alpha v+\beta w)=\alpha Nv+\beta Nw$. If the image class is zero, $Nv\in K_{j-1}$, and therefore $N^jv=0$, so the original class in $K_{j+1}/K_j$ is zero. Injectivity implies $\dim(K_{j+1}/K_j)\le\dim(K_j/K_{j-1})$, which is $d_{j+1}-d_j\le d_j-d_{j-1}$.

S03. The increments are $3,3,2,1$, followed by zero because $K_4=V$. The exact-size counts are $3-3=0$ of size one, $3-2=1$ of size two, $2-1=1$ of size three, and $1-0=1$ of size four. Thus the sizes are $4,3,2$. They reproduce the dimensions as $1+1+1=3$, $2+2+2=6$, $3+3+2=8$, and $4+3+2=9$.

S04. These values would give $\dim(K_1/K_0)=2$ and $\dim(K_2/K_1)=3$. The injective linear map from the latter quotient to the former would inject a three-dimensional space into a two-dimensional space, contradicting rank–nullity. Hence the data are impossible, independently of an assumed Jordan classification.

S05. Write $E=e_{-1}(T)$ and $F=e_1(T)$. Their sum is $I$ by the polynomial identity. The polynomial product is

$$e_{-1}(t)e_1(t)=\frac{(t-1)^2(3-t)(t+1)}{16},$$

which is a multiple of the characteristic polynomial. Cayley–Hamilton gives $EF=FE=0$. Since $F=I-E$, the equality $E(I-E)=0$ gives $E^2=E$. Similarly $(I-F)F=0$ gives $F^2=F$. Thus they are complementary projections with no choice of basis required.

S06. Since $N^2=0$, every block has size one or two. Each size-two block contributes rank one and each size-one block contributes rank zero, so there are $r$ size-two blocks. The remaining dimension is $n-2r$, giving that many size-one blocks. Independently, $N^2=0$ implies $NV\subseteq\ker N$, so $r\le n-r$ by rank–nullity and hence $2r\le n$. This proves the nonnegativity of the proposed size-one count rather than presupposing it.

S07. Choose the top $v_3=e_3$. Its images are $v_2=e_1+e_2$ and $v_1=2e_1$, with $Nv_1=0$. Append $w=e_4$ as a length-one chain. The matrix with these columns is

$$P=\begin{pmatrix}2&1&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix},\qquad\det P=2\ne0.$$

The equations $Tv_1=2v_1$, $Tv_2=v_1+2v_2$, $Tv_3=v_2+2v_3$, and $Tw=2w$ give $TP=P(J_3(2)\oplus[2])$. This verifies the form and the chain ordering without computing $P^{-1}$.

S08. The rotation matrix of Worked example 4 has no real eigenvalues. The statement about every real eigenvalue is therefore vacuously true, while no real eigenbasis exists. The characteristic polynomial must split over the field, in addition to any conditions on dimensions of its actual eigenspaces. Over $\mathbb C$ this matrix does split and is diagonalizable, showing why the field must be named.

S09. Put $N=J_r(\lambda)-\lambda I$. It satisfies $N^r=0$, so $(t-\lambda)^r$ annihilates the block. Divide any polynomial $p$ by this monic polynomial and write its remainder in powers of $t-\lambda$ as $\sum_{j=0}^{r-1}b_j(t-\lambda)^j$. If $p(J_r(\lambda))=0$, applying it to the top vector $v_r$ gives $\sum_{j=0}^{r-1}b_jv_{r-j}=0$. The chain is independent, so all $b_j=0$. Hence every annihilating polynomial is divisible by $(t-\lambda)^r$, proving it is the minimal polynomial.

S10. If $\operatorname{char}F\ne3$, the two blocks for eigenvalue one contribute $\min(k,2)+\min(k,1)$, so the requested dimensions are $2,3,3$. On the $-2$ block, $T-I=-3I+R$ is invertible by the finite geometric inverse because $-3\ne0$. It contributes no kernel: only the three-dimensional generalized eigenspace for eigenvalue one is counted. If $\operatorname{char}F=3$, then $-2=1$. All three blocks now belong to eigenvalue one, and the kernel contribution is $\min(k,2)+1+\min(k,2)$ for $k\ge1$. The dimensions are therefore $3,5,5$, and the stabilized generalized eigenspace is all of $V$. Calling the last block a different primary eigenvalue or dividing by $-3$ in this case would be invalid.

Summary

The injective maps between kernel quotients ensure that chain choices can be made compatibly from the top downward. Polynomial projections supply the separate generalized eigenspaces, and the nilpotent construction supplies a basis in each. Kernel increments then prove uniqueness and identify exact block counts. Splitting over the stated field remains an essential hypothesis throughout.