Computing Jordan Forms and Using Their Nilpotent Parts

Learning goals and prerequisites

A characteristic polynomial does not by itself specify a matrix up to similarity. A complete Jordan computation must determine the block sizes and produce a basis that actually satisfies the chain equations. After doing that, powers and polynomial expressions can be calculated without multiplying long products of the original matrix.

We use the split Jordan existence and uniqueness theorem from the preceding chapter, exact elimination, polynomial evaluation and the minimal polynomial. Matrices in this chapter have real or complex entries; real matrices may be treated over $\mathbb C$ when their characteristic polynomials do not split over $\mathbb R$. Polynomial derivatives below are ordinary derivatives in characteristic zero. Discrete recurrences and bounded powers are applications of canonical forms.

1. An exact workflow, with a basis certificate

For each eigenvalue $\lambda$, compute successive kernels of $A-\lambda I$. Their dimensions $d_k$ give increments $q_k=d_k-d_{k-1}$, the numbers of blocks of size at least $k$. The differences $q_k-q_{k+1}$ give exact-size counts. Continue until the relevant generalized eigenspace has been reached; that may have dimension smaller than the whole space.

Then construct compatible chains. A long-chain top must belong to the required kernel but not the preceding one, and chosen chains must remain jointly independent. The kernel-filtration construction is a systematic way to ensure this. Solving $Nv_{j+1}=v_j$ separately for arbitrary previously chosen bottoms is not always possible.

Finally form the matrix $P$ with all chain vectors as columns, each chain ordered bottom to top. There are two distinct checks:

1. $P$ is invertible, because its columns form a basis.
2. $AP=PJ$, where $J$ is the proposed block matrix.

The second identity verifies every chain equation at once. Once both are established, multiplying by $P^{-1}$ proves $P^{-1}AP=J$. Checking $AP=PJ$ for a singular $P$ would only describe an invariant subspace, not a similarity on the full space.

Worked example 1: remove a coupling between distinct eigenvalues

Consider

$$A=\begin{pmatrix} 2&1&0&-3\\ 0&2&1&0\\ 0&0&2&0\\ 0&0&0&-1 \end{pmatrix}.$$

Triangularity gives $\chi_A(t)=(t-2)^3(t+1)$. For $N=A-2I$, the first three coordinate vectors form a length-three chain: $Ne_1=0$, $Ne_2=e_1$, $Ne_3=e_2$. On the fourth coordinate, $Ne_4=-3e_1-3e_4$. Consequently $\ker N$, $\ker N^2$, and $\ker N^3$ have dimensions $1,2,3$; the nonzero fourth diagonal entries of these powers exclude a fourth-coordinate contribution to each kernel. Thus the eigenvalue-two part is exactly one size-three block.

For eigenvalue $-1$, solving $(A+I)w=0$ gives $w=e_1+e_4$ up to scalar. Indeed $A(e_1+e_4)=-(e_1+e_4)$. Hence

$$P=(e_1\ e_2\ e_3\ e_1+e_4) =\begin{pmatrix}1&0&0&1\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}.$$

Its determinant is one, and the four displayed action equations give

$$AP=P\bigl(J_3(2)\oplus[-1]\bigr).$$

The characteristic and minimal polynomials are both $(t-2)^3(t+1)$, because the largest block sizes are three for eigenvalue two and one for eigenvalue minus one. The original upper-right entry is not a new Jordan link between different eigenvalues; the new eigenvector removes it.

Worked example 2: two chains rather than one long chain

Let

$$B=\begin{pmatrix} 3&1&0&1\\ 0&3&0&0\\ 0&1&3&-1\\ 0&0&0&3 \end{pmatrix},\qquad N=B-3I.$$

The images are $Ne_2=e_1+e_3$, $Ne_4=e_1-e_3$, and $Ne_1=Ne_3=0$. The two nonzero images are independent, so $\operatorname{rank}N=2$, $\dim\ker N=2$, and $N^2=0$. The block counts are therefore two blocks of size two, not a single block of size four.

Choose chains $(e_1+e_3,e_2)$ and $(e_1-e_3,e_4)$. The column matrix

$$P_B=\begin{pmatrix}1&0&1&0\\0&1&0&0\\1&0&-1&0\\0&0&0&1\end{pmatrix}$$

has determinant $-2$. Applying $B$ to each column verifies $BP_B=P_B(J_2(3)\oplus J_2(3))$. Its characteristic polynomial is $(t-3)^4$, but its minimal polynomial is only $(t-3)^2$.

2. Polynomial evaluation retains derivative data

Let $J=\lambda I+N$ with $N^r=0$. For a polynomial $p$, its finite Taylor expansion at $\lambda$ is an exact polynomial identity. Substituting $N$ into that identity gives

$$p(J)=\sum_{j=0}^{r-1}\frac{p^{(j)}(\lambda)}{j!}N^j.$$

To verify the polynomial identity directly, write $p(t)=\sum_{\ell=0}^d a_\ell t^\ell$. The finite binomial theorem gives

$$p(\lambda+z)=\sum_{j=0}^d\left(\sum_{\ell=j}^d a_\ell\binom{\ell}{j}\lambda^{\ell-j}\right)z^j.$$

In characteristic zero, the coefficient in parentheses is exactly $p^{(j)}(\lambda)/j!$, by differentiating each monomial $j$ times. Substitute the commuting operator $z=N$ and discard powers $N^j$ with $j\ge r$. This proves the displayed finite formula and explains precisely why factorial denominators are valid here.

Terms above degree $r-1$ vanish because they contain $N^r$. This is finite polynomial algebra; no convergence or interchange of infinite series is needed. In particular,

$$p(J_3(\lambda))= \begin{pmatrix} p(\lambda)&p'(\lambda)&p''(\lambda)/2\\ 0&p(\lambda)&p'(\lambda)\\ 0&0&p(\lambda) \end{pmatrix}.$$

Eigenvalues alone would capture the diagonal entries but miss all derivative information above them.

If $A=PJP^{-1}$, then $A^k=PJ^kP^{-1}$ for all nonnegative integers $k$, by cancellation of neighboring factors $P^{-1}P$. Taking finite linear combinations gives $p(A)=Pp(J)P^{-1}$.

Worked example 3: a polynomial and an inverse in the original coordinates

For the matrix $A$ of Worked example 1, take $p(t)=t^4-3t+2$. Its relevant data are

$$p(2)=12,\quad p'(2)=29,\quad p''(2)/2=24,\quad p(-1)=6.$$

Thus its size-three block becomes $\begin{pmatrix}12&29&24\\0&12&29\\0&0&12\end{pmatrix}$, and the remaining scalar block becomes $[6]$. Conjugating with the displayed $P$ gives

$$p(A)=\begin{pmatrix}12&29&24&-6\\0&12&29&0\\0&0&12&0\\0&0&0&6\end{pmatrix}.$$

The last column can also be checked without multiplying inverses: $p(A)e_1=12e_1$ and $p(A)(e_1+e_4)=6(e_1+e_4)$ imply $p(A)e_4=-6e_1+6e_4$.

For any block with $\lambda\ne0$, the finite geometric identity gives

$$(\lambda I+N)^{-1}=\frac1\lambda\sum_{j=0}^{r-1}\left(-\frac N\lambda\right)^j.$$

Indeed multiplying cancels all interior powers and leaves $I-(-N/\lambda)^r=I$. Applying this to the blocks and conjugating gives

$$A^{-1}=\begin{pmatrix} 1/2&-1/4&1/8&-3/2\\ 0&1/2&-1/4&0\\ 0&0&1/2&0\\ 0&0&0&-1 \end{pmatrix}.$$

An eigenvalue zero would prevent this inverse formula for that block and would make the full matrix singular.

3. Matching a matrix function by finite polynomial data

Suppose the largest block size at $\lambda_i$ is $r_i$, so

$$m_A(t)=\prod_i(t-\lambda_i)^{r_i}.$$

For polynomials $p,q$,

$$p(A)=q(A)\quad\Longleftrightarrow\quad p^{(j)}(\lambda_i)=q^{(j)}(\lambda_i) \quad\text{for all }i\text{ and }0\le j<r_i.$$

To prove this, the minimal-polynomial division theorem gives $p(A)=q(A)$ exactly when $m_A$ divides $p-q$. Expanding a polynomial at $\lambda_i$ shows that divisibility by $(t-\lambda_i)^{r_i}$ is equivalent to the first $r_i$ Taylor coefficients, or those derivatives, being zero. Distinct linear factors are relatively prime, so the separate divisibilities combine to divisibility by their product. This proves both implications.

Consequently a polynomial matching the values alone is sufficient only for size-one blocks. Larger blocks require matching derivatives as well. Polynomial interpolation with repeated-node derivative conditions is often called Hermite interpolation. Here the criterion is proved; no general interpolation existence theorem is needed for the concrete computations below.

4. Powers, discrete recurrences, and the zero-eigenvalue case

Since $\lambda I$ commutes with $N$, the finite binomial theorem gives

$$J_r(\lambda)^k=\sum_{j=0}^{\min(k,r-1)}\binom{k}{j}\lambda^{k-j}N^j\qquad(k\ge0).$$

The coefficient convention is $\lambda^0=1$, including when $\lambda=0$, as in polynomial evaluation. The formula does not use negative powers of zero because $j\le k$. For a zero-eigenvalue block it reduces to $J_r(0)^k=N^k$, which vanishes once $k\ge r$. For $\lambda\ne0$, the first superdiagonal is $k\lambda^{k-1}$ and the second is $\binom{k}{2}\lambda^{k-2}$. These polynomial factors record the nontrivial chains.

Worked example 4: an exact recurrence with a quadratic factor

For the matrix $A$ of Worked example 1, the block formula and the same $P$ yield, for every integer $k\ge0$,

$$A^k=\begin{pmatrix} 2^k&k2^{k-1}&\binom{k}{2}2^{k-2}&(-1)^k-2^k\\ 0&2^k&k2^{k-1}&0\\ 0&0&2^k&0\\ 0&0&0&(-1)^k \end{pmatrix}.$$

Here $\binom{k}{2}=0$ for $k=0,1$, and the base two is nonzero, so all scalar powers displayed are well-defined even at those small indices. The formula gives $A^0=I$ and $A^1=A$.

The recurrence $x_{k+1}=Ax_k$ with $x_0=e_3+e_4$ has the unique solution $x_k=A^kx_0$: induction verifies the recurrence, and each next value is determined by the preceding one. Explicitly,

$$x_k=\begin{pmatrix} \binom{k}{2}2^{k-2}+(-1)^k-2^k\\ k2^{k-1}\\ 2^k\\ (-1)^k \end{pmatrix}.$$

At $k=0$ this gives $(0,0,1,1)^T$, and at $k=1$ it gives $(-3,1,2,-1)^T$, agreeing with direct multiplication. The quadratic-in-$k$ factor is caused by a length-three chain, not by a third distinct eigenvalue.

5. What the invariants do and do not classify

For a split matrix, the characteristic polynomial records the total block sizes at each eigenvalue. The minimal polynomial records the largest such size. The eigenspace dimension records the number of blocks. These data can leave several possible partitions; the entire sequence of kernel increments resolves the ambiguity.

Worked example 5: same characteristic and minimal polynomials, different similarity classes. Let

$$N_1=J_3(0)\oplus J_3(0)\oplus J_2(0),\qquad N_2=J_3(0)\oplus J_2(0)\oplus J_2(0)\oplus J_1(0).$$

Both are eight-dimensional, have characteristic polynomial $t^8$, and have minimal polynomial $t^3$. But $\dim\ker N_1=3$ while $\dim\ker N_2=4$. Similarity preserves kernels, because $B=P^{-1}AP$ implies $\ker B=P^{-1}(\ker A)$. The different nullities therefore prove that these matrices are not similar. Their fuller nullity sequences are $3,6,8$ and $4,7,8$.

For two split matrices, matching all eigenvalues and every kernel-increment sequence gives the same Jordan blocks up to order. Their Jordan matrices are then related by a permutation of whole block bases, so the original matrices are similar. This converse uses the existence and uniqueness theorem, not only invariance of the listed data.

6. Two precise conclusions about powers

Fix a basis and call the powers bounded if every entry of $A^k$ is bounded as $k$ varies. A fixed similarity preserves this property in both directions: each entry of $PJP^{-1}$ is a finite linear combination of entries of $J$, and conversely for $J=P^{-1}AP$. The same reasoning preserves entrywise convergence to zero.

Theorem. Over $\mathbb C$:

1. $A^k\to0$ if and only if every eigenvalue has magnitude strictly below one.
2. The powers are bounded if and only if every eigenvalue has magnitude at most one and every Jordan block belonging to an eigenvalue of magnitude one has size one.

Proof. If $0<\rho<1$, then $k^j\rho^k\to0$ for each fixed nonnegative integer $j$. For $j>0$, the ratio of successive positive terms is $\rho(1+1/k)^j\to\rho<1$; choose a number strictly between $\rho$ and one to bound this ratio eventually, and compare the remaining terms with a decaying geometric sequence. The case $j=0$ is already geometric. Since $\binom{k}{j}\le k^j/j!$, every entry in a block with $0<|\lambda|<1$ tends to zero by the power formula. A block with $\lambda=0$ vanishes after finitely many powers. Similarity then proves sufficiency in statement 1.

For necessity, if $Av=\lambda v$ with $v\ne0$, then $A^kv=\lambda^kv$. An eigenvalue with $|\lambda|\ge1$ prevents this vector sequence from tending to zero, whereas entrywise $A^k\to0$ would make every component of $A^kv$ tend to zero. This proves statement 1.

For statement 2, blocks strictly inside the unit circle tend to zero and therefore have bounded entries; size-one blocks on the unit circle have entries of magnitude one. This proves sufficiency. If $|\lambda|>1$, the eigenvector argument produces an unbounded component, incompatible with bounded matrix entries. If $|\lambda|=1$ and a block has size at least two, its first superdiagonal in the $k$th power has magnitude $|k\lambda^{k-1}|=k$. Its entries are unbounded, and similarity preserves that failure. These are exactly the two obstructions in the statement. $\square$

For real matrices the same conclusions follow by viewing the entries over $\mathbb C$; their real powers are unchanged. In particular “all eigenvalues have magnitude at most one” is not, by itself, a boundedness criterion for defective matrices.

7. Graded exercises

P01 — Computation. For $C=\begin{pmatrix}4&2&0\\0&4&3\\0&0&4\end{pmatrix}$, construct a chain basis, give $P$, and verify $CP=PJ_3(4)$. Find the characteristic and minimal polynomials.

P02 — Basis audit. For $B$ and $P_B$ of Worked example 2, verify the four column equations and compute $\det P_B$. Explain why four independent generalized eigenvectors with no checked chain relations would not by themselves identify the displayed Jordan matrix.

P03 — Classification. A nilpotent operator in dimension ten has kernel dimensions $4,7,9,10$. Determine its Jordan form, characteristic polynomial, and minimal polynomial.

P04 — Diagnosis. Can nilpotent kernel dimensions start $3,5,8$? Identify the first violated constraint and explain why nondecreasing nullities are not a sufficient test.

P05 — Powers. For $J=J_3(-1)$, derive a formula for $J^k$ for all $k\ge0$, explicitly showing the entries on its first two superdiagonals. Decide whether these powers are bounded.

P06 — Polynomial. Compute $p(J_3(2))$ for $p(t)=t^3-2t+1$ using derivative data. Verify the constant diagonal value and each superdiagonal coefficient.

P07 — Inverse. Find $J_3(2)^{-1}$ as a polynomial in $N=J_3(2)-2I$. Multiply the two expressions to verify the answer without an elimination algorithm.

P08 — Similarity. Give two non-similar nine-dimensional nilpotent matrices with characteristic polynomial $t^9$, minimal polynomial $t^4$, and the same kernel dimension. Prove non-similarity using a higher-power nullity invariant, not a visual comparison of their written matrices.

P09 — Repeated-node conditions. Suppose $A$ has minimal polynomial $(t-1)^2(t-2)$. Find a polynomial $q$ of degree at most two with $q(A)=A^{-1}$. Derive and use the necessary value/derivative conditions, then verify the inverse algebraically modulo the minimal polynomial.

P10 — Power behavior. Let $D=J_2(1/2)\oplus[-1/3]$. Prove directly that $D^k\to0$, giving a geometric-ratio bound for its off-diagonal entry. Contrast this with $J_2(1)$, whose eigenvalues also have magnitude at most one.

8. Full solutions

S01. Set $N=C-4I$. Choose $v_3=e_3$, $v_2=3e_2$, and $v_1=6e_1$. Then $Nv_3=3e_2=v_2$, $Nv_2=6e_1=v_1$, and $Nv_1=0$. Thus $P=\operatorname{diag}(6,3,1)$ has determinant $18\ne0$, and these equations give $CP=PJ_3(4)$. Triangularity gives $\chi_C=(t-4)^3$. Since $N^2e_3=6e_1\ne0$ while $N^3=0$, the minimal polynomial is $(t-4)^3$ by the chain-block theorem.

S02. The columns are $u_1=e_1+e_3$, $u_2=e_2$, $v_1=e_1-e_3$, and $v_2=e_4$. The defining entries of $B$ give $Bu_1=3u_1$, $Bu_2=u_1+3u_2$, $Bv_1=3v_1$, and $Bv_2=v_1+3v_2$. These are exactly the columns of $P_B(J_2(3)\oplus J_2(3))$. Expanding the determinant along the second and fourth rows leaves the $2\times2$ determinant $\det\begin{pmatrix}1&1\\1&-1\end{pmatrix}=-2$. Independence proves a basis, but the actual coefficients of the operator in that basis still depend on how each vector maps. Generalized eigenvector status alone gives no prescribed adjacent-chain images; the checked equations supply them.

S03. With $d_0=0$ and $d_5=10$, the increments are $4,3,2,1,0$. Their consecutive differences give one block each of sizes one, two, three, and four. Hence the Jordan form is $J_4(0)\oplus J_3(0)\oplus J_2(0)\oplus J_1(0)$ up to block order. The size sum is ten, so the characteristic polynomial is $t^{10}$. The largest block has size four, so the minimal polynomial is $t^4$.

S04. The increments would be $3,2,3$. The third exceeds the second, violating the injection $K_3/K_2\to K_2/K_1$. The quotient dimensions would be three and two, respectively, which cannot admit that injection. Although the raw nullities increase, their increments must be nonincreasing; that stronger condition is the failed one here.

S05. Write $J=-I+N$, where $N^3=0$. The binomial formula gives

$$J^k=(-1)^k\left(I-kN+\binom{k}{2}N^2\right).$$

This includes $k=0,1$ by the convention that $\binom{k}{2}=0$ there. The diagonal entries are $(-1)^k$, the first superdiagonal entries are $(-1)^{k+1}k$, and the second superdiagonal entry is $(-1)^k\binom{k}{2}$. Already the first superdiagonal magnitudes grow as $k$, so the powers are not bounded.

S06. We have $p(2)=8-4+1=5$, $p'(2)=3(2)^2-2=10$, and $p''(2)/2=6(2)/2=6$. Therefore

$$p(J_3(2))=5I+10N+6N^2=\begin{pmatrix}5&10&6\\0&5&10\\0&0&5\end{pmatrix}.$$

The constant diagonal is the value of $p$ at the eigenvalue. The first and second superdiagonals come from $N$ and $N^2$, so their coefficients are respectively the first derivative and half the second derivative, not additional eigenvalues.

S07. Since $N^3=0$,

$$(2I+N)^{-1}=\frac12I-\frac14N+\frac18N^2.$$

Multiplication gives $I+(-1/2+1/2)N+(1/4-1/4)N^2+(1/8)N^3=I$. The same product in the reverse order is identical because both factors are polynomials in $N$. Thus it is a two-sided inverse.

S08. Take $M_1=J_4(0)\oplus J_4(0)\oplus J_1(0)$ and $M_2=J_4(0)\oplus J_3(0)\oplus J_2(0)$. Each has total size nine, largest block size four, and three blocks. Hence they have the required characteristic and minimal polynomials and both have kernel dimension three. However, $\dim\ker M_1^2=2+2+1=5$, whereas $\dim\ker M_2^2=2+2+2=6$. If the matrices were similar, their squares would be similar by the same basis change, and the square kernels would be isomorphic. Their different dimensions contradict this. Even matching the characteristic polynomial, minimal polynomial, and eigenspace dimension has not classified similarity.

S09. The minimal polynomial has nonzero constant term, so $A$ is invertible. To make $q(A)$ its inverse it is sufficient and necessary that $tq(t)-1$ be divisible by $(t-1)^2(t-2)$. At the double root one this requires $q(1)=1$ and $q(1)+q'(1)=0$, hence $q'(1)=-1$. At two it requires $2q(2)=1$. Write $q(t)=a+b(t-1)+c(t-1)^2$. The conditions give $a=1$, $b=-1$, and $c=1/2$. Thus

$$q(t)=\tfrac12t^2-2t+\tfrac52.$$

Indeed

$$tq(t)-1=\tfrac12(t^3-4t^2+5t-2)=\tfrac12(t-1)^2(t-2).$$

Evaluation at $A$ makes this zero, so $Aq(A)=I$. Since $q(A)$ commutes with $A$, it is also a left inverse. Matching $q(1)$ and $q(2)$ alone would not impose the derivative condition associated with the length-two block.

S10. The size-two block power has diagonal entries $2^{-k}$ and off-diagonal entry $k(1/2)^{k-1}$. Let $b_k=k2^{1-k}$ for $k\ge1$. Its ratio is $b_{k+1}/b_k=(k+1)/(2k)\le3/4$ for $k\ge2$. Hence $b_k\le b_2(3/4)^{k-2}\to0$ for $k\ge2$. The remaining scalar entry $(-1/3)^k$ also tends to zero. Thus every entry of $D^k$ tends to zero. In contrast $J_2(1)^k=I+kN$ has off-diagonal entry $k$, which is unbounded. The distinction is the nontrivial chain at an eigenvalue on the unit circle, not merely the inequality $|\lambda|\le1$.

Summary

A full Jordan computation supplies block counts, actual chains, an invertible basis matrix, and the identity $AP=PJ$. Defective blocks retain derivative information in polynomial evaluation and polynomial factors in their powers. Nullity sequences distinguish similarity classes that characteristic and minimal polynomials may fail to distinguish. The same block formulas give exact convergence and boundedness criteria for powers, with unit-circle chains treated separately.