3.6 Cayley–Hamilton, Schur, and the Jordan form

Not every matrix can be diagonalized, but every matrix can be made triangular by a unitary change of basis (Schur), satisfies its own characteristic polynomial (Cayley–Hamilton), and is similar to an almost-diagonal matrix of Jordan blocks. The Jordan form says exactly how diagonalization fails.

Chapter 2.8 ended with a warning: the shear $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ has the eigenvalue 1 twice but only one line of eigenvectors, so it cannot be diagonalized. What is the best we can do for such matrices? This chapter gives three answers of increasing strength. Every complex matrix can be made upper triangular by a rotation of axes (Schur). Every matrix satisfies its own characteristic polynomial (Cayley–Hamilton). And every matrix is similar to a block-diagonal matrix of simple pieces called Jordan blocks, which records exactly how far from diagonalizable it is.

Before you start

Chapter 2.8 (eigenvalues, characteristic polynomial, diagonalization) and 3.1 (the induction in the spectral theorem’s proof).

Quick checkWhy can't the shear

\begin{bmatrix}1&1\\0&1\end{bmatrix}

be diagonalized?

Quick checkIn the spectral theorem's proof, after finding one unit eigenvector

q_1

, what happened next?

The picture

Turn on eigenlines: the shear has only the horizontal axis. Every other direction is tilted towards it. The map is “the identity plus a push along the eigenline”: $\begin{bmatrix}1&1\\0&1\end{bmatrix} = I + N$ where $N = \begin{bmatrix}0&1\\0&0\end{bmatrix}$ sends $(0, 1)$ to $(1, 0)$ and $(1, 0)$ to $O$ . That pair of vectors, one sent onto the other and the other to zero, is a Jordan chain, and $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ is already a Jordan block.

Where this came from· Cayley, Hamilton, Jordan

Arthur Cayley’s 1858 memoir on matrices was the first paper to treat matrices as objects with their own algebra. In it he stated that a matrix satisfies its own characteristic equation, proved this for $2 \times 2$ matrices and said he had checked $3 \times 3$ by direct computation. He did not attempt the general case. William Rowan Hamilton had found a special case for quaternions in the 1850s, which is why both names are attached. Ferdinand Georg Frobenius published a general proof in 1878. Camille Jordan gave the normal form now named after him in his 1870 treatise on substitutions (Traité des substitutions et des équations algébriques), and Karl Weierstrass’s work on pairs of quadratic forms around the same time contains a closely related result.

Invariant subspaces and Schur’s theorem

Definition B

Invariant subspace

A subspace $W$ is invariant under $A$ if $Aw \in W$ for every $w \in W$ . Eigenspaces, the null space and the range of $A$ are invariant.

In one lineA subspace W is invariant under A if A maps W into itself: w in W implies Aw in W.

Why it existsOn an invariant subspace, A can be studied as a smaller map, which is how every structure theorem in this chapter is proved.

Definition T

Schur decomposition

A Schur decomposition of a complex square matrix $A$ is $A = UTU^*$ with $U$ unitary (3.3) and $T$ upper triangular. The diagonal of $T$ lists the eigenvalues of $A$ .

In one lineA = UTU* with U unitary and T upper triangular: every complex matrix is triangular in a suitable orthonormal basis.

Why it existsTriangular matrices show their eigenvalues on the diagonal, and a unitary change of basis is numerically safe.

Theorem T Spagnolie Lecture 18 (stated); Axler 6.38

Schur's theorem

Every complex $n \times n$ matrix $A$ can be written $A = UTU^*$ with $U$ unitary and $T$ upper triangular.

In one lineEvery complex square matrix has a Schur decomposition A = UTU*.

Why it is truePeel off one unit eigenvector at a time, as in the spectral theorem; without symmetry the rest only stays 'above the diagonal', which is exactly triangular.

ProofAxler 6.38

The idea: Induction on n. Take a unit eigenvector u₁ (one exists over ℂ). Extend to an orthonormal basis. In that basis the first column of A is (λ₁, 0, …, 0). Apply the induction hypothesis to the remaining (n − 1) × (n − 1) block.

For

n = 1

there is nothing to prove. Assume the theorem for size

n - 1

Base case: a

1 \times 1

matrix is already triangular.

A

has an eigenvalue

\lambda_1

with a unit eigenvector

u_1

The characteristic polynomial has a complex root (fundamental theorem of algebra, 0.6), and every eigenvalue has an eigenvector.

Let

U_1 = [u_1\ w_2 \cdots w_n]

be unitary with first column

u_1

Any unit vector extends to an orthonormal basis (extend, then Gram–Schmidt, 2.9).

U_1^*AU_1 = \begin{bmatrix}\lambda_1 & * \\ 0 & B\end{bmatrix}

for some

(n-1)\times(n-1)

matrix

B

The first column of

U_1^*AU_1

U_1^*Au_1 = \lambda_1 U_1^*u_1 = \lambda_1 e_1

, because

u_1

is the first column of

U_1

B = U_2T_2U_2^*

with

U_2

unitary and

T_2

upper triangular.

Induction hypothesis on

B

U^*AU = \begin{bmatrix}\lambda_1 & * \\ 0 & T_2\end{bmatrix} = T

, upper triangular. So

A = UTU^*

\blacksquare

Combine:

U = U_1\begin{bmatrix}1&0\\0&U_2\end{bmatrix}

is a product of unitary matrices, hence unitary, and conjugating by it triangularizes the lower block while keeping the first column.

Worked example

Find a Schur decomposition of

A = \begin{bmatrix}4&1\\-2&1\end{bmatrix}

Solution.

Find one eigenvalue. Trace $5$ , determinant $4 + 2 = 6$ , so $\lambda^2 - 5\lambda + 6 = (\lambda - 2)(\lambda - 3)$ . Take $\lambda_1 = 2$ . The proof starts with any eigenvalue.
Find a unit eigenvector. $A - 2I = \begin{bmatrix}2&1\\-2&-1\end{bmatrix}$ gives $(1, -2)$ , so $u_1 = \frac1{\sqrt5}(1, -2)$ . Divide by the length $\sqrt5$ .
Complete to an orthonormal basis. $u_2 = \frac1{\sqrt5}(2, 1)$ , perpendicular to $u_1$ . Let $U = [u_1\ u_2] = \frac1{\sqrt5}\begin{bmatrix}1&2\\-2&1\end{bmatrix}$ . In $\mathbb R^2$ , the perpendicular direction is the rotated vector.
Compute $T = U^*AU$ . $Au_1 = 2u_1$ gives the first column $(2, 0)$ . For the second: $Au_2 = \frac1{\sqrt5}(9, -3)$ , with $u_1\cdot Au_2 = \frac{9 + 6}5 = 3$ and $u_2\cdot Au_2 = \frac{18 - 3}5 = 3$ . So $T = \begin{bmatrix}2&3\\0&3\end{bmatrix}$ . Entry $(i, j)$ of $U^*AU$ is $u_i\cdot Au_j$ . The $0$ below the diagonal is the proof’s key step: $u_1$ is an eigenvector.
Check. The diagonal $2, 3$ lists the eigenvalues of $A$ . A triangular matrix shows its eigenvalues on the diagonal.

Answer. $A = UTU^*$ with $U = \frac1{\sqrt5}\begin{bmatrix}1&2\\-2&1\end{bmatrix}$ and $T = \begin{bmatrix}2&3\\0&3\end{bmatrix}$ . Here $A$ is not symmetric, so $T$ cannot be made diagonal by an orthogonal $U$ , and the $3$ in the corner is what is left.

Proof workshopRebuild the proof of Schur's theorem

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Same idea elsewhere· Spectral theorem as a special case

If $A$ is Hermitian, then $T = U^*AU$ is Hermitian too, and an upper-triangular Hermitian matrix must be diagonal (the entries above the diagonal equal the conjugates of those below, which are 0). So Schur’s theorem contains the spectral theorem of 3.1.

Cayley–Hamilton

Theorem T Apostol 4.x exercises; Frobenius 1878; Strang 6.2

Cayley–Hamilton theorem

Let $p(\lambda) = \det(\lambda I - A) = \lambda^n + c_{n-1}\lambda^{n-1} + \dots + c_0$ . Then $p(A) = A^n + c_{n-1}A^{n-1} + \dots + c_0 I$ is the zero matrix.

In one lineEvery square matrix satisfies its own characteristic polynomial: if p(λ) = det(λI − A), then p(A) = 0.

Why it is trueThe adjugate identity (λI − A)·adj(λI − A) = p(λ)I holds for every number λ; comparing coefficients of each power of λ and telescoping gives p(A) = 0.

Watch out· The tempting one-line proof is wrong

“Substitute $\lambda = A$ : $p(A) = \det(AI - A) = \det(0) = 0$ .” This confuses a number with a matrix: $p(A)$ is a matrix, while $\det(\cdots)$ is a number, and substituting a matrix into $\lambda I$ inside a determinant is not defined. The real proof is below.

ProofStrang 6.2, the adjugate proof

The idea: Use the cofactor matrix: (λI − A)·adj(λI − A) = det(λI − A)·I = p(λ)I for every λ. The adjugate's entries are polynomials in λ of degree ≤ n − 1, so write adj(λI − A) = B_{n−1}λ^{n−1} + … + B_0 with constant matrices B_k. Compare coefficients of each power of λ, multiply the k-th equation by A^k, and add: the left side telescopes to 0.

For every number

\lambda

(\lambda I - A)\operatorname{adj}(\lambda I - A) = p(\lambda)\,I

The cofactor formula for the inverse (2.7):

M\,\operatorname{adj}(M) = \det(M)\,I

for any square

M

, where

\operatorname{adj}M

, the adjugate, is the transpose of the cofactor matrix. This holds even when

M

is not invertible.

Write

\operatorname{adj}(\lambda I - A) = B_{n-1}\lambda^{n-1} + \dots + B_1\lambda + B_0

, with constant matrices

B_k

Each entry of the adjugate is a determinant of an

(n-1)\times(n-1)

submatrix of

\lambda I - A

, so it is a polynomial in

\lambda

of degree at most

n - 1

. Collect equal powers.

B_{n-1} = I

\ B_{k-1} - AB_k = c_k I

for

1 \le k \le n - 1

\ -AB_0 = c_0 I

Expand the left side and match coefficients of

\lambda^n, \lambda^{n-1}, \dots, \lambda^0

with

p(\lambda)I

. Two polynomials agreeing for every

\lambda

have equal coefficients.

A^nB_{n-1} = A^n

\ A^kB_{k-1} - A^{k+1}B_k = c_kA^k

\ -AB_0 = c_0 I

Multiply the equation for

\lambda^k

A^k

on the left (and the first one by

A^n

0 = A^n + c_{n-1}A^{n-1} + \dots + c_1A + c_0I = p(A)

\blacksquare

Add all these equations. On the left each

A^kB_{k-1}

appears once with

+

and once with

-

, so everything cancels (a telescoping sum).

Worked example

Verify Cayley–Hamilton for

A = \begin{bmatrix}1&2\\3&4\end{bmatrix}

and use it to find

A^{-1}

Solution.

Characteristic polynomial. $p(\lambda) = \lambda^2 - (\operatorname{tr}A)\lambda + \det A = \lambda^2 - 5\lambda - 2$ . For $2 \times 2$ : trace 5, determinant $4 - 6 = -2$ (2.8).
Compute $A^2$ . $A^2 = \begin{bmatrix}7&10\\15&22\end{bmatrix}$ . Row times column.
Check $p(A)$ . $A^2 - 5A - 2I = \begin{bmatrix}7 - 5 - 2 & 10 - 10\\ 15 - 15 & 22 - 20 - 2\end{bmatrix} = 0$ . The theorem in action.
Solve for $A^{-1}$ . $A^2 - 5A = 2I$ , so $A(A - 5I) = 2I$ and $A^{-1} = \frac12(A - 5I) = \frac12\begin{bmatrix}-4&2\\3&-1\end{bmatrix}$ . Factor out $A$ ; possible because the constant term $-2 \ne 0$ , which is the same as $\det A \ne 0$ .

Answer. $A^{-1} = \begin{bmatrix}-2&1\\1.5&-0.5\end{bmatrix}$ . In general Cayley–Hamilton writes $A^{-1}$ , and every power $A^k$ , as a polynomial in $A$ of degree less than $n$ .

Your answerFor

A = \begin{bmatrix}1&2\\3&4\end{bmatrix}

, Cayley–Hamilton gives

A^2 = 5A + 2I

. Use it (twice) to find the top-left entry of

A^3

Definition B

Minimal polynomial

The minimal polynomial $m_A$ is the monic polynomial (leading coefficient 1) of smallest degree with $m_A(A) = 0$ . By Cayley–Hamilton it has degree at most $n$ , and it divides every polynomial that kills $A$ , including the characteristic polynomial (divide and look at the remainder, which kills $A$ and has smaller degree, so it is 0).

In one lineThe monic polynomial of smallest degree with m(A) = 0; it divides the characteristic polynomial and has the same roots.

Why it existsIt detects diagonalizability: A is diagonalizable exactly when its minimal polynomial has no repeated roots.

Quick checkWhat is the minimal polynomial of the shear

\begin{bmatrix}1&1\\0&1\end{bmatrix}

The shear shows the pattern. The criterion we will use is: over $\mathbb C$ , a matrix is diagonalizable exactly when its minimal polynomial has no repeated roots. One direction is quick. If $A = P\operatorname{diag}(\mu_1, \dots, \mu_n)P^{-1}$ and $\lambda_1, \dots, \lambda_s$ are its distinct eigenvalues, then $(A - \lambda_1I)\cdots(A - \lambda_sI) = P\,\operatorname{diag}(\dots)\,P^{-1}$ has a zero in every diagonal slot (each $\mu_i$ equals some $\lambda_j$ , which zeroes that factor), so this product kills $A$ . The minimal polynomial divides it, and the product has no repeated roots, so neither does the minimal polynomial. The other direction comes from the Jordan form below: a block $J_k(\lambda)$ has minimal polynomial $(t - \lambda)^k$ , because $J_k(\lambda) - \lambda I$ is a shift that needs exactly $k$ applications to reach 0. So a block of size 2 or more forces a repeated root.

Generalized eigenvectors and the Jordan form

Definition B

Generalized eigenvector

A non-zero $v$ is a generalized eigenvector of $A$ for $\lambda$ if $(A - \lambda I)^k v = O$ for some $k \ge 1$ . A Jordan chain of length $k$ is $v_1, \dots, v_k$ with $(A - \lambda I)v_1 = O$ and $(A - \lambda I)v_j = v_{j-1}$ for $j \ge 2$ : each vector is pushed onto the previous one, and the first onto $O$ .

In one lineA vector v with (A − λI)^k v = O for some k ≥ 1: killed by A − λI after enough repetitions, though maybe not after one.

Why it existsWhen there are too few eigenvectors, generalized eigenvectors fill the gap, and together they always form a basis.

Definition B

Jordan block

The Jordan block $J_k(\lambda)$ is the $k \times k$ matrix

J_k(\lambda) = \begin{bmatrix}\lambda & 1 & & \\ & \lambda & \ddots & \\ & & \ddots & 1 \\ & & & \lambda\end{bmatrix}.

In the basis of a Jordan chain $v_1, \dots, v_k$ , the map $A$ has exactly this matrix: $Av_1 = \lambda v_1$ and $Av_j = \lambda v_j + v_{j-1}$ .

In one lineJ_k(λ) is the k×k matrix with λ on the diagonal, 1 directly above it, and 0 elsewhere; it is λI plus a 'push' along a chain.

Why it existsIt is the simplest possible non-diagonalizable matrix, and every matrix is built from these.

Definition B

Jordan form

The Jordan form of $A$ is a block-diagonal matrix $J = \operatorname{diag}\big(J_{k_1}(\lambda_1), \dots, J_{k_s}(\lambda_s)\big)$ with $A = PJP^{-1}$ for some invertible $P$ , whose columns are Jordan chains.

In one lineEvery complex square matrix is similar to a block-diagonal matrix of Jordan blocks, unique up to the order of the blocks.

Why it existsIt is the complete answer to 'how close to diagonal can A be made': diagonal exactly when every block is 1×1.

Theorem B Axler 8.60; Strang 8.3

Jordan normal form theorem

Every complex square matrix $A$ is similar to a matrix in Jordan form. The number and sizes of the blocks for each eigenvalue are determined by $A$ : the number of blocks for $\lambda$ equals $\dim N(A - \lambda I)$ (the number of independent eigenvectors), and the number of blocks of size at least $j$ equals $\dim N(A - \lambda I)^j - \dim N(A - \lambda I)^{j-1}$ .

In one lineEvery complex square matrix is similar to a Jordan form, unique up to reordering the blocks.

Why it is trueℂⁿ splits into generalized eigenspaces, one per eigenvalue; on each, A − λI is nilpotent, and a nilpotent map always has a basis of chains.

Beyond the course: for understanding· About the proof

The full proof takes a chapter of Axler (chapter 8) and is a core part of MATH 540. Its two steps: (1) $\mathbb C^n$ is the direct sum of the generalized eigenspaces $G(\lambda) = N\big((A - \lambda I)^n\big)$ , one for each eigenvalue, each invariant under $A$ (this uses the factorisation of the characteristic polynomial and Cayley–Hamilton); (2) on each $G(\lambda)$ , the map $N = A - \lambda I$ is nilpotent ( $N^m = 0$ ), and every nilpotent map has a basis made of chains $v, Nv, N^2v, \dots$ . The dimension formula in the theorem comes from counting chains by length. Here we use the theorem and compute with it.

Worked example

Find the Jordan form of

A = \begin{bmatrix}2&1&0\\0&2&0\\0&0&3\end{bmatrix}

Solution. The method: read off eigenvalues, count eigenvectors for each, then build chains. The step-through shows each stage.

Answer. $J = J_2(2) \oplus J_1(3)$ : a block of size 2 for the eigenvalue 2 and a block of size 1 for 3, where $\oplus$ means “place these blocks along the diagonal”. Here $P = I$ because $A$ was already in Jordan form.

Worked example

5 \times 5

matrix has the single eigenvalue 4, with

\dim N(A - 4I) = 2

\dim N(A - 4I)^2 = 4

and

\dim N(A - 4I)^3 = 5

. Find its Jordan form.

Solution.

Number of blocks. $\dim N(A - 4I) = 2$ , so there are 2 blocks. The theorem: blocks for $\lambda$ = number of independent eigenvectors.
Blocks of size at least 2. $\dim N(A - 4I)^2 - \dim N(A - 4I) = 4 - 2 = 2$ . Every block of size 2 or more contributes one new vector to the null space when we go from the first power to the second.
Blocks of size at least 3. $\dim N(A - 4I)^3 - \dim N(A - 4I)^2 = 5 - 4 = 1$ . Only one block is as big as 3.
Blocks of size at least 4. The null space has dimension 5 at power 3, which is the whole space, so the next difference is $0$ . No block is of size 4 or more.
Read off sizes. Two blocks in total, both of size at least 2, exactly one of size at least 3 and none of size 4: sizes $3$ and $2$ . Check: $3 + 2 = 5$ .

Answer. $J = J_3(4) \oplus J_2(4)$ : a block of size 3 and a block of size 2, each with 4 on the diagonal.

Your answerIn the example above, what is the size of the largest Jordan block?

Worked example

Find the Jordan form of

A = \begin{bmatrix}3&1\\-1&1\end{bmatrix}

Solution.

Eigenvalues. $p(\lambda) = \lambda^2 - 4\lambda + 4 = (\lambda - 2)^2$ . Trace 4, determinant $3 + 1 = 4$ .
Eigenvectors. $A - 2I = \begin{bmatrix}1&1\\-1&-1\end{bmatrix}$ has rank 1, so only one eigenvector, $v_1 = (1, -1)$ . Fewer eigenvectors than the repetition: one block of size 2.
Chain. Solve $(A - 2I)v_2 = v_1$ : $x + y = 1$ , so $v_2 = (1, 0)$ works. Any solution will do; it is not unique.
Assemble. $P = \begin{bmatrix}1&1\\-1&0\end{bmatrix}$ , $J = \begin{bmatrix}2&1\\0&2\end{bmatrix}$ , and $AP = PJ$ . Check: $Av_1 = 2v_1$ and $Av_2 = (3, -1) = 2v_2 + v_1$ .

Answer. $A$ is similar to $J_2(2)$ : it is ” $2I$ plus a push”, a scaled shear in the basis $v_1, v_2$ .

Worked example

Use Cayley–Hamilton to write

A^5

\alpha A + \beta I

for

A = \begin{bmatrix}0&1\\-1&2\end{bmatrix}

Solution.

Characteristic polynomial. Trace $2$ , determinant $0 + 1 = 1$ , so $p(\lambda) = \lambda^2 - 2\lambda + 1$ . $2 \times 2$ formula (2.8).
Apply the theorem. $A^2 - 2A + I = 0$ , so $A^2 = 2A - I$ . Cayley–Hamilton: $p(A) = 0$ .
Climb one power at a time. Multiply by $A$ and substitute $A^2$ again: $A^3 = 2A^2 - A = 2(2A - I) - A = 3A - 2I$ . Then $A^4 = 3A^2 - 2A = 3(2A - I) - 2A = 4A - 3I$ . Then $A^5 = 4A^2 - 3A = 4(2A - I) - 3A = 5A - 4I$ . Each time, the $A^2$ that appears is replaced by $2A - I$ , so the degree never exceeds 1.
Check the pattern and the numbers. $A^k = kA - (k-1)I$ , and $5A - 4I = \begin{bmatrix}-4&5\\-5&6\end{bmatrix}$ . Direct multiplication of $A$ five times gives the same matrix.

Answer. $A^5 = 5A - 4I = \begin{bmatrix}-4&5\\-5&6\end{bmatrix}$ .

Worked example

What is the minimal polynomial of

\operatorname{diag}(2, 2, 3)

, and of

\begin{bmatrix}2&1&0\\0&2&0\\0&0&3\end{bmatrix}

Solution.

Candidates. Both matrices have characteristic polynomial $(\lambda - 2)^2(\lambda - 3)$ , so by the division argument the minimal polynomial divides it. The only possibilities with roots 2 and 3 are $(\lambda - 2)(\lambda - 3)$ and $(\lambda - 2)^2(\lambda - 3)$ . The minimal polynomial has the same roots as the characteristic polynomial, and divides it.
Test the short one on the diagonal matrix. $(D - 2I)(D - 3I) = \operatorname{diag}(0,0,1)\cdot\operatorname{diag}(-1,-1,0) = 0$ . Diagonal matrices multiply entry by entry, and each entry is 0.
Test the short one on the other matrix. $(A - 2I)(A - 3I) = \begin{bmatrix}0&1&0\\0&0&0\\0&0&1\end{bmatrix}\begin{bmatrix}-1&1&0\\0&-1&0\\0&0&0\end{bmatrix} = \begin{bmatrix}0&-1&0\\0&0&0\\0&0&0\end{bmatrix} \ne 0$ . The block $J_2(2)$ survives one factor of $A - 2I$ .
So the longer one is needed. $(A - 2I)^2(A - 3I) = 0$ . It is the characteristic polynomial, which kills $A$ by Cayley–Hamilton.

Answer. First matrix: $(\lambda - 2)(\lambda - 3)$ , no repeated root, diagonalizable. Second: $(\lambda - 2)^2(\lambda - 3)$ , a repeated root, not diagonalizable.

Quick checkWhich matrix is NOT diagonalizable?

Your answerA

5 \times 5

matrix has the single eigenvalue 4 and

\dim N(A - 4I) = 2

. How many Jordan blocks does it have?

Worked example

Prove that the eigenvalues of an upper triangular matrix are its diagonal entries, and use Schur's theorem to show that

\operatorname{tr}A

is the sum of the eigenvalues.

Solution.

Subtract $\lambda$ on the diagonal. If $T$ is upper triangular, so is $\lambda I - T$ , with diagonal entries $\lambda - t_{11}, \dots, \lambda - t_{nn}$ . $\lambda I$ only changes the diagonal.
Determinant of a triangular matrix. $\det(\lambda I - T) = (\lambda - t_{11})\cdots(\lambda - t_{nn})$ . The determinant of a triangular matrix is the product of its diagonal entries (2.7).
Roots. The roots of this polynomial, the eigenvalues, are $t_{11}, \dots, t_{nn}$ . A product is zero exactly when a factor is.
Use Schur. $A = UTU^*$ with $T$ triangular, and $U^* = U^{-1}$ , so $A$ is similar to $T$ and has the same characteristic polynomial. Similar matrices share the characteristic polynomial (2.8).
Compare traces. $\operatorname{tr}A = \operatorname{tr}(UTU^*) = \operatorname{tr}(TU^*U) = \operatorname{tr}T$ . $\operatorname{tr}(XY) = \operatorname{tr}(YX)$ , with $X = U$ and $Y = TU^*$ .
Read off. $\operatorname{tr}T = t_{11} + \cdots + t_{nn}$ , the sum of the eigenvalues counted with repetition. Step 3.

Answer. Eigenvalues of a triangular matrix sit on its diagonal, and $\operatorname{tr}A$ is the sum of the eigenvalues. The same argument gives $\det A$ as their product. $\blacksquare$

Your answerA

3 \times 3

matrix has a Schur form

T

with diagonal

2, 5, -1

. What is

\det A

Worked example Axler 8

Let

N = J_4(0)

. Compute

N^j

for

j = 1, \dots, 4

, find

\dim N(N^j)

, and explain how this counts Jordan blocks.

Solution.

What $N$ does. $N$ has 1s directly above the diagonal, so $Ne_1 = O$ and $Ne_j = e_{j-1}$ for $j \ge 2$ . Column $j$ of $N$ is $e_{j-1}$ .
Powers. $N^j$ sends $e_i$ to $e_{i-j}$ if $i > j$ and to $O$ if $i \le j$ . So $N^2$ has 1s two above the diagonal, $N^3$ has a single 1 in the top-right corner, and $N^4 = 0$ . Apply $N$ repeatedly: each application moves the basis vector one step down the chain.
Null spaces. $N(N^j) = \operatorname{span}(e_1, \dots, e_j)$ , of dimension $j$ , for $j \le 4$ . Exactly $e_1, \dots, e_j$ reach $O$ in $j$ steps.
Count the blocks. For one block of size 4, $\dim N(N^j) - \dim N(N^{j-1}) = 1$ for $j = 1, 2, 3, 4$ and $0$ afterwards. Each power adds one new vector until the chain runs out.
Many blocks. For a sum of blocks the dimensions add. Going from power $j - 1$ to $j$ , a block of size at least $j$ adds exactly 1 and a smaller block adds 0. So the difference counts the blocks of size at least $j$ , which is the formula in the Jordan theorem.

Answer. $\dim N(N^j) = j$ for $j \le 4$ , and the differences count blocks by size. $\blacksquare$

Quick checkIn a Schur decomposition

A = UTU^*

, what does the diagonal of

T

list?

Proof workshopRebuild the Cayley–Hamilton proofStrang 6.2

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Where it is used

Python labA Jordan block is fragile

import numpy as np

# a Jordan block J_2(1) = [[1, 1], [0, 1]], with a tiny change eps in the bottom-left corner
for eps in (0.0, 1e-12, 1e-8, 1e-4):
    A = np.array([[1.0, 1.0], [eps, 1.0]])
    vals = np.linalg.eigvals(A)
    print(f"eps = {eps:.0e}   eigenvalues = {np.round(vals, 8)}   shift from 1 = {abs(vals[0] - 1):.1e}")

# the same change to a diagonalizable matrix barely matters
D = np.array([[1.0, 0.0], [1e-8, 3.0]])
print("diagonalizable:", np.linalg.eigvals(D))

Try this. Run it. A change of size $10^{-12}$ in one entry moves the eigenvalues by $10^{-6}$ , a million times more: the shift is about $\sqrt{\varepsilon}$ , because the eigenvalues of the perturbed block solve $(\lambda - 1)^2 = \varepsilon$ . The last line shows a diagonalizable matrix shrugging off a change of $10^{-8}$ . Try changing eps in the list to 1e-16 and see that the computer's own rounding errors already do this to any Jordan block.

Used inside maths· Inside maths: matrix exponentials and repeated roots (Unit 7)

For a Jordan block, $J = \lambda I + N$ with $N^k = 0$ , so $e^{tJ} = e^{\lambda t}\big(I + tN + \frac{t^2}{2}N^2 + \dots\big)$ stops after $k$ terms. That is why repeated roots of a linear ODE’s characteristic equation produce solutions $e^{rt}, te^{rt}, t^2e^{rt}, \dots$ : the powers of $t$ count positions along a Jordan chain. MATH 376 used this rule; 7.4 derives it.

Used in the real world· Engineering: defective systems and resonance-like growth

In control engineering, a system $\dot x = Ax$ whose matrix has a Jordan block for an eigenvalue with zero real part grows like $t$ instead of staying bounded, even though no eigenvalue has a positive real part. Stability analyses must therefore look at Jordan structure (or at the minimal polynomial), not only at eigenvalues. Numerically, the Jordan form is never computed for real data: an arbitrarily small perturbation can change it (it is badly conditioned, 3.5), which is why software uses the Schur form instead.

Traps

Watch out· Repeated eigenvalue does not mean not diagonalizable

$2I$ has the eigenvalue 2 three times and is diagonal. What matters is whether there are as many independent eigenvectors as the repetition. Count $\dim N(A - \lambda I)$ .

Watch out· Cayley–Hamilton is not 'plug A into det'

$p(A) = 0$ is a statement about a matrix polynomial; the proof needs the adjugate (or Schur), not substitution into a determinant.

Problems

check

Write the Jordan form of a matrix whose characteristic polynomial is $(\lambda - 1)^2(\lambda - 5)$ and which has exactly two independent eigenvectors.

Solution

Two eigenvectors: one for $\lambda = 5$ and only one for $\lambda = 1$ . So $J = \operatorname{diag}\big(J_2(1), J_1(5)\big) = \begin{bmatrix}1&1&0\\0&1&0\\0&0&5\end{bmatrix}$ .

check

What is the minimal polynomial of $\operatorname{diag}(2, 2, 3)$ ?

Solution

$(\lambda - 2)(\lambda - 3)$ : it kills $A$ , and no polynomial of degree 1 does. No repeated roots, matching “diagonalizable”.

use

Use Cayley–Hamilton to express $A^4$ as $\alpha A + \beta I$ for $A = \begin{bmatrix}0&1\\-2&3\end{bmatrix}$ .

Solution

$p(\lambda) = \lambda^2 - 3\lambda + 2$ , so $A^2 = 3A - 2I$ . $A^3 = 3A^2 - 2A = 7A - 6I$ . $A^4 = 7A^2 - 6A = 15A - 14I$ .

use

Find the Jordan form of $\begin{bmatrix}5&4\\-1&1\end{bmatrix}$ .

Solution

$p = \lambda^2 - 6\lambda + 9 = (\lambda - 3)^2$ ; $A - 3I = \begin{bmatrix}2&4\\-1&-2\end{bmatrix}$ has rank 1, so one eigenvector: $J = \begin{bmatrix}3&1\\0&3\end{bmatrix}$ .

prove

Prove that the eigenvalues of an upper triangular matrix are its diagonal entries.

Solution

$\lambda I - T$ is upper triangular with diagonal $\lambda - t_{ii}$ , and the determinant of a triangular matrix is the product of its diagonal (2.7). So $\det(\lambda I - T) = \prod(\lambda - t_{ii})$ , whose roots are the $t_{ii}$ .

prove

Prove that the trace of $A$ is the sum of its eigenvalues using Schur’s theorem.

Solution

$A = UTU^*$ is similar to $T$ , and similar matrices have equal traces ( $\operatorname{tr}(PMP^{-1}) = \operatorname{tr}(M)$ , since $\operatorname{tr}(XY) = \operatorname{tr}(YX)$ ). The trace of $T$ is the sum of its diagonal, which lists the eigenvalues with multiplicity.

stretchAxler 8

Let $N = J_k(0)$ . Compute $N^j$ for $j = 1, \dots, k$ and show $\dim N(N^j) = j$ for $j \le k$ . Explain how this gives the block-counting formula in the Jordan theorem.

Solution

$N$ sends $e_1 \mapsto O$ , $e_j \mapsto e_{j-1}$ . So $N^j$ kills $e_1, \dots, e_j$ and shifts the rest down by $j$ ; $N^k = 0$ . Hence $N(N^j) = \operatorname{span}(e_1, \dots, e_j)$ , dimension $j$ . For a sum of blocks, each block of size $\ge j$ adds exactly 1 to $\dim N(N^j) - \dim N(N^{j-1})$ , and smaller blocks add 0, which is the formula.

Summary

Schur: every complex matrix is $UTU^*$ with $U$ unitary and $T$ upper triangular (eigenvalues on the diagonal). Hermitian matrices give diagonal $T$ : the spectral theorem.
Cayley–Hamilton: $p(A) = 0$ , proved with the adjugate and a telescoping sum. It writes $A^{-1}$ and all powers of $A$ as polynomials of degree $< n$ .
Minimal polynomial: the smallest monic polynomial killing $A$ ; $A$ is diagonalizable iff it has no repeated roots.
Jordan form: $A = PJP^{-1}$ with $J$ made of blocks $\lambda I + N$ ; the number of blocks for $\lambda$ is $\dim N(A - \lambda I)$ .
Jordan blocks explain the $te^{rt}$ solutions of ODEs; in numerical work the Schur form is used instead.

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

3.6 Cayley–Hamilton, Schur, and the Jordan form

The picture

Invariant subspaces and Schur’s theorem

Invariant subspace

Schur decomposition

Schur's theorem

Cayley–Hamilton

Cayley–Hamilton theorem

Minimal polynomial

Generalized eigenvectors and the Jordan form

Generalized eigenvector

Jordan block

Jordan form

Jordan normal form theorem

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere