Triangularisation and Jordan Form: Examples and Counterexamples

Subtracting $(x,0,0)\in W$ shows $(x,y,z)+W=(0,y,z)+W$; if two such representatives agree, their difference $(0,y-y^{\mathrm{\prime }},z-z^{\mathrm{\prime }})$ lies in $W$, forcing $y=y^{\mathrm{\prime }}$ and $z=z^{\mathrm{\prime }}$.

The standard list $((1,0,0),(0,1,0),(0,0,1))$ is a basis of $F^3$, so [L2] makes the last two cosets a quotient basis.

The displayed formula for $\pi$, its dimension, and $\ker \pi =W$ now follow from steps 1.1-1.2 and [L1].

Solving $x+y=0$ and $y+z=0$ gives $(x,y,z)=y(-1,1,-1)$, so the kernel is the stated line.

For every $(a,b)\in F^2$, $T(a,0,b)=(a,b)$, so $T$ is surjective.

Fact [L1] gives the induced isomorphism with the displayed formula, and step 1.2 shows that sending $(a,b)$ to $(a,0,b)+\ker T$ is its two-sided inverse.

The matrix is strictly upper triangular, so direct determinant computation gives ${\chi }_A=x^3$; also $A\mathrm{\ne }0$ but $A^2=0$, so ${\mu }_A=x^2$.

The polynomial $x^3$ splits, so [L1] confirms triangularisability, while the repeated factor in $x^2$ makes [L2] rule out diagonalisability.

The formulas show $T⟨e_1⟩\subseteq ⟨e_1⟩$ and $T⟨e_1,e_2⟩\subseteq ⟨e_1,e_2⟩$, while the final space is automatically invariant.

The coordinate columns of $T{e}_1,T{e}_2,T{e}_3$ give the displayed matrix, and [L1] identifies this upper-triangular form with the verified complete flag.

The six displayed images verify [L1] separately for the three listed strings, and their concatenation is the standard basis.

In that order the matrix has blocks $J_3(0)$, $J_2(0)$, and $J_1(0)$; direct iteration gives $N^3=0$ and $N^2{e}_3=e_1\mathrm{\ne }0$, consistently with [L2].

The successive differences ${\rho }_{k-1}-{\rho }_k$ are $3,2,1,0$ for $k=1,2,3,4$, so there are respectively three, two, one, and zero blocks of size at least those values.

Taking successive differences again gives one block of each exact size $1,2,3$ and none larger; their sizes sum to $6$, as required.

At $\lambda =0$, the $J_2(1)$ summand stays invertible while the ranks of powers of $J_3(0)$ are $3,2,1,0,0,\dots$, giving $5,4,3,2,2,\dots$. At $\lambda =1$, the $J_3(0)-I$ summand stays invertible while the nilpotent part of $J_2(1)$ has ranks $2,1,0,0,\dots$, giving $5,4,3,3,\dots$.

Applying [L1] gives one exact size-three block at $0$, one exact size-two block at $1$, and zero other blocks; [L2] then gives both displayed polynomials.

Direct computation gives ${\chi }_R=x^2+1$, which has no real root and hence does not split over $\mathbb{R}$; [L1] rules out real Jordan form.

Over $\mathbb{C}$, $R(1,-i)=i(1,-i)$ and $R(1,i)=-i(1,i)$; these eigenvectors are independent, so their basis gives the displayed diagonal Jordan form, as [L1] permits.

Direct determinant computation gives ${\chi }_R=x^2+1$. It has no root in $\mathbb{R}$, so it does not split over the base field.

Fact [L1] therefore says that $R$ is not triangularisable over $\mathbb{R}$, refuting the universal claim.

The characteristic polynomial of $R$ is $x^2+1$, which has no real root and does not split over $\mathbb{R}$.

By [L1], $R$ has no Jordan form over its real base field, so the claim is false; scalar extension to $\mathbb{C}$ is essential in this example.

Both matrices have total size four and largest block size two, so [L1] gives ${\chi }_A={\chi }_B=(x-\lambda {)}^4$ and ${\mu }_A={\mu }_B=(x-\lambda {)}^2$.

Nevertheless $\mathrm{rank}(A-\lambda I)=2$ while $\mathrm{rank}(B-\lambda I)=1$, and their block multisets are respectively $\{2,2\}$ and $\{2,1,1\}$.

Fact [L2] therefore says that $A$ and $B$ are not similar, refuting the claim.

Both $A$ and $B$ are Jordan matrices with the same block multiset, and a permutation matrix that moves the one-dimensional block past the two-dimensional block conjugates one to the other.

Their diagonal sequences are $(0,0,1)$ and $(1,0,0)$, so $A\mathrm{\ne }B$ as literal matrices.

This is precisely the block-order freedom retained in [L1], and it refutes literal uniqueness without an additional ordering convention.

Each matrix has exactly two $\lambda$-blocks, so [L1] gives geometric multiplicity two for both.

Yet $(A-\lambda I{)}^2$ has rank one, contributed by $J_3(0{)}^2$, while $(B-\lambda I{)}^2=0$; [L2] therefore distinguishes the block multisets $\{3,1\}$ and $\{2,2\}$.

Equal geometric multiplicity has not determined the sizes, so the claim is false.

For $h(x)=3x^2-2x^3$, one has $h(0)=h^{\mathrm{\prime }}(0)=0$ and $h(1)=1$, $h^{\mathrm{\prime }}(1)=0$; evaluating on the two size-two blocks therefore gives $h(J_2(0))=0$ and $h(J_2(1))=I_2$. These identities remain valid in characteristics $2$ and $3$ by direct reduction.

Hence $D=h(T)$ is the scalar part on the two generalised eigenspaces, while $N=T-D$ is the direct sum of their nilpotent parts as in [L2].

The displayed blocks give $N^2=0$ and $DN=ND$; commutation also follows because both $D$ and $N=T-h(T)$ are polynomials in $T$.

Suppose $R=D+N$ as claimed. By [L2], ${\chi }_D$ splits, while [L3] makes ${\chi }_N=x^2$ split; commutation and [L1] give one real basis in which both $D$ and $N$ are upper triangular.

Their sum $R$ is upper triangular in that basis, so [L4] would make ${\chi }_R$ split over $\mathbb{R}$.

But direct computation gives ${\chi }_R=x^2+1$, which has no real root. This contradiction refutes the claimed decomposition over the base field.

Matrix multiplication gives $C{e}_1=e_2$ and $C^2{e}_1=e_3$, so the three power vectors are the standard basis and $e_1$ is cyclic.

The columns show $C{e}_3=-e_1-e_2$, so $C^3{e}_1=C{e}_3=-e_1-e_2=-e_1-C{e}_1$ and hence $f(C)e_1=(C^3+C+I)e_1=0$. Since $f(C)$ commutes with $C$, $f(C)C^j{e}_1=C^{j}f(C)e_1=0$ for $j=0,1,2$, and step 1.1 makes $e_1,C{e}_1,C^2{e}_1$ a basis, so $f(C)=0$.

By step 1.1 the vector $e_1$ is cyclic, so [L2] gives ${\mu }_C={\chi }_C$, and [L3] makes ${\chi }_C$ monic of degree $3$; thus ${\mu }_C$ is monic of degree $3$. By step 2.1, ${\mu }_C$ divides the monic degree-three polynomial $f$, so ${\mu }_C=f$ and therefore ${\mu }_C={\chi }_C=f$. Since $m_{C,e_1}$ divides $f$ and $(e_1,C{e}_1,C^2{e}_1)$ is a basis of $Z(e_1;C)=F^3$, $m_{C,e_1}=f$, and [L1] in that basis is exactly the displayed matrix, with last column $(-a_0,-a_1,-a_2)=(-1,-1,0)$.

For any $v\in F^2$, every power $I^{k}v$ equals $v$, so $Z(v;I)=Fv$ has dimension at most one and cannot equal $F^2$.

Equivalently, ${\mu }_I=x-1$ while ${\chi }_I=(x-1{)}^2$, so [L1] also rules out a cyclic vector. Thus the identity on $F^2$ refutes the universal claim.