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Over a field, a cyclic vector gives a companion matrix, whose characteristic relation annihilates its generator and hence every iterate. If there is no cyclic vector spanning the whole space, a nonzero proper cyclic subspace is invariant. Induction annihilates the restriction and the quotient; the quotient polynomial first sends the whole space into the invariant subspace, and the restriction polynomial then kills it. The block-triangular characteristic polynomial is the product of these two polynomials.

On a $d$-dimensional vector space, a cyclic vector $v$ gives a basis $v,Tv,\dots ,T^{d-1}v$. Write $T^{d}v=-{\sum }_{j<d}{c}_j{T}^{j}v$ and set $f(t)=t^d+{\sum }_{j<d}{c}_j{t}^j$. Then $f(T)v=0$. Since polynomials in $T$ commute with $T$, $f(T)$ kills every basis vector, so $f(T)=0$. No nonzero polynomial of degree less than $d$ can kill $v$, by linear independence. Hence the minimal polynomial is $f$. The matrix in this basis is $C(f)$, and the characteristic polynomial of a companion matrix is $f$ as well.

For the state on a C-star algebra $a\mapsto \mathrm{Tr}(\rho a)$ on $M_d(\mathbb{C})$, the GNS construction has the displayed Hilbert-space dimension. Here $\rho$ is a density matrix. Its null space for the GNS seminorm consists of $b$ with $b{\rho }^{1/2}=0$, because $\mathrm{Tr}(\rho b^{\ast }b)=\Vert b{\rho }^{1/2}{\Vert }_{\mathrm{HS}}^2$. Thus the GNS space is the space of matrices $b{\rho }^{1/2}$ with the Hilbert-Schmidt inner product. If $\rho$ has rank $r$, diagonalization identifies these matrices with $r$ arbitrary columns, giving dimension $dr$. Left multiplication is the representation and ${\rho }^{1/2}$ is its cyclic vector. In particular a pure rank-one matrix state gives the defining $d$-dimensional representation, while a full-rank state gives a $d^2$-dimensional space.

The minimal polynomial $m_T$ is the unique monic polynomial of least degree with $m_T(T)=0$. Existence follows from a linear dependence among powers of $T$ in the finite-dimensional vector space $\mathrm{End}(V)$; polynomial division proves uniqueness and that it divides every annihilating polynomial. The characteristic polynomial is ${\chi }_T(t)=\det (tI-[T{]}_{\mathcal{B}})$, independent of basis because a change of basis conjugates the matrix.

WriteIn the cyclic basis, the companion matrix isFor $q(t)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$, the displayed vector relation gives $q(T)v=0$. Polynomial evaluation commutes with $T$, so $q(T)T^{j}v=T^{j}q(T)v=0$ for every basis vector. Thus $q(T)=0$.

No nonzero polynomial of degree below $n$ can annihilate $T$, since applying it to the cyclic vector $v$ would contradict independence of $v,Tv,\dots ,T^{n-1}v$. Hence $m_T=q$. Finally, direct expansion of $\det (tI-C)$, or induction along the last column, gives ${\chi }_T(t)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0=q(t)$. ThereforeThe argument computes the determinant and annihilator directly and does not invoke the Cayley-Hamilton theorem.

Since $v$ is a cyclic vector and $V\ne 0$, $v\ne 0$. Finite dimension of a vector space forces a first dependence in the sequence $v,Tv,T^{2}v,\dots$. Let $k\ge 1$ be the smallest index with $T^{k}v$ in the linear span of its predecessors. Thenis linearly independent. Its linear span $W$ is invariant under $T$: all but the last basis candidate map to the next, and the last maps to $T^{k}v\in W$. Induction therefore puts every $T^{j}v$ in $W$. Cyclicity gives $W=V$, so $B$ is a basis and $k=n$.

WriteThe matrix of $T$ in this basis is the companion matrixFor $p(t)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$, the defining relation says $p(T)v=0$. Since $p(T)$ commutes with $T$, it also annihilates every $T^{j}v$ in $B$, hence $p(T)=0$. No nonzero polynomial of degree less than $n$ can annihilate $v$, because that would contradict linear independence of $B$. Thus $m_T=p$.

Finally the characteristic polynomial of a companion matrix is $p$, by direct determinant expansion. Explicitly, expansion along the first row of $tI-C$ gives the recurrencestarting from $D_1=t+a_0$. It yields $D_n=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$. ConsequentlyThis proves that a cyclic operator has equal minimal and characteristic polynomials by the cyclic basis and a direct determinant calculation, with no appeal to the Cayley-Hamilton theorem.

For $T_1$, choose the third standard basis vector $v=e_3$. Its first three orbit vectors areThey are linearly independent: their matrix of columns has determinant $-1$. They therefore span ${\mathbb{R}}^3$, proving $T_1$ has a cyclic vector, namely $e_3$.

For $T_2$, write $T_2=I+N$ where $N^2=0$. The binomial theorem gives $T_2^{j}v=v+jNv$ for every integer $j\ge 0$. The entire orbit of any $v$ lies in $\mathrm{span}\{v,Nv\}$, which has dimension of a vector space at most two. Therefore $T_2$ has no cyclic vector in the three-dimensional space. Equivalently, its minimal polynomial is $(t-1{)}^2$, while its characteristic polynomial is $(t-1{)}^3$.

The $n$ spanning vectors form a basis. WriteIn this cyclic basis the matrix is the companion matrixExpanding $\det (tI-C)$ along its last column givesFor example, the term containing $a_j$ has minor $t^j$ and the signs from the subdiagonal cancel the cofactor sign; the diagonal term supplies $t^n$. The defining relation says ${\chi }_T(T)x=0$. Every polynomial in $T$ commutes with $T$, soIt therefore vanishes on a basis, provingThis direct argument for a cyclic vector does not assume the Cayley-Hamilton theorem.

Induct on $n$. The one-dimensional assertion is immediate, and the zero-dimensional case is vacuous. If $T$ has a cyclic vector spanning all of $V$, the direct proof in part (ii) applies. Otherwise choose $x\ne 0$ and let $W$ be its cyclic subspace. Part (i) makes $W$ invariant; the absence of a cyclic vector makes it proper and nonzero.

By induction, the restriction to $W$ is annihilated by ${\chi }_A$, and the induced map on $V/W$ is annihilated by ${\chi }_D$. The quotient assertion means ${\chi }_D(T)V\subset W$. Applying the restriction assertion next givesBy the characteristic factorization in part (iii), this is precisely ${\chi }_T(T)V=0$. HenceThis is the Cayley-Hamilton theorem from cyclic subspaces; using the quotient map is what makes the argument valid even when the block $B$ is nonzero.

For the standard basis vector $e_j$, matrix multiplication selects the $j$th column, soConsequentlyand the columns of $PM$ are $P{c}_1,\dots ,P{c}_n$.

Suppose first that $v,Av,\dots ,A^{n-1}v$ are linearly independent, and defineThen $S$ is invertible. The first $n-1$ columns of $AS$ are the last $n-1$ columns of $S$. By the Cayley-Hamilton theorem,This is exactly the last-column rule for the companion matrix $C$, so

Conversely, suppose $S^{-1}AS=C$, or $AS=SC$, and write the columns of $S$ as $s_1,\dots ,s_n$. Comparing the first $n-1$ columns givesHenceSince $S$ is invertible, its columns are linearly independent. Taking $v=s_1$ therefore makeslinearly independent. Thus $A$ is similar to $C$ exactly when it has the stated cyclic vector.

If $V=V_1\oplus V_2$ were a nontrivial invariant decomposition, each restricted minimal polynomial would be a power of $x-\alpha$ of degree at most $\dim V_i<n$. Their least common multiple is the minimal polynomial of $\varphi$, contradicting its degree $n$.

Since $m_{\varphi }$ has degree $n$, $\varphi$ has a cyclic vector $b_1$. Put $b_k={\varphi }^{k-1}{b}_1$. Then $\varphi (b_k)=b_{k+1}$ for $k<n$. In this basis the matrix is the companion matrix with ones immediately below the diagonal and last columnobtained by expanding $(x-\alpha {)}^n$.

Write $d=\dim V$. A cyclic vector for $S$ is a vector $v$ whose successive iterates span $V$; equivalently, when $d>0$,are a basis. The equivalence follows because the first linear dependence in the iterates makes their preceding span invariant under $S$, so no later iterate can enlarge it.

In this basis there are unique coefficients $c_0,\dots ,c_{d-1}$ with $S^{d}v=-{\sum }_{j<d}{c}_j{S}^{j}v$. Put $f(t)=t^d+{\sum }_{j<d}{c}_j{t}^j$. Then $f(S)v=0$, and commutation with powers of $S$ gives $f(S)S^{j}v=S^{j}f(S)v=0$. Therefore $f(S)$ kills an entire basis and is the zero map. No nonzero polynomial of degree less than $d$ annihilates $v$, because that would be a linear dependence of the displayed basis. Thus $m_S=f$.

The matrix of $S$ in the cyclic basis is the companion matrix $C(f)$ with ones on its subdiagonal and last column $(-c_0,\dots ,-c_{d-1}{)}^{\mathsf{T}}$. Expanding $\det (tI-C(f))$ along its last column givesthe cofactors are $1,t,\dots ,t^{d-1}$. This proves the characteristic polynomial of a companion matrix formula here. HenceThis proves that a cyclic operator has equal minimal and characteristic polynomials. If $V=\{0\}$, both polynomials are one by convention, so the equality remains valid.

The rational canonical form of $T$ is a block diagonal matrixsimilar to a matrix of $T$, where the $f_j\in F[t]$ are monic nonconstant polynomials and the sum of their degrees is $\dim V$. They are the invariant factors of a linear operator. The block $C(f_j)$ is the companion matrix for $f_j$, representing multiplication by $t$ on $F[t]/(f_j)$. The invariant factors are uniquely determined by $T$.

Every companion block has a cyclic vector and therefore has both minimal polynomial and characteristic polynomial equal to $f_j$, by part (a). The determinant of a block diagonal matrix is the product of its block determinants, givingA polynomial $h$ annihilates the whole matrix exactly when it annihilates every block. On block $j$, dividing $h$ by $f_j$ leaves a remainder of smaller degree; if that remainder annihilated the block without being zero, it would contradict minimality of $f_j$. Thus the annihilation condition is exactly $f_j\mid h$ for every $j$. The monic polynomial of least degree with this property is their least common multiple. The divisibility chain makes this $f_r$, soThese are the minimal and characteristic polynomials from invariant factors formulae. On the zero vector space the product is empty and both polynomials are defined to be one.

Let $J=\left(\begin{array}{cc}0& -2\\ 1& 0\end{array}\right)$, so the displayed rational canonical form of $T$ is $J\oplus J$. Since $J^2=-2I_2$, we have $T^2=-2I_4$.

If $a=0$, then $U=b{I}_4$ is a scalar matrix. Its invariant factors of a linear operator are four copies of $t-b$, so its rational canonical form isHere $m_U=t-b$ and $c_U=(t-b{)}^4$, including when $b=0$.

Now let $a\ne 0$ and putOn each two-dimensional invariant block, $A=aJ+b{I}_2$ satisfies $q(A)=0$. The vectors $e_1,A{e}_1$ form a cyclic vector basis, since their coordinate matrix is $P=\left(\begin{array}{cc}1& b\\ 0& a\end{array}\right)$ with determinant $a\ne 0$. Directly,Applying this basis change to both blocks givesThus $m_U=q$ and $c_U=q^2$. In particular there are two invariant factors $q,q$, rather than a single block for $q^2$. Also $q(t)>0$ for every real $t$, so it is an irreducible polynomial over $\mathbb{R}$. This is an instance of rational canonical form under an affine change.

Finally $T^2=-2I_4$ implies $T^{-1}=-T/2$. Substituting $a=-1/2$, $b=0$ in the preceding result gives $q(t)=t^2+1/2$, and henceIts minimal polynomial is $t^2+1/2$ and its characteristic polynomial is $(t^2+1/2{)}^2$. The normalized reciprocal of $t^2+2$ is also $t^2+1/2$, agreeing with rational canonical form of an inverse.

Let $U=aT+bI$ on a nonzero finite-dimensional vector space over a field, with $a\ne 0$. An invariant block for $T$ is also invariant for $U$. If $v,Tv,\dots ,T^{d-1}v$ is its cyclic vector basis, then $v,Uv,\dots ,U^{d-1}v$ is another basis: the binomial theorem gives a triangular change-of-basis matrix with diagonal $1,a,\dots ,a^{d-1}$. The monic polynomial $g(t)=a^{d}f((t-b)/a)$ annihilates this block, and its degree is $d$, so it is the block's minimal polynomial. In the new cyclic basis the block is $C(g)$.

The substitution preserves divisibility of the monic invariant factors of a linear operator. Apply it to every block to obtain the rational canonical form of $U$. If $a=0$, then $U=bI$ is a scalar matrix instead: all invariant factors are $t-b$, one for each dimension. This exceptional case cannot be obtained by dividing by $a$.