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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 $(x-\alpha {)}^n$.

The prime ideals of $\mathbb{R}[x]$ are $(0)$ and the maximal ideals generated by the monic irreduciblesFor a nonzero prime $I=(p)$, use the residue classes of $1,x,\dots ,x^{N-1}$, where $N=n\mathrm{deg}p$. Multiplication by $x$ on $\mathbb{R}[x]/(p^n)$ is represented by the companion matrix of the monic polynomial $p^n$. In particular, for $p=x-a$ this may equivalently be written as one size-$n$ Jordan block with eigenvalue $a$. The ideal $(0)$ does not give a finite-dimensional quotient.

With $f(x,y)=-x{y}^2-y^5$, write $t=x_{n+1}=x_n+h$ and $z=y_{n+1}$. The Backward Euler method requiresor the monic quintic polynomial equationConstruct its $5\times 5$ companion matrixIts characteristic polynomial is $\det (zI-C_n)=p(z)$. Thus the polynomial roots needed for this update are exactly the eigenvalues of $C_n$.

At each step, assemble $C_n$, compute its eigenvalues, for example with the QR algorithm, and choose its unique positive real eigenvalue as $y_{n+1}$. To justify this selection, letIf $y_n>0$, then $F(0)=-y_n<0$, $F(y_n)=h{y}_n^5+ht{y}_n^2>0$, andThe intermediate value theorem and strict monotonicity prove there is exactly one positive polynomial root, located in $(0,y_n)$. Starting at $y_0=1$ therefore makes every update well defined, positive and decreasing. In numerical arithmetic, select the nearly real eigenvalue in this interval and check the polynomial residual $|F(z)|$; a final Newton root-finding iteration can refine it. This reduces each update to the requested eigenvalue problem.

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.

The nonconstant invariant factors of a linear operator satisfy $d_1\mid \cdots \mid d_k=m_A$. By part (a)(i), $m_A=X^2+1$ is an irreducible polynomial, so every $d_j$ must equal $X^2+1$. Their polynomial degrees sum to $n$, giving $n=2k$.

The companion matrix of $X^2+1$, in the basis $(1,X)$ of $\mathbb{Q}[X]/(X^2+1)$, isTherefore, for each fixed even $n$, every such $A$ has rational canonical formThere are no such matrices when $n$ is odd. This is an instance of the rational canonical form with irreducible minimal polynomial principle; equality of minimal polynomials alone does not determine rational canonical form in general.

In the basis $(v,Av)$, the linear map given by $A$ sends $v$ to the second basis vector and sends $Av$ to $A^{2}v=-v$. Equivalently,Multiplying on the left by the matrix inverse $E^{-1}$ givesThis change of basis produces the companion matrix found in part (a)(ii).

An integer-entry example isDirect matrix multiplication gives $A^2=\left(\begin{array}{cc}-1& 1\\ -1& 0\end{array}\right)$, so adding $A$ and $I_2$ gives the zero matrix. Equivalently, this is the companion matrix of $t^2+t+1$ and has the two nonreal cube roots of unity as eigenvalues. The direct multiplication is already a complete proof. It also shows why part (e) does not extend to all real square matrices: with this $A$, choosing $B=0$ satisfies the original product identity.

A quintic polynomial is a polynomial of polynomial degree five. Its polynomial roots can be found as the eigenvalues of a $5\times 5$ companion matrix.

The rational canonical form of a linear operator on a finite-dimensional vector space over a field is a direct sum $C(f_1)\oplus \cdots \oplus C(f_r)$ of companion matrices, where the monic nonconstant invariant factors of a linear operator satisfy $f_1\mid \cdots \mid f_r$. Their degrees sum to the dimension of the space. It exists over any field and classifies similar matrices, without requiring their eigenvalues to lie in that field. It comes from the structure theorem for finitely generated modules over a principal ideal domain, applied to the module in which the indeterminate acts as the given operator.

Let an invertible matrix $T$ have monic invariant factors of a linear operator $f_j$ of degrees $d_j$. Their constant coefficients are nonzero, since their product is the characteristic polynomial and $\det T\ne 0$. On a cyclic block, $f(T)=0$ implies $T^{-d}f(T)=0$, so $f^{\vee }(T^{-1})=0$, where $f^{\vee }(t)=t^{d}f(1/t)/f(0)$ is monic of degree $d$. The vector $v$ remains cyclic for $T^{-1}$: multiplying $v,T^{-1}v,\dots ,T^{-(d-1)}v$ by the invertible map $T^{d-1}$ gives the reversed original cyclic basis. Thus this polynomial is the minimal polynomial of the inverse block.

Normalized reciprocation preserves divisibility because $(fg{)}^{\vee }=f^{\vee }{g}^{\vee }$ when both constant coefficients are nonzero. Consequently the rational canonical form of $T^{-1}$ is the direct sum of the companion matrices $C(f_j^{\vee })$. The normalization by $f_j(0)$ is essential to make each reciprocal polynomial monic.

Let $A$ act on a nonzero finite-dimensional vector space over a field $F$, with minimal polynomial an irreducible polynomial $f$ of polynomial degree $d$. The nonconstant invariant factors of a linear operator all divide $f$, so each equals $f$. Their degrees sum to the dimension of a vector space, giving $d\mid n$ and rational canonical form consisting of $n/d$ copies of the companion matrix $C_f$. Equivalently, make the vector space a module over the field $F[X]/(f)$ by letting $X$ act as $A$; a basis over this field gives the same block decomposition.

For $u_t=u_x+u_y$ with equal spatial mesh sizes, the centered leapfrog recurrence has amplification polynomial of a multilevel finite difference scheme $G^2-2i\mu (\sin \xi +\sin \eta )G-1$. Uniform stability for arbitrary two-level starting perturbations holds exactly for $0<\mu <1/2$. At $\mu =1/2$ the phase pair $(\pi /2,\pi /2)$ gives $(G-i{)}^2$, and its companion matrix has a nontrivial Jordan block, producing growth proportional to the number of steps. For larger $\mu$ one polynomial root has modulus greater than one. Bounds on eigenvalue moduli alone therefore give a misleading non-strict endpoint.