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Let $A$ be a companion matrix of a monic irreducible polynomial $f\in F[x]$ of degree $n$. Form the $N$-by-$N$ block matrix $B$ with $A$ on its diagonal, $I_n$ immediately above the diagonal, and zero elsewhere. If the field characteristic is $p>0$, write $f(x)=h(x^q)$ with $q=p^r$ maximal, so that $h^{\prime }\ne 0$; in characteristic zero take $q=1$ and $h=f$. ThenThus a separable polynomial gives $m_B=f^N$, but an irreducible polynomial with zero formal derivative can give a smaller exponent. For example, over the rational function field ${\mathbb{F}}_2(t)$, the irreducible polynomial $f(x)=x^2-t$ has $q=2$, so a two-block chain still has minimal polynomial $f$.

To prove the formula, write $B=D+K$, where $D$ has diagonal blocks $A$ and $K$ is the block shift. These are commuting operators, and $K^j$ vanishes exactly when $j\ge N$. The binomial theorem in positive field characteristic gives $B^q=D^q+K^q$. Expanding $h(D^q+K^q)$ gives $f(B)=K^{q}H$, with $H$ commuting with $K$. The diagonal blocks of $H$ are $h^{\prime }(A^q)$, which are invertible by Bézout's identity for polynomials, since $h$ is irreducible with nonzero formal derivative and $h(A^q)=0$. Therefore $f(B{)}^s=K^{qs}{H}^s$ vanishes exactly when $qs\ge N$. Finally, the block upper triangular matrix determinant gives the displayed characteristic polynomial, while the Cayley-Hamilton theorem and irreducibility force the minimal polynomial to be a power of $f$.

Write $f=m_T$. Part (ii) identifies the first superdiagonal block exactly:where $f^{\prime }$ is the formal derivative. Thus the candidate is $g=f^{\prime }$. If the irreducible polynomial $f$ has $f^{\prime }\ne 0$, its polynomial degree makes $f$ and $f^{\prime }$ coprime. Equivalently, $f$ is a separable polynomial. This includes every irreducible polynomial over a field of characteristic zero, and every irreducible polynomial over a perfect field.

Under this separability hypothesis, Bézout's identity for polynomials gives $a(x)f(x)+b(x)f^{\prime }(x)=1$. Evaluating at $A$ proves that $C=f^{\prime }(A)$ is an invertible matrix. Moreover, $f(B)$ has zero diagonal blocks, because $f(A)=0$. It is therefore a nilpotent matrix with $f(B{)}^N=0$. The only product of $N-1$ strictly upward block steps that reaches block $(1,N)$ uses the first superdiagonal every time, soFor $N=1$, the corresponding assertion is simply $f(B{)}^0=I\ne 0$. The block upper triangular matrix $B$ hasThe minimal polynomial divides this characteristic polynomial. Since $f$ is an irreducible polynomial, it must be a power of $f$; the exact nilpotency index just found forces

The arbitrary-field qualification is necessary. In the rational function field $F={\mathbb{F}}_2(t)$, take $f(x)=x^2-t$. This is an irreducible polynomial: a square of a rational function has even order of vanishing at $t=0$, while $t$ has odd order. Its formal derivative is zero. With $N=2$,The off-diagonal block in $f(B)$ is zero, and $m_U=f$, rather than $f^2$. No alternative polynomial $g$ coprime to $f$ could represent that block: $g(A)=0$ would imply $m_T=f\mid g$. Thus the printed coprimality claim fails as stated over a general field.

There is also a precise general replacement. In characteristic $p>0$, let $q=p^r$ be the largest power for which $f(x)=h(x^q)$, so that $h^{\prime }\ne 0$; set $q=1$ in characteristic zero. Irreducibility of $f$ implies that $h$ is an irreducible polynomial, so $h$ and $h^{\prime }$ are coprime. The binomial theorem in characteristic $p$ gives $B^q=D^q+K^q$. Expanding $h(D^q+K^q)$, with the coefficients equivalently supplied by the Hasse derivatives, giveswhere $H$ commutes with $K$ and is a block upper triangular matrix whose diagonal blocks are $h^{\prime }(A^q)$. These blocks are invertible by Bézout's identity for polynomials, since $h(A^q)=f(A)=0$. Thus $H$ is invertible, andThe same characteristic polynomial and divisibility argument now prove the corrected arbitrary-field answer:This is the minimal polynomial of a companion block chain; $q=1$ recovers the intended separable case.

In the determinant expansion of the block upper triangular matrix, a nonzero product must assign each of the bottom $n$ rows to one of the last $n$ columns, because the lower-left block is zero. Those rows therefore occupy all the last $n$ columns, leaving the first $n$ columns for the top rows. The contributing permutations are precisely independent permutations within the two blocks. Their signs multiply, so the sum factors:The off-diagonal block contributes to no surviving term.

View $X$ as the matrix of a linear map $T:F^m\to F^n$ of matrix rank $r$. Choose vectors $u_1,\dots ,u_r$ whose images form a basis of $\mathrm{im}T$, extend them by a basis of $\ker T$ to a basis of $F^m$, and extend $T{u}_1,\dots ,T{u}_r$ to a basis of $F^n$. In these two bases the matrix of $T$ is the rank normal formThe two changes of basis give invertible $P,Q$ with the displayed matrix equal to $Q^{-1}XP$.

For the block upper triangular matrix $A=\left(\begin{array}{cc}B& D\\ 0& C\end{array}\right)$, every nonzero term in the Leibniz formula for determinants sends the rows belonging to $C$ into the columns belonging to $C$; bijectivity then sends the remaining rows into the $B$ columns. The permutation sum consequently factors into the determinant sums for $B$ and $C$, proving

Finally suppose $L(X)=0$, so $AX=XB$: the map $X:{\mathbb{C}}^m\to {\mathbb{C}}^n$ is an intertwining operator. If $v$ lies in the generalized eigenspace of $B$ for $\lambda$, then for some $k$,Because $A$ and $B$ have no common eigenvalue, $A-\lambda I$ is invertible, hence $Xv=0$. The generalized eigenspaces of $B$ span ${\mathbb{C}}^m$, so $X=0$. Thus the Sylvester equation operator $L$ is injective.

Since $S\mathbf{w}=c{\mathbf{e}}^{(1)}$ and $A\mathbf{w}={\lambda }_1\mathbf{w}$,Thus the first column of $\hat{A}$ is $({\lambda }_1,0,\dots ,0{)}^T$, and $\hat{A}$ is a block upper triangular matrix:where $B$ is its bottom-right $(n-1)\times (n-1)$ submatrix. ThereforeA similarity transformation preserves the characteristic polynomial, sowith algebraic multiplicities included.

To construct $S$, use a Householder transformation. With $\alpha =\Vert \mathbf{w}{\Vert }_2$, choose the sign of $\alpha$ to avoid cancellation and setThen $S$ is an orthogonal matrix and maps $\mathbf{w}$ to $\alpha {\mathbf{e}}^{(1)}$ up to the chosen sign. If $\mathbf{w}$ already lies on the first coordinate axis, take a suitable diagonal sign matrix. This is the one-vector case of orthogonal coordinate reduction of a subspace.

Let $A_n$ denote the given tridiagonal matrix and put $D_n=\det A_n$, including $D_0=1$ for the empty determinant. The starting values areFor $n\ge 3$, apply the following elementary row operations in the indicated order, always using the updated rows:The first three rows now beginwith the fourth column omitted if $n=3$. When $n\ge 4$, also use $R_4\leftarrow R_4-R_3$. The resulting block upper triangular matrix has a $3\times 3$ upper triangular matrix with diagonal entries all one in its top left corner, and $A_{n-3}$ in its bottom right corner. All elementary row operations preserved the determinant except the single row interchange. ThereforeApplying this recurrence relation to the three starting values proves by mathematical induction thatEquivalently, the sequence is periodic with period six, with values $1,1,0,-1,-1,0$ starting at $n=0$. A compact closed form is $D_n=\sin ((n+1)\pi /3)/\sin (\pi /3)$.