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A block matrix is block diagonal if all off-diagonal blocks vanish. It represents independent actions on a direct sum of vector spaces. Its determinant is the product of the diagonal-block determinants.

A block upper triangular matrix is a block matrix with square diagonal blocks and zero blocks below them. Its determinant is the product of the determinants of its diagonal blocks.

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 $f$.

Proceed by induction on $n=\dim V$. The zero-dimensional case is immediate. Over $\mathbb{C}$, the fundamental theorem of algebra supplies a root $\lambda$ of the characteristic polynomial. Since $\det (T-\lambda I)=0$, there is a nonzero vector $v_1$ with $T{v}_1=\lambda v_1$. The line $W=\mathbb{C}{v}_1$ is invariant, so $T$ induces a linear map $\overline{T}$ on $V/W$. By induction, choose an upper triangular basis for that quotient and lift its vectors to $v_2,\dots ,v_n$. Then $T{v}_j\in \mathrm{span}(v_1,\dots ,v_j)$ for every $j$. Thus $v_1,\dots ,v_n$ is the required basis and $\boxed{T\text{ is triangularizable}}$.

The same conclusion holds over $F$ when the characteristic polynomial splits over $F$. Indeed an eigenvalue then lies in $F$, and the block matrix arising from the invariant line gives ${\chi }_T(X)=(X-\lambda ){\chi }_{\overline{T}}(X)$. The quotient characteristic polynomial therefore also splits, so the induction works over $F$ without requiring algebraic closure.

Proceed by induction on $n=\dim V$. The zero-dimensional case is immediate. Over $\mathbb{C}$, the fundamental theorem of algebra supplies a root $\lambda$ of the characteristic polynomial. Since $\det (T-\lambda I)=0$, there is a nonzero vector $v_1$ with $T{v}_1=\lambda v_1$. The line $W=\mathbb{C}{v}_1$ is invariant, so $T$ induces a linear map $\overline{T}$ on $V/W$. By induction, choose an upper triangular basis for that quotient and lift its vectors to $v_2,\dots ,v_n$. Then $T{v}_j\in \mathrm{span}(v_1,\dots ,v_j)$ for every $j$. Thus $v_1,\dots ,v_n$ is the required basis and $\boxed{T\text{ is triangularizable}}$.

The same conclusion holds over $F$ when the characteristic polynomial splits over $F$. Indeed an eigenvalue then lies in $F$, and the block matrix arising from the invariant line gives ${\chi }_T(X)=(X-\lambda ){\chi }_{\overline{T}}(X)$. The quotient characteristic polynomial therefore also splits, so the induction works over $F$ without requiring algebraic closure.

Write $B=D+K$, where $D$ has $A$ in each diagonal block and zero elsewhere, while $K$ has $I$ immediately above the diagonal and zero elsewhere. These block matrices satisfy $DK=KD$: either product has $A$ immediately above the diagonal. Also, $K^i$ has $I$ on its $i$th superdiagonal and is zero elsewhere.

The binomial theorem for commuting matrices therefore yieldsOnly $r=i$ can contribute to block $(k,k+i)$, provided that block lies in the matrix. HenceThus ${\beta }_{i,j}=\left(\genfrac{}{}{0ex}{}{j}{i}\right)$, interpreted as an element of the field. The formula remains valid in positive field characteristic: some of these coefficients can then vanish. This possibility matters in the next part.

Use block matrix elimination with the invertible square block $A$:The left multiplier is block triangular with identity diagonal blocks, so its determinant is $1$. Taking the determinant of the block-triangular product provesThe matrix $D-C{A}^{-1}B$ is the Schur complement of $A$. The proof needs $A$ invertible, but does not assume the Schur complement or the full matrix is invertible.

A common symmetric saddle-point system couples primal unknowns with constraint multipliers through the displayed block matrix, with symmetric $H,C$. At $C=0$, full row rank of $A$ and a nonsingular reduced Hessian are exactly the nonsingularity conditions. Such matrices arise from equality-constrained quadratic models and mixed finite element methods. A regularized block $C=\mu I$ also arises when an augmented linear system replaces a badly scaled quadratic-penalty Newton system.

For a block matrix $M=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$ with $A$ an invertible matrix, the Schur complement of $A$ in $M$ is $S=D-C{A}^{-1}B$. It is the effective coefficient matrix for the remaining variables after block elimination. The definition requires invertibility of $A$, but not of $S$ or $M$. Its size is that of $D$, even when $B$ and $C$ are rectangular matrices.

In the original example under block matrix inverse, $D=5$ becomes $S=19/6$ because changing the remaining variable also changes the two variables solved through $A$. This is what makes a Schur complement more informative than simply deleting rows and columns. It retains their mediated effect, and its matrix inverse is the corresponding block of $M^{-1}$ when the block matrix inverse exists.

The same elimination appears when minimizing a quadratic form. The Schur complement as a minimized quadratic form gives a geometric example; the Schur complement determinant identity gives an exact test for an invertible matrix.