Diagonalisation and the Minimal Polynomial

The zero polynomial lies in $\mathrm{Ann}(T)$. If $p(T)=q(T)=0$, then $(p-q)(T)=0$; and if $h\in F[x]$ and $p(T)=0$, distributivity of the finite polynomial sums and associativity of composition give $(hp)(T)=h(T)p(T)=0$. Hence $\mathrm{Ann}(T)$ is an ideal.

By [L2], ${\chi }_T\in \mathrm{Ann}(T)$. Fix an ordered basis $\mathcal{B}$ of the $n$-dimensional $V$; by [L3], ${\chi }_T={\chi }_{[T{]}_{\mathcal{B}}^{\mathcal{B}}}$ with $[T{]}_{\mathcal{B}}^{\mathcal{B}}\in M_n(F)$, so ${\chi }_T$ is monic of degree $n$ and in particular nonzero. Thus the ideal is nonzero, including when $V=0$, where ${\chi }_T=1$.

If $u$ and $v$ are monic generators of this nonzero ideal, then $u=av$ and $v=bu$ for some polynomials $a,b$. Degrees force $a,b$ to be nonzero constants, and monicity forces both constants to be $1$; hence $u=v$.

By [L1], write $\mathrm{Ann}(T)=(g)$ with $g\mathrm{\ne }0$. Multiplying $g$ by the inverse of its leading coefficient gives a monic generator ${\mu }_T$ by [L4], and step 1.3 shows that this monic generator is unique.

Finally, $p(T)=0$ means $p\in \mathrm{Ann}(T)=({\mu }_T)$, which is equivalent to $p={\mu }_{T}q$ for some $q\in F[x]$, that is, ${\mu }_T\mid p$. When $V=0$, step 1.2 gives $(1)=F[x]$, so its monic generator is $1$.

By induction on $k$, [L1] gives $[T^k{]}_{\mathcal{B}}^{\mathcal{B}}=A^k$ for every $k\ge 0$; taking the same finite linear combination on both sides yields $[p(T){]}_{\mathcal{B}}^{\mathcal{B}}=p(A)$ for every $p\in F[x]$.

A linear map is zero exactly when its matrix in a basis is zero, so step 1.1 and [L4] give $p(T)=0$ if and only if $p(A)=0$. The annihilator ideals coincide, hence their unique monic generators coincide by [L3].

If $B=P^{-1}AP$, induction gives $B^k=P^{-1}{A}^{k}P$, and therefore $p(B)=P^{-1}p(A)P$. Thus $p(B)=0$ exactly when $p(A)=0$, so [L3] again gives ${\mu }_B={\mu }_A$. This also follows from [L2].

Every polynomial over $F$ that annihilates $A$ still annihilates it over $K$. Hence the $K$-minimal polynomial divides the $F$-minimal polynomial and has no larger degree.

Conversely, let $q={\sum }_{i=0}^r{c}_i{x}^i\in K[x]$ be a nonzero annihilator of $A$. Choose a maximal $F$-linearly independent sublist $d_1,\dots ,d_s$ from the finite list of nonzero coefficients $c_i$; maximality makes it span all the $c_i$. Write $c_i={\sum }_j{a}_{ij}{d}_j$ with $a_{ij}\in F$.

The equality $0=q(A)={\sum }_j{d}_j({\sum }_i{a}_{ij}{A}^i)$ holds entrywise. Since every entry of each inner matrix lies in $F$ and the $d_j$ are $F$-independent, every matrix ${\sum }_i{a}_{ij}{A}^i$ is zero. At least one corresponding polynomial $q_j={\sum }_i{a}_{ij}{x}^i$ is nonzero and has degree at most $r$.

Applying step 2.1 to the $K$-minimal polynomial gives a nonzero $F$-annihilator of no larger degree. Thus the two minimal polynomials have equal degree; step 1.1 and monicity then force equality. For $n=0$, [L3] gives $1$ over either field.

By [L1], ${\chi }_T$ annihilates $T$.

Therefore [L2] gives ${\mu }_T\mid {\chi }_T$. In dimension zero both polynomials are $1$, and the same argument applies.

Suppose $Av=\lambda v$ for some nonzero $v\in K^n$. Induction gives $A^{k}v={\lambda }^{k}v$, so [L4] gives $p(A)v=p(\lambda )v$ for every $p\in K[x]$. Taking $p={\mu }_A$ and using [L1] yields $0={\mu }_A(A)v={\mu }_A(\lambda )v$, hence ${\mu }_A(\lambda )=0$.

Conversely, if ${\mu }_A(\lambda )=0$, then [L2] gives ${\chi }_A(\lambda )=0$. The determinant formula for $xI-A$ is unchanged after embedding $F$ in $K$, so [L3] applied over $K$ says $\lambda$ is an eigenvalue of $A$ on $K^n$.

When $n=0$, ${\mu }_A=1$ by [L1], so it has no roots, while the zero space has no nonzero eigenvector.

If $q\mid {\mu }_T$, then [L1] immediately gives $q\mid {\chi }_T$.

Conversely suppose $q\mid {\chi }_T$. Choose an ordered basis of $V$, let $A$ be the matrix of $T$, and form $K$ and $a$ as in [L2]. Since $q(a)=0$, also ${\chi }_T(a)=0$; by [L8] this is the characteristic polynomial of $A$, whose determinant formula is unchanged after scalar extension. Thus [L4] applied to the resulting endomorphism of $K^n$ gives a nonzero $K$-eigenvector with eigenvalue $a$.

By [L7], $A$ has minimal polynomial ${\mu }_T$, and [L3] says its scalar extension has the same minimal polynomial. Applying [L6] to the eigenvector from step 1.2 gives ${\mu }_T(a)=0$.

The monic irreducible minimal polynomial of $a$ divides $q$ because $q(a)=0$, and it is nonconstant; irreducibility of $q$ makes it equal to $q$. Now [L5] and step 2.1 give $q\mid {\mu }_T$.

Steps 1.1 and 3.1 prove both directions. In the zero-dimensional case ${\mu }_T={\chi }_T=1$, so neither has an irreducible factor.

By [L1] and associativity of composition, $Tp(T)=p(T)T$.

If $v\in \ker p(T)$, then $p(T)(Tv)=T(p(T)v)=0$, so $Tv\in \ker p(T)$.

If $v\in \mathrm{im}p(T)$, write $v=p(T)u$; then $Tv=p(T)(Tu)$ by step 1.1, so $Tv\in \mathrm{im}p(T)$.

Invariance makes $T{\mathrm{\mid }}_W$ an endomorphism of $W$, and for every polynomial $p$ one has $p(T{\mathrm{\mid }}_W)=p(T){\mathrm{\mid }}_W$ by induction on powers.

Since ${\mu }_T(T)=0$, step 1.1 gives ${\mu }_T(T{\mathrm{\mid }}_W)=0$. Applying [L1] to $T{\mathrm{\mid }}_W$ yields ${\mu }_{T{\mathrm{\mid }}_W}\mid {\mu }_T$. This includes $W=0$, where ${\mu }_{T{\mathrm{\mid }}_W}=1$.

Choose $a,b$ as in [L1]. Evaluating the identity gives $a(T)f(T)+b(T)g(T)=I$.

For $v\in V$, write $v=a(T)f(T)v+b(T)g(T)v$. The first summand lies in $\ker g(T)$ and the second in $\ker f(T)$ because polynomial evaluations commute and $(fg)(T)=0$. Thus the two kernels span $V$.

If $v$ lies in both kernels, step 1.1 gives $v=a(T)f(T)v+b(T)g(T)v=0$. Hence their intersection is zero, and [L2] proves the direct sum. Unit factors and the zero space satisfy the same calculation.

By [L1], the distinct powers $f_i=q_i^{e_i}$ are pairwise coprime. Since $({\prod }_i{f}_i)(T)={\mu }_T(T)=0$, repeated application of [L2] gives $V={⨁}_{i<r}\ker f_i(T)$.

Each summand is $T$-invariant because $T$ commutes with $f_i(T)$. The restriction $T_i=T{\mathrm{\mid }}_{V_i}$ is annihilated by $f_i$, so [L3] gives ${\mu }_{T_i}\mid f_i$.

Suppose for some $i$ that ${\mu }_{T_i}$ is a proper divisor of $f_i$. Put $h={\mu }_{T_i}{\prod }_{j\mathrm{\ne }i}{f}_j$. On $V_i$, the first factor annihilates; on $V_j$ with $j\mathrm{\ne }i$, the factor $f_j$ annihilates. Step 1.1 therefore gives $h(T)=0$.

The polynomial $h$ has smaller degree than ${\mu }_T$ and so cannot be divisible by ${\mu }_T$, contradicting [L3]. Hence ${\mu }_{T_i}=f_i$ by monicity. If $r=0$, then ${\mu }_T=1$, [L3] gives $I_V=0$, and $V=0$, which is precisely the empty direct sum.

The polynomials $f_i$ and $g_i$ are coprime, so choose $a_i,b_i$ with $a_i{f}_i+b_i{g}_i=1$ by [L2], and set $E_i=b_i(T)g_i(T)$ using [L3].

On $V_i=\ker f_i(T)$ the evaluated Bézout identity gives $E_i=I$. On $V_j$ for $j\mathrm{\ne }i$, the polynomial $g_i$ is divisible by $f_j$, so $E_i=0$.

By the unique decomposition in [L1], the operator described in step 2.1 is exactly projection onto $V_i$ along the sum of the other components.

By [L1], the irreducible factors are the distinct linear polynomials $x-{\lambda }_i$. Applying [L2] gives the displayed direct sum, and [L4] identifies its $i$-th summand with $G_{{\lambda }_i}^{(e_i)}(T)$.

Fix $i$ and $k\ge e_i$. The kernel of $(T-{\lambda }_{i}I{)}^k$ contains the $i$-th summand. On every other primary summand, $(x-{\lambda }_i{)}^k$ is coprime to $(x-{\lambda }_j{)}^{e_j}$; evaluating a Bézout identity from [L3] shows $(T-{\lambda }_{i}I{)}^k$ is invertible there. Hence its kernel contains no vector from the other summands.

Thus the kernel at every $k\ge e_i$ is exactly $G_{{\lambda }_i}^{(e_i)}(T)$. If the factorisation is empty, [L2] gives $V=0$.

Suppose $T$ is diagonalisable and fix an eigenbasis. Only finitely many eigenvalues occur among its finitely many vectors; group the basis vectors by those distinct eigenvalues. If $v\in E_{\lambda }(T)$ is expanded in the eigenbasis, comparing the coefficients in $Tv=\lambda v$ forces every coefficient at a basis vector with eigenvalue different from $\lambda$ to vanish. Thus the group carrying $\lambda$ is a basis of $E_{\lambda }(T)$. The groups exhaust the eigenbasis, so the eigenspaces sum to $V$; and a nonzero $v\in E_{\lambda }(T)$ lying in the sum of the other eigenspaces would have one eigenbasis expansion supported on the $\lambda$-group and another supported off it, contradicting uniqueness of coordinates in a basis. Both conditions of [L3] therefore hold.

Conversely, suppose the displayed direct sum holds. Choose a basis of each eigenspace using [L4] and concatenate the finite lists. The first condition of [L3] makes the concatenation spanning. It is independent: in a vanishing combination, group the terms by eigenspace, so each group sums to a vector of its $E_{\lambda }(T)$ and one such vector equals minus the sum of the others; the second condition of [L3] forces every group to sum to $0_V$, and independence inside each chosen basis then kills every coefficient. So it is a basis of eigenvectors and [L1] makes $T$ diagonalisable.

The two constructions prove both implications. When $V=0$, the empty basis and empty direct sum satisfy them.

Suppose $T$ is diagonalisable and $\dim V\ge 1$, and choose an eigenbasis. Its matrix is diagonal, so $xI-[T]$ is triangular of positive size and [L5] and [L7] show that ${\chi }_T$ is the product of $x-\lambda$ over the diagonal entries. For each $\lambda$, the basis vectors carrying that diagonal entry form a basis of $E_{\lambda }(T)$; hence their number is both multiplicities in [L1].

Conversely, suppose ${\chi }_T$ splits and the multiplicities agree. By [L8] the roots of ${\chi }_T$ in $F$ are exactly the eigenvalues of $T$, so by [L4] the sum of the algebraic multiplicities of the eigenvalues is the degree of ${\chi }_T$, and [L6] identifies that degree with $\dim V$. Hence the sum of the dimensions of the distinct eigenspaces is $\dim V$.

Choose a basis of each eigenspace. By [L3] their concatenation is independent, and step 1.2 says it has $\dim V$ vectors, so it is a basis. Therefore [L2] makes $T$ diagonalisable.

Steps 1.1 and 2.1 prove the equivalence when $\dim V\ge 1$. If $\dim V=0$, the empty basis is an eigenbasis, so $T$ is diagonalisable by [L2]; [L6] gives ${\chi }_T=1$, which splits as an empty product by [L4]; and $T$ has no eigenvalue, so the multiplicity condition holds vacuously. Both sides therefore hold, which proves the equivalence in every finite dimension.