Coprimality gives polynomials $a,b\in \mathbb{Q}[x]$ with $ag+bh=1$, so the ideals $(g)$ and $(h)$ are comaximal. AlsoIndeed, if $u=gr=hs$, multiplying $ag+bh=1$ by $u$ writes $u=aghs+bhgr$, a multiple of $gh$; the reverse inclusion is immediate. Part (b), with $f=gh$, givesThe explicit isomorphism takes the class of a polynomial $r$ to its classes modulo $g$ and $h$.

Define $\Phi :R\to R/I\times R/J$ by $\Phi (r)=(r+I,r+J)$. It is a ring homomorphism withSince $I+J=R$, choose $i\in I,j\in J$ such that $i+j=1$. For arbitrary $a,b\in R$, the element $r=aj+bi$ satisfies $r-a=i(b-a)\in I$ and $r-b=j(a-b)\in J$. Hence $\Phi (r)=(a+I,b+J)$, proving surjectivity. The first isomorphism theorem now givesThis is the two-ideal Chinese remainder theorem for ideals. The hypothesis is comaximality of the two ideals; no assertion that every pair of ideals has this property is needed.

The first isomorphism theorem for rings states that the mapis a ring isomorphism. It is well defined because two representatives differ by an element of the kernel exactly when their images agree. It preserves addition and multiplication, is onto its stated codomain, and has trivial kernel. In particular, if $f$ is surjective then $R/\ker f\cong S$.

Equality in part (a) requires $a$ to lie in the span of the two chosen unit vectors. Equivalently, there are real constants $r,s$ such thatOne can also see the necessity directly by writing the nonnegative deficit aswhere ${\bar{a}}_e$ and ${\bar{a}}_o$ are the means of the even and odd entries. The deficit vanishes exactly when each group is constant; the two group constants need not be equal.

Let $p={\sum }_{j=1}^m⟨u,v_j⟩v_j$. Orthonormality gives $⟨u-p,v_k⟩=0$ for every $k$, so $u-p$ is orthogonal to $p$. The Pythagorean theorem for inner product spaces givesSince a squared norm is nonnegative,This is Bessel's inequality, with equality exactly when $u$ belongs to the span of the orthonormal set.

Part (b) requires $m_T$ to be a monic divisor of $x^2(x-1)$. The constant divisor one cannot annihilate a nonzero vector space. Every remaining divisor can occur in dimension three:Here are corresponding matrices, all satisfying $T^3-T^2=0$:The last matrix contains a size-two nilpotent Jordan block, requiring the factor $x^2$, and an eigenvector with eigenvalue one, requiring $x-1$. Thus its minimal polynomial is exactly $f$. The zero matrix has minimal polynomial $x\ne f$, providing the other requested example.

Write $m_1,m_2$ for the minimal polynomials of the two restrictions. Since $p(T)$ vanishes on $U$ and $q(T)$ vanishes on $W$, part (b) givesHence $m_1,m_2$ are coprime. By the direct-sum decomposition, a polynomial annihilates $T$ exactly when it annihilates both restrictions, which happens exactly when it is divisible by both $m_1$ and $m_2$. ConsequentlyIf $m_1=p$ and $m_2=q$, this immediately gives $m_T=pq=f$. Conversely, if $m_T=pq$, the monic polynomials $p/m_1$ and $q/m_2$ have product one, so each is one. ThusThis also handles a zero summand, whose minimal polynomial is one; a nonconstant factor cannot be its minimal polynomial.

Since $p$ and $q$ are coprime, Bézout's identity for polynomials gives $a,b\in F[x]$ with $ap+bq=1$. For every $v\in V$,The first summand belongs to $U$: applying $p(T)$ gives $b(T)p(T)q(T)v=0$. The second belongs to $W$ by the same calculation. Here all polynomials in $T$ commute, and $p(T)q(T)=f(T)=0$. Thus $V=U+W$.

If $v\in U\cap W$, both terms on the right of the identity vanish, so $v=0$. ThereforeFinally, $p(T)Tv=Tp(T)v=0$ whenever $v\in U$; hence $T(U)\subseteq U$. Replacing $p$ by $q$ proves $T(W)\subseteq W$. These are invariant subspaces; this is the coprime-factor case of the primary decomposition theorem for linear maps.

Apply polynomial division in $F[x]$:Evaluation at $T$ gives $r(T)=f(T)-q(T)m_T(T)=0$. A nonzero remainder could be rescaled to a monic polynomial annihilating $T$ of smaller degree, contradicting the defining minimality. Hence $r=0$ andIn particular, the annihilating polynomials form the principal ideal $(m_T)$.

The minimal polynomial of a linear transformation $T$ is the unique monic polynomial of least degree among the nonzero polynomials $p\in F[x]$ satisfying $p(T)=0$:Such a polynomial exists because the endomorphism space is finite-dimensional, so sufficiently many powers $I,T,T^2,\dots$ are linearly dependent. Monic normalization is possible over the field $F$. Two monic polynomials annihilating $T$ of the same least degree have a difference of smaller degree annihilating $T$, so their difference is zero. For the zero vector space the convention is $m_T=1$.