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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 $m_T=1$.