Lagrange interpolation and minimal polynomials of vector spaces

Question

$V$ is a vector space over the field $F$, and $\def\End{\operatorname{End}} \def\Hom{\operatorname{Hom}}T \in\End(V)=\Hom(V,V)$ is a linear transformation with minimal polynomial $p(x)=(x-a_1)(x-a_2)...(x-a_k)$, where $a_k\in F$ and are distinct, with $k\ge 2$. Also let $p_i(x)=p(x)/(x-a_i)$, and $q_i(x)=p_i(x)/p_i(a_i)$.

Now I want to express $1$ and $x$ as a linear combination respectively of $q_i$ using Lagrange Interpolation to the basis of the space $P_k$ of all polynomials with degree less than $k$. This is dual to the set of evaluations $\{E_{a_1},...,E_{a_k}\}\subset P_k^*$. Then this should help me write $I=T_1+\cdots+T_k$, where $T_i=p_i(T)\in\End(V)$, wehre $p_i$ is given as above. Prove that $T_iT_j=0, i\ne j$ and that $T_i^2=T_i$ (T is idempotent).

Answer 1

First, I believe you meant $q_i(x)=\frac{p_i(x)}{p_i(a_i)}$.

Define $m(x)=\sum_{i=1}^k q_i(x)-1$ and $n(x)=\sum_{i=1}^{k}a_iq_i(x)-x$. Prove that the degree of $m(x)$ and $n(x)$ are smaller or equal to $k-1$ and $a_1,\ldots,a_k$ are k distinct roots of $m(x)$ and $n(x)$. Thus $m(x)=n(x)=0$.

Answer 2

The situation is as follows: the linear operator $T$ is diagonalisable (since it has a minimal polynomial that is split with simple roots) with eigenvalues $a_1,\ldots,a_k$. For any polynomial $P$, the action of $P[T]$ on the eigenspace for$~a_i$ is multiplication by the scalar $P[a_i]$, and since the (direct) sum of the eigenspaces is the whole space, this determines the action of $P[T]$ completely.

Now the (Lagrange interpolation) polynomials $q_i$ are defined so as to be dual to the system of evaluation functions $E_j$ in $a_j$, in other words $E_j(q_i)=q_i[a_j]=\delta_{i,j}$ for $i,j\in\{1,\ldots,k\}$. This means that $q_i[T]$ is the projection onto the eigenspace for $a_i$, parallel to the sum of the other eigenspaces.

Now for any system of $k$ scalars $c_1,\ldots,c_k$, one can realise the linear operator that acts on the eigenspace for$~a_i$ as multiplication by the scalar$~c_i$, as a polynomial (of degree less than$~k$, and as such unique) in$~T$, namely using the polynomial $c_1q_1+\cdots+c_kq_k$. In particular one can realise the identity (with $c_i=1$ for all$~i$) as $I=(q_1+\cdots+q_k)[T]$, and the operator $T$ itself (taking $c_i=a_i$ for all$~i$) as $T=(a_1q_1+\cdots+a_kq_k)[T]$. Given that $k\geq2$ (so that the polynomial$~X$ is of degree${}<k$), this implies that $a_1q_1+\cdots+a_kq_k=X$.

I don't really understand what the last question is, but I believe the above should answer it (with $T_i=q_i[T]$ the projection on the eigenspace for$~a_i$).