Unary polynomial space with a number of times less than or equal to n-1

The main object of the study of advanced algebra is the linear space, the $\mathbb{f}$ of all the times less than or equal to $n-1$ in the number domain constitute a linear space, recorded as $v=\mathbb{f}[x]_{n}$, then obviously $\dim v=n$, and it is easy to know that there is a set of bases for $$1,x,x^{2},\dots,x^{n-1}$$

In fact, according to the definition of linear space base it is easy to verify that there are also base $$1, (X-a), (x-a) ^{2},\dots, (x-a) ^{n-1}$$ and lagerange interpolation base $$ G_{i} (x) =\frac{1}{x-x_{i}}\prod_{j= 1}^{n} (X-x_{j}), i=1,2,\dots,n.$$

Share the following 2015 Sichuan University advanced Algebra Examination Questions in the exam, the use of Lagerange interpolation basis function will become very ingenious.

The $v$ is a $n$ dimensional linear space on a number of fields $\mathbb{f}$, $T $ is a linear transformation on $v$, $\lambda_{1},\lambda_{2},..., \lambda_{n}\in \mathbb{f}$ are not the same, and are not $t$ $I $ is an identity transform on the $v$.

1. Proof: For each \le i \le n,t-\lambda_{i}i$ is a reversible linear transformation on $v$.
2. Proof: Exists $\alpha_{1},\alpha_{2},..., \alpha_{n}\in \mathbb{f}$ makes $$ \sum_{k=1}^{n}\alpha_{k} (T-\lambda_{k}i) ^{-1}=I. $$

Proof: 1. For any $i=1,2,..., n$, $T-\lambda_{i}$ irreversible $\iff \left| T-\lambda_{i}\right|=0\iff \lambda_{i}$ is the characteristic value of $t$, and by the condition of the question set, $\lambda_{i}$ is not the characteristic value of $t$, so $t-\lambda_{i}$ is reversible.

2. Make $ $F (x) =\prod_{k=1}^{n} (X-\lambda_{k}) $$ because $t-\lambda_{i}$ reversible, so $f (T) =\prod_{k=1}^{n} (t-\lambda_{k}i) $ reversible, and in $\ mathbb{f}[x]_{n}$, $ $g _{i} (x) =\frac{1}{x-x_{i}}\prod_{j=1}^{n} (X-x_{j}), i=1,2,\dots,n$$ is $\mathbb{f}[x]_{n}$ A set of bases, so $f (x) $ can be shown by its linear form, i.e.: presence of $\alpha_{k}\in\mathbb{f}$ makes

$$ F (x) =\sum_{k=1}^{n}\alpha_{k}g_{k} (x) $$

The indefinite element $x$ in the above formula is still established with the transformation $t$, namely:

$$ F (t) =\sum_{k=1}^{n}\alpha_{k}g_{k} (t) $$

Both sides simultaneously function $f (T) ^{-1}$, with $$ \sum_{k=1}^{n}\alpha_{k} (T-\lambda_{k}i) ^{-1}=i. $$

The certificate is completed.

Unary polynomial space with a number of times less than or equal to n-1