linear combination of some matrices is identity matrix 
Assume $T$ is a $n\times n$ matrix over number field $\mathbb{F}$. If $\lambda$ is not an eigenvalue of $T$, we know $T-\lambda E$ is invertible matrix where $E$ is the identity matrix. Now if we have $n$ different numbers $\lambda_1,\cdots,\lambda_n\in\mathbb{F}$ and each one is not an eigenvalue of $T$, how to prove there exist $n$ numbers $a_1,\cdots,a_n\in\mathbb{F}$ which satisfy $$\sum_{k=1}^na_k(T-\lambda_k E)^{-1}=E\ ?$$

I don't have much idea. I figured that it's sufficient to prove the case $\mathbb{F}=\mathbb{C}$. Because once we have $n$ complex numbers satisfying the equation, consider $\mathbb{C}$ as a linear space over $\mathbb{F}$ we can get $n$ numbers in $\mathbb{F}$ satisfying the equation. And if we set the characteristic polynomial of $T$ is $p(x)$ and $$p(x)=g_k(x)(x-\lambda_k)+p(\lambda_k),$$ then $$(T-\lambda_k E)^{-1}=\frac{g_k(T)}{p(\lambda_k)}.$$ But i don't know how to continue. Any help would be appreciated.
 A: Let $p(x)=\det(xI-T)$ be the characteristic polynomial of $T$.  Let $q(x)$ denote $\prod_{i=1}^n(x-\lambda_i)$.  Then,
$$f(x)=q(x)-p(x)$$
is a polynomial of degree at most $n-1$.  That is,
$$\frac{f(x)}{q(x)}=\sum_{i=1}^n\frac{a_i}{x-\lambda_i}$$
for some $a_1,a_2,\ldots,a_n\in \Bbb F$.  To be precise, $$a_i=\frac{f(\lambda_i)}{ \prod_{j\neq i}(\lambda_i-\lambda_j)}=-\frac{p(\lambda_i)}{\prod_{j\neq i}(\lambda_i-\lambda_j)}=-\frac{p(\lambda_i)}{q'(\lambda_i)}.$$
Therefore,
$$q(T)=q(T)-p(T)=f(T)=\left(\sum_{i=1}^na_i(T-\lambda_i E)^{-1}\right)\ q(T).$$
Since $q(T)$ is invertible, so
$$E=\sum_{i=1}^na_i(T-\lambda_iE)^{-1}=-\sum_{i=1}^n\frac{p(\lambda_i)}{\prod_{j\neq i}(\lambda_i-\lambda_j)}(T-\lambda_i E)^{-1}.$$
A: The accepted answer is simply excellent. Partial fractions - not just for calculus!
But the accepted answer doesn't say anything about why $f/q$ has a partial-fraction decomposition as claimed. It's possible to give a proof that "partial fractions work" in $\Bbb C(x)$ using a little bit of complex analysis; in fact I'm guilty of publishing such a proof, in Complex Made Simple. Given that and the fact that the OP specifies that $\Bbb F$ is a subfield of  $\Bbb C$ and says "I figured that it's sufficient to prove the case $\Bbb F=\Bbb C$", it seems possible that some readers might get the idea that the argument is specific to complex numbers.
No, it works over any field. And here,  since the $\lambda_j$ are distinct, it turns out that it's much simpler than I realized until yesterday - I thought I'd share the argument.
Notation is as above, except that $\Bbb F$ is an arbitrary field. Define $$q_k(x)=\frac{q(x)}{x-\lambda_k}=\prod_{j\ne k}(x-\lambda_j).$$ We need to show that there exist scalars $a_j$ with $$f=\sum a_jq_j.$$ Letting $V$ be the space of polynomials of degree no larger than $n-1$, it's enough to show that


$q_1,\dots,q_n$ span $V$.


Since $\dim(V)=n$ this is the same as 


$q_1,\dots,q_n$ are independent.


And that's more or less obvious: Say $$\sum c_jq_j=0.$$
Noting that $q_j(\lambda_k)=0$ for $j\ne k$ this shows that $$0=\sum_jc_jq_j(\lambda_k)=c_kq_k(\lambda_k);$$hence $c_k=0$, since $q_k(\lambda_k)\ne0$.
