Htykuut

Assume $T$ is a $ntimes n$ 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 $lambda_1,cdots,lambda_ninmathbb{F}$ and each one is not an eigenvalue of $T$, how to prove there exist $n$ numbers $a_1,cdots,a_ninmathbb{F}$ 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.

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,
$$frac{f(x)}{q(x)}=sum_{i=1}^nfrac{a_i}{x-lambda_i}$$
for some $a_1,a_2,ldots,a_nin Bbb F$. To be precise, $$a_i=frac{f(lambda_i)}{ prod_{jneq i}(lambda_i-lambda_j)}=-frac{p(lambda_i)}{prod_{jneq i}(lambda_i-lambda_j)}=-frac{p(lambda_i)}{q'(lambda_i)}.$$
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
$$E=sum_{i=1}^na_i(T-lambda_iE)^{-1}=-sum_{i=1}^nfrac{p(lambda_i)}{prod_{jneq i}(lambda_i-lambda_j)}(T-lambda_i E)^{-1}.$$

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 $$q_k(x)=frac{q(x)}{x-lambda_k}=prod_{jne k}(x-lambda_j).$$ 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 $jne k$ this shows that $$0=sum_jc_jq_j(lambda_k)=c_kq_k(lambda_k);$$hence $c_k=0$, since $q_k(lambda_k)ne0$.