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Let $A$ be a $n \times n$ matrix. Then further let $p(\lambda)$ be characteristic polynomial of $A$ given by $$p(\lambda) = (\lambda - \lambda_1)^{r_1}(\lambda - \lambda_2)^{r_2} \ldots (\lambda - \lambda_k)^{r_k}$$ Define $$q_i(\lambda) = \frac{p_i(\lambda)}{(\lambda - \lambda_i)^{r_i}} \ for \ each \ i = 1 \ to \ k$$ Now consider the following factorization $$\frac{1}{p(\lambda)} = \frac{a_1(\lambda)}{(\lambda - \lambda_1)^{r_1}} + \frac{a_2(\lambda)}{(\lambda - \lambda_2)^{r_2}} + \ldots + \frac{a_k(\lambda)}{(\lambda - \lambda_k)^{r_k}}$$ where $a_i(\lambda)'s$ are functions of $\lambda$.

using above two equations we get $$1 = a_1(\lambda)q_1(\lambda)+ a_2(\lambda)q_2(\lambda) + \ldots + a_k(\lambda)q_k(\lambda)$$

Now i strucked after this point. It is written in my notes that above polynomial equation is satisfied by $A$ i.e. $$I_n = a_1(A)q_1(A)+ a_2(A)q_2(A) + \ldots + a_k(A)q_k(A)$$ where $I_n$ is an identity matrix.

How this result holds? As i know that $A$ can satisfy only those polynomials which are multiple of its minimal polynomial. Moreover degree of the polynomial is $n-1$.

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    well, the equation you've written holds for every $\lambda$, since it is an identity of polynomials. So if you substitute $A$ into it, it has to hold – 2017-01-23
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    No this is not true. i know that $A$ cannot satisfy any polynomial. Eg Take $A$ to be a nilpotent matrix with $A^2 = 0$, then $A$ donot satisy $\lambda^2=1$ – 2017-01-23
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    Hmm, what is not true? If $A$ is nilpotent, the equation becomes $1=1$, which is obviously satisfied by all eigenvalues of $A$. – 2017-01-24
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    @user1551 How? Equation is true? it becomes $0 = 1$ – 2017-01-29
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    When $A$ is nilpotent, we have $k=1,\ q_1=1$ and $a_1=1$. So, the equation $1=a_1q_1$ becomes $1=1$. – 2017-01-29
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    @user1551 I am talking about any general polynomial, not which is given in the question, My second comment's reply is for first comment. – 2017-01-29

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