# Partial fraction of complex polynomials : a linear algebra proof

I will show a proof I think I devised which I will explain in details.

I will prove that

Suppose that $$(P,Q) \in (\mathbb{C}[X])^2$$ such that $$\deg(P)<\deg(Q), \deg(Q)>1\\ \forall x \in \mathbb{C}, Q(x)=c\prod_{i=1}^n(x-\alpha_i)^{m_i},$$

where $$\deg(Q) = \sum_{i=1}^nm_i$$, and $$\alpha_i$$ denotes the $$i^\text{th}$$ root of $$Q$$ whose multiplicty is $$m_i$$. Then, there exists a unique set of values $$\{a_{i,j}\}_{\substack{1\leq i\leq n \\1\leq j \leq m_i}}$$ such that $$\frac{P(x)}{Q(x)}=\sum_{i=1}^n\sum_{j=1}^{m_i}\frac{a_{i,j}}{(x-\alpha_i)^j}.$$

Here is the proof outline

Define $$E=\left\{\frac{P}{Q}\,\middle|\, P\in \mathbb{C}[X] \wedge \deg(P)<\deg(Q)\right\}\\ F=\left\{\sum_{i=1}^n\sum_{j=1}^{m_i}\frac{a_{i,j}}{(x-\alpha_i)^j}\,\middle|\, a_{i,j}\in{\mathbb C}\mbox{ for all }i,j\right\}.$$ First, we prove that $$E$$ and $$F$$ are both vector spaces with equal dimension $$d=\deg(Q)$$. Then, proving that $$F\subseteq E$$ implies that $$E=F$$, (dimension theory on finite vector spaces) then the uniqueness of vector decomposition in a base yields the result.

Thus, the proof steps are the following :

• Prove that E=$$\DeclareMathOperator{\spann}{span} \spann{(\frac{1=x^0}{Q(X)}},\dots,\frac{x^{d-1}}{Q(x)})$$, and therefore $$E$$ is a vector space (done, will post soon)
• Prove that the elements of the set are linearly independent, and thus, $$\dim{E}=d$$ (done as well)
• Do the same work with $$F=\spann{(\frac{1}{(x-\alpha_1)},\dots,\frac{1}{(x-\alpha_1)^{m_1}},\dots,\frac{1}{(x-\alpha_n)^{m_n}})}$$ (need help to do this rigourously)
• Show that $$F\subset E$$ (help needed)

Can anyone help me to write those steps?

Thanks!

Hint : To write $$\spann$$ in math put the following at the top of your answer:

$$\DeclareMathOperator{\spann}{span}$$


then use \spann{something} in math mode

• You can also use \operatorname{span} each time you need it. If you are only using it once or maybe twice, that is easier. But it you use it multiple times, \DeclareMathOperator is more convenient. (Of course, you can also just use \text{span}, but then you don't get the correct spacing.) Commented Jun 1, 2019 at 5:53

Step 3: It is easy to see that $$F$$ is spanned by the set $$\left\{\frac{1}{(x-\alpha_i)^j}\right\}_{\substack{j=1,2\ldots,m_i\\i=1,2,\ldots,n}}$$. Next, we show that the elements of the set are linearly independent by looking for a non-trivial solution for the following equation: \begin{align} \sum_{i=1}^n\sum_{j=1}^{m_i}\frac{a_{ij}}{(x-\alpha_i)^j}=0&\implies \sum_{i=1}^n\sum_{j=1}^{m_i}\frac{a_{ij}(x-\alpha_i)^{m_i-j}}{(x-\alpha_i)^{m_i}}=0\\ &\implies \sum_{i=1}^n\sum_{j=1}^{m_i}\frac{a_{ij}(x-\alpha_i)^{m_i-j}\prod_{k=1,k\neq i}^{n}(x-\alpha_k)^{m_k}}{\prod_{k=1}^n(x-\alpha_k)^k}=0\\ &\implies \sum_{i=1}^n\sum_{j=1}^{m_i}a_{ij}(x-\alpha_i)^{m_i-j}\prod_{k=1,k\neq i}^{n}(x-\alpha_k)^{m_k}=0. \end{align} Substituting $$x=\alpha_l$$ in the above expression gives $$a_{lm_l}=0$$, for all $$l=1,2,\ldots,n$$. Thus, we get $$$$\sum_{i=1}^n\sum_{j=1}^{m_i-1}a_{ij}(x-\alpha_i)^{m_i-j}\prod_{k=1,k\neq i}^{n}(x-\alpha_k)^{m_k-1}=0.$$$$ After repeated substitution of $$x=\alpha_l$$ in the above simplified expression, we obtain that $$a_{ij}=0$$, for $$j=1,2\ldots,m_i$$ and $$i=1,2,\ldots,n$$. Thus, the elements of the set are linearly independent.
Step 4: Using a simplication similar to that in step 3, we get that any element of $$F$$ can be expresed as the ratio of two polynomials: $$\frac{\sum_{i=1}^n\sum_{j=1}^{m_i}a_{ij}(x-\alpha_i)^{m_i-j}\prod_{k=1,k\neq i}^{n}(x-\alpha_k)^{m_k}}{\prod_{k=1}^n(x-\alpha_k)^k}= \frac1c \frac{\sum_{i=1}^n\sum_{j=1}^{m_i}a_{ij}(x-\alpha_i)^{m_i-j}\prod_{k=1,k\neq i}^{n}(x-\alpha_k)^{m_k}}{Q(x)}$$ Clearly, the degree of the numerator at most $$d-1$$ and thus, it is less than the degree of denominator. Hence, $$F\subseteq E$$.