Algebraic method to see this finite sum is equal to 1

Question

Let $(\lambda_j)_{j=1}^r\subset\mathbb{C}^\times$ be pairwise disjoint. Then with these non-zero complex numbers chosen, we have$\sum_{j=1}^rf_j(x)=1$, where $$f_j(x)=\frac{(x-\lambda_1)\cdots(x-\lambda_{j-1})(x-\lambda_{j+1})\cdots(x-\lambda_r)}{(\lambda_j-\lambda_1)\cdots(\lambda_j-\lambda_{j-1})(\lambda_j-\lambda_{j+1})\cdots(\lambda_j-\lambda_r)}.$$

I wanted to try and deduce this algebraically and I've shown it holds for $r=2$ and $r=3$, which made me think of induction, but it got very complex since I couldn't see a way to simplify things while proving it holds for $r=n$.

This equality is derived from having a square matrix $A$ satisfying $f(A)\equiv0$, where $f(x)=(x-\lambda_1)\cdots(x-\lambda_r)$, so I believe it should hold for all natural numbers $r$.

Answer 1

It's easy to see that the polynomial $$P(x)=\sum_{j=1}^rf_j(x)-1$$ is of degree $r-1$. Let's calculate the value of this polynomial at all $\lambda_k$. If $k\ne j$ $f_j(\lambda_k)$ contains the term $\lambda_k-\lambda_k$ in the numerator, so it is $0$. If $k=j$ then all the terms in the denominator are the same as in the numerator, so $f_j(\lambda_j)=1$. Therefore all $r$ $\lambda_k$ values are roots of the polynomial $P(x)$. By the fundamental theorem of algebra, for a polynomial of degree $r-1$ you should have $r-1$ roots. Since you have $r$ roots it means that the polynomial is $0$.

Answer 2

I’m assuming that $\lambda_j$’s are all pairwise distinct so that $f_j(x)$’s are well-defined. Consider the polynomial $$ p(x) = 1-\sum_{j=1}^r f_j(x). $$ It is clear that it has degree $r-1$ because each of the terms in the sum has degree $r-1$. However, $p(\lambda_i) = 0$ for all $i=1,\ldots,r$ because the $f_j(\lambda_i)=1$ for $i=j$ and $0$ for $i\ne j$. This is impossible because the number of roots of a nonzero polynomial is at most its degree. Therefore, $p(x)$ must be the zero polynomial.

Answer 3

Here's a proof by induction, but it's a bit more involved, and probably only interesting for the sake of knowing that it's possible.

To prove this, we have to assume (and prove) a stronger induction hypothesis: that for any polynomial $g$ of degree at most $r-1$, $$ \sum_{j=1}^r g(\lambda_j) f_j(x) = g(x). $$ For small $r$, this can still be verified by brute force, giving us the base case of our induction.

Having assumed the induction hypothesis for some $r\ge 1$, pick an arbitrary $\lambda_{r+1}$, define $F_j(x) = \frac{x - \lambda_{r+1}}{\lambda_j - \lambda_{r+1}} f_j(x)$ for $j=1, \dots, r$, and define $F_{r+1}(x) = \prod_{j=1}^r \frac{x - \lambda_j}{\lambda_{r+1} - \lambda_j}$. To successfully induct, we must use our induction hypothesis to prove the corresponding statement for $F_1, \dots, F_{r+1}$ and for polynomials $g$ of degree at most $r$.

Well, for any polynomial $h$ of degree at most $r-1$, we have \begin{align} \sum_{j=1}^r (\lambda_j - \lambda_{r+1})h(\lambda_j) F_j(x) &= \sum_{j=1}^r (x - \lambda_{r+1})h(\lambda_j) f_j(x) \\ &= (x - \lambda_{r+1})\sum_{j=1}^r h(\lambda_j) f_j(x) \\ &= (x - \lambda_{r+1}) h(x) \end{align} where the last step uses our induction hypothesis. This proves the claim we want for every polynomial $g$ of degree at most $r$ which has the form $(x - \lambda_{r+1})h(x)$: those that have a root at $\lambda_{r+1}$.

But there's nothing special about excluding $\lambda_{r+1}$: we can equally well exclude $\lambda_1$. That way, we can prove the claim we want for every polynomial of degree at most $r$ with a root at $\lambda_1$.

Every polynomial can be written as a linear combination of polynomials with a root at $\lambda_1$ and polynomials with a root at $\lambda_{r+1}$: for any polynomial $g$ of degree at most $r$, we have $$ g(x) = \underbrace{g(x) - g(\lambda_1)}_{\text{root at }\lambda_1} + \underbrace{\frac{g(\lambda_1)}{\lambda_{r+1} - \lambda_1} (x - \lambda_1)}_{\text{root at }\lambda_1} - \underbrace{\frac{g(\lambda_1)}{\lambda_{r+1} - \lambda_1} (x - \lambda_{r+1})}_{\text{root at }\lambda_{r+1}}. $$ The claim we want to prove is linear in the polynomial $g$, so now we've proven it for all polynomials of degree at most $r$, completing the induction.

Answer 4

It is simply the Lagrange interpolation of $\,f(\lambda_j)=1,\, j=1,\ldots, r,\,$ which is derivable as a special case of the CRT formula for solution of the congruences $\,\{\,f(x)\equiv 1 \pmod{\!x\!-\!\lambda_j}\,\}_{\,j=1}^{\,r}.\,$ The above linked answer gives a very simple proof which yields the sought result.