3 Texified since this was on the front-page anyway

My favorite elementary application of linear algebra is proving that the decomposition used in Calculus of rational functions into partial fractions works.

Start with a polynomial Q(x)=(x-r1)(x-r2)...(x-rn). $Q(x)=(x-r_1)(x-r_2)\cdots(x-r_n)$. Then the space of P(x)/Q(x) $P(x)/Q(x)$ with $deg P < deg Q Q$ is n-dimensional $n$-dimensional since it has a basis {1/Q(x), x/Q(x), x2/Q(x)..., xn-1/Q(x)}. $\frac{1}{Q(x)}, \frac{x}{Q(x)}, \frac{x^2}{Q(x)}, \dots, \frac{x^{n-1}}{Q(x)}$}. But {1/(x-r1),1/(x-r2),...,1/(x-rn)} $\frac{1}{(x-r_1)},\frac{1}{(x-r_2)},\dots,\frac{1}{(x-r_n)}$} are linearly independent vectors in the space and thus a basis.

Hence, P(x)/Q(x)=A1/(1-x-r1)+A2/(x-r2)+...+An/(x-rn) $\frac{P(x)}{Q(x)}=\frac{A_1}{(1-x-r_1)}+\frac{A_2}{(x-r_2)}+\dots+\frac{A_n}{(x-r_n)}$ for some constants {A_1,...,A_n}, $A_1,\dots,A_n$}, which then we can furthermore find by taking the limit of (x-ri)P(x)/Q(x) $(x-r_i)\frac{P(x)}{Q(x)}$ as x $x$ goes to ri.$r_i$.

2 fixed typo

My favorite elementary application of linear algebra is proving that the decomposition used in Calculus of rational functions into partial fractions works.

Start with a polynomial Q(x)=(x-r1)(x-r2)...(x-rn). Then the space of P(x)/Q(x) with deg P < deg Q is n-dimensional since it has a basis {1/Q(x), x/Q(x), x2/Q(x)..., xnn-1/Q(x)}. But {1/(x-r1),1/(x-r2),...,1/(x-rn)} are linearly independent vectors in the space and thus a basis.

Hence, P(x)/Q(x)=A1/(1-x-r1)+A2/(x-r2)+...+An/(x-rn) for some constants {A_1,...,A_n}, which then we can furhtermore furthermore find by taking the limit of (x-ri)P(x)/Q(x) as x goes to ri.