# How to show that there exist a m such that $a_0I+a_1T+dots+a_mT^m=0$?

Mathematics Asked by Sunit das on December 20, 2020

Prove that for any transformation $$T:Vrightarrow V$$ , there exist a m and $$a_0$$,$$a_1$$,….,$$a_m$$ not all zero such that $$a_0I+a_1T+dots+a_mT^m=0$$.

Honestly, I don’t understand this question. And didn’t understand how to begin. Why there has to be such m?. Its clearly indicate such an $$m$$ exists for which $${I,T,……,T^m}$$ is linearly dependent. Is there anything special since the domain and codomain of T are the same?. Please I need your help. Thank you.

Note that V is finite dimensional.

Let $$V$$ be a finite dimensional vector space over a field $$mathbb{F}$$, with a basis $${ b_1,b_2,cdots,b_m}$$. If $$T$$ is a linear transformation on $$V$$, then $$T$$ is completely determined by the unique constants $${ a_{r,s} }$$, for $$1 le r,s le m$$ such that $$T(b_1) = a_{1,1}b_1+a_{1,2}b_2+cdots+a_{1,m}b_m \ T(b_2) = a_{2,1}b_1+a_{2,2}b_2+cdots+a_{2,m}b_m \ vdots \ T(b_m) = a_{m,1}b_1+a_{m,2}b_2+cdots+a_{m,m}b_m.$$ So $$T$$ is represented by an $$mtimes m$$ matrix over $$mathbb{F}$$. The set of $$mtimes m$$ matrices over $$mathbb{F}$$ is an $$m^2$$ dimensional vector space. Therefore $${ I,T,T^2,cdots,T^{mtimes m}}$$ is a linearly-depenedent set of matrices, which gives constants $$alpha_0,alpha_1,cdots,alpha_m$$ that are not all $$0$$ such that $$alpha_0 I + alpha_1 T + cdots + alpha_m T^{m^2} = 0.$$