Let $T : V\rightarrow V$ a linear map, where $\dim V=2$. Then there exist $a_1-a_5 \in \Bbb R$, not all $0$, s.t $a_1T +… + a_4T^4 = a_5I_V$ .

Question

Prove or give counter exemple:

Let $T : V \rightarrow V $ be a linear map, where $\dim V = 2$. Then there exist $a_1 a_2, a_3, a_4, a_5 \in \Bbb R$, not all equal to $0$, such that $a_1T + a_2T^2 + a_3T^3 + a_4T^4 = a_5I_V$ .

Note:if $n$ is a positive integer, then $T^n=T\circ T\circ T\circ \cdots\circ T$ $_{(n-times)}$

I think the statement is true and wanted to use the identity transformation to show that if $a_1=a_2=a_3=a_4=a_5=1$ , and $T(x,y)=(x,y)$ I get $T\circ T=T(T(x,y))=(x,y)$, but I feel that I'm missing something.

Any help is greatly appreciated.

Answer 1

Yes, it is true. The space of endomorphisms of $V$ has dimension $4$. So, since $\operatorname{Id},T,T^2,T^3,T^4$ are $5$ elements of that space, they are linearly dependent.