Vtyjkyuk

Let V is a complex vector space and if $A$ is a linear transformation such that spectar of $A=5,6$. Prove that $(A-5I)^n-1circ(A-6I)^n-1=O$. And $n=dim V$.

My teacher said that when you do not know on $n$-dimensional space you put down on 2 space that is easy, and I use that $n=2$. Now I know the spectrum of $(A-5I)=0,1$ and the spectrum of $(A-6I)=-1,0$ now I define linear transformation for some vector $(A-5I)(x,y)=(0,y)$ and $(A-6I)(x,y)=(-x,0)$ from here we can see that $operatornameim(A-6I)=ker(A-5I)$ so than $(A-5I)circ(A-6I)=O$, but what do you think is that enough?

No, you cannot define the linear transformation. Every square matrix satisfies its characteristic equaltion. Note that $5$ cannot be a root of order $n$ since $6$ is also a root. Hence $5$ is a root of order at most $n-1$. same thing is true of $6$. Hence $(lambda -5)^n-1 (lambda -6)^n-1$ is divisible by the characteristic polynomial. Hence $(A -5I)^n-1 (A-6I)^n-1=0$.

Since the spectrum of $A$ is $5,6$, the char. polynomial $p$ of $A$ is given by

$p(z)=(z-5)^k(z-6)^l$,

with $k,l ge 1$ and $k+l=n$. By Cayley - Hamilton:

$0=p(A)=(A-5I)^k(A-6I)^l$.

Can you proceed ?