nvm I solved it. We see Ti*Tj=0 for i!=j, and Ti^2=Ti. Then we get that T1+T2+...+Tk=I or Ti. But if it equals Ti, then T1+T2+...+Ti-1+Ti+1+...Tk=0, which is false, so T1+T2+...+Tk=I
Homework Statement
Prove I=T1+T2+...+Tk
Where Ti=pi(T)
Homework Equations
T is kxk
pi(x)=(x-c1)...(x-ck) is the minimal polynomial of T.
pi=\pii(x)/\pii(ci)
\pii=\pi(x)/(x-ci)
To evaluate these functions at a matrix, simply let ci=ciI
The Attempt at a Solution
From lagrange interpolation...
We cannot define f(kv)=1 and f(w)=0 for w≠v, since if a is a multiple of v and b is a multiple of v, a+b is a multiple of v, so 1=f(a+b)≠f(a)+f(b)=2. So f won't be a linear transformation.
Homework Statement
Let V be a finite dimensional vector field over F. Let T:V→V
Let c be a scalar and suppose there is v in V such that T(v)=cv, then show there exists a non-zero linear functional f on V such that Tt(f)=cf.
Tt denotes T transpose.
Homework Equations
Tt(f)=f°T...