Vietnam Undergraduate Mathematics Competition 1994 B-5

[Namdung]

Let A is square real matrix of order n such that A^2 = E. Show that
rank(A+E) + rank(A-E) = n

Source VUMC 1994

[Moubinool]

I like this one A^2=E this mean that A is a symetry

A vanihes the polynomial x^2-1=(x-1)(x+1)

x+1, x-1 are prime => kernel decomposition theorem =>

R^n = ker(A-E)@ker(A+E) where @ is direct sum of sub vector space

take dimension

(*) n = dimker(A-E)+dimker(A+E)

rank theorem => dimker(A-E)=n-rg(A-E) , dimker(A+E)=n-rg(A+E) we plug in (*)

rg(A-E)+rg(A+E)=n

I don't know the name of decomposition theorem in english,
in Paris people say
"th�or�me de d�composition des noyaux" or " lemme des noyaux"

Theorem: E an R-vector space finite dimension, A,B polynomial with
real coefficients gcd(A,B) = 1, P=A.B product of A ,B
u:E->E linear map
then kerP(u)=kerA(u)@kerB(u) where @ is direct sum

How do you say this theorem in vietnam langage ?
In romanian ?