wisvuze said:show that the intersection is 0 and then use the rank-nullity theorem
wisvuze said:show that the intersection is 0
The notation T:V->V indicates that T is a linear transformation from a vector space V to itself. In other words, T takes in a vector from V and returns another vector in V.
To prove that T^2 = T, we need to show that T^2(v) = T(v) for all vectors v in V. This can be done by expanding T^2(v) using the definition of a linear transformation and simplifying it to match T(v).
When ker(T) + im(T) = V, it means that every vector in V can be written as a sum of a vector in the kernel of T and a vector in the image of T. In other words, every vector in V can be written as the output of T acting on another vector in V.
To prove that ker(T) and im(T) are subspaces of V, we need to show that they both contain the zero vector, are closed under vector addition, and are closed under scalar multiplication. This can be done using the definitions of a kernel and image of a linear transformation.
Yes, the identity transformation on any vector space V satisfies T^2 = T and ker(T) + im(T) = V. This is because the identity transformation maps every vector to itself, satisfying T^2 = T, and every vector in V can be written as a sum of itself (in im(T)) and the zero vector (in ker(T)), satisfying ker(T) + im(T) = V.