91Ó°ÊÓ

Before we start the proof, let's recall some definitions. 1. A linear map T: V -> W is a function that takes vectors from V and maps them to vectors in W, such that the following properties hold: - T(u+v) = T(u) + T(v) for all u, v in V - T(cu) = cT(u) for all c in the field F and u in V 2. The null space (kernel) of a linear map T: V -> W, denoted as N(T), is the set of all vectors in V that are mapped to the zero vector in W: - N(T) = {v ∈ V : T(v) = 0} 3. The range (image) of a linear map T: V -> W, denoted as Im(T), is the set of all vectors in W that can be the result of mapping vectors from V: - Im(T) = {w ∈ W : w = T(v) for some v ∈ V} 4. A vector space is called finite dimensional if it has a finite basis. Now, let's proceed with the proof.

We are given that there is a linear map T: V -> W on the vector space V such that its null space N(T) and range Im(T) are both finite dimensional. We want to show that the vector space V itself is finite dimensional.

The Dimension Theorem states that for any linear map T: V -> W, the sum of the dimensions of the null space N(T) and the range Im(T) is equal to the dimension of the domain (V): - dim(N(T)) + dim(Im(T)) = dim(V) Since we are given that dim(N(T)) and dim(Im(T)) are both finite numbers, their sum is also a finite number. From the dimension theorem, this finite sum is equal to the dimension of V: - dim(V) = dim(N(T)) + dim(Im(T)) Since we have shown that dim(V) is a finite number, V is finite dimensional.