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# 一個線性代數(linear algebra)問題(20點)

Suppose that V is a finite-dimensional vector space and T: V--->V is a linear transformation.

Prove that if the intersection of R(T) and N(T) = { 0 } (This is the zero vector of V), then V is the direct sum of R(T) and N(T).

Sorry, 很難打符號，所以都用敘述的。><

### 1 Answer

- 9 years agoFavorite Answer
V is finite dimensional. We assume that the dimension of V is n. By the rank-nullity theorem, dim N(T)+dim R(T)=n. Since T is a self-map, N(T) and R(T) are both subspaces V. Let B={v_1,...,v_m} be a basis for N(T) and C={u_1,...,u_k} be a basis for R(T). We know that m+k=n. Since the intersection of N(T) and R(T) is the zero subspace of V, we find that the union of B and C={v_1,...,v_m,u_1,...,u_k} is linearly independent. Since m+k=n, we find that the union of B and C forms a basis for V. This implies that V is the direct sum of R(T) and N(T). More precisely, since {v_1,..,v_m,u_1,...,u_k} is a basis for V, for any vector x in V, there exist a set of numbers, a_1,...,a_m, b_1,...,b_k so that x=a_1v_1+...+a_mv_m+b_1u_1+...+b_ku_k. Denote y=a_1v_1+...+a_mv_m and z=b_1u_1+...+a_ku_k. Then x=y+z for y in N(T) and z in R(T). Hence V is a sum of N(T) and R(T). Since the intersection of N(T) and R(T) is zero, we conclude that V is the direct sum of R(T) and N(T).