- #1
Buri
- 273
- 0
I would like to have someone go over my proof and see if its correct or at least on the right track. Here's the problem:
Let V be a finite-dimensional vector space, and let T: V → V be linear. If rank(T) = rank(T²), prove that R(T) ∩ N(T) = {0}.
PROOF:
Lemma: Let V be a finite-dimensional vector space, and let T: V → V be linear. If rank(T) = rank(T²), then N(T) = N(T²).
Proof: By the Nullity-Rank Theorem (i.e. Let V and W be vector spaces, and let T: V → W be linear. If V is finite-dimensional, then nullity(T) + ran(T) = dim(V)) I have,
dim(V) = rank(T) + nullity(T)
dim(V) = rank(T²) + nullity(T²)
This implies that nullity(T) = nullity(T²). Furthermore, N(T) and N(T²) are both subspaces of V and as a matter of fact, they are both vector spaces. Now it is easily shown that N(T) ⊂ N(T²) and that N(T) is a subspace of N(T²). Therefore, by the Dimension Theorem (i.e. Let W be subspace of a finite-dimensional vector space V. Then W is finite-dimensional and dim(W) ≤ dim(V). Moreover, if dim(W) = dim(V), then V = W) we have N(T) = N(T²). □
Seeking a contradiction, suppose that R(T) ∩ N(T) ≠ {0}. Therefore, there is an x ≠ 0 ∈ R(T) ∩ N(T). This implies that x ∈ R(T) and x ∈ N(T). Since x ∈ R(T) then T(y) = x for some y ∈ V. Note that T²(y) = T(T(y)) = T(x) = 0 which means that y ∈ N(T²) and hence by the lemma y ∈ N(T) also. However, this implies that T(y) = 0 and hence a contradiction since x = T(y) ≠ 0. ♦
I'd appreciate if someone could look at it. Thanks a lot!
Let V be a finite-dimensional vector space, and let T: V → V be linear. If rank(T) = rank(T²), prove that R(T) ∩ N(T) = {0}.
PROOF:
Lemma: Let V be a finite-dimensional vector space, and let T: V → V be linear. If rank(T) = rank(T²), then N(T) = N(T²).
Proof: By the Nullity-Rank Theorem (i.e. Let V and W be vector spaces, and let T: V → W be linear. If V is finite-dimensional, then nullity(T) + ran(T) = dim(V)) I have,
dim(V) = rank(T) + nullity(T)
dim(V) = rank(T²) + nullity(T²)
This implies that nullity(T) = nullity(T²). Furthermore, N(T) and N(T²) are both subspaces of V and as a matter of fact, they are both vector spaces. Now it is easily shown that N(T) ⊂ N(T²) and that N(T) is a subspace of N(T²). Therefore, by the Dimension Theorem (i.e. Let W be subspace of a finite-dimensional vector space V. Then W is finite-dimensional and dim(W) ≤ dim(V). Moreover, if dim(W) = dim(V), then V = W) we have N(T) = N(T²). □
Seeking a contradiction, suppose that R(T) ∩ N(T) ≠ {0}. Therefore, there is an x ≠ 0 ∈ R(T) ∩ N(T). This implies that x ∈ R(T) and x ∈ N(T). Since x ∈ R(T) then T(y) = x for some y ∈ V. Note that T²(y) = T(T(y)) = T(x) = 0 which means that y ∈ N(T²) and hence by the lemma y ∈ N(T) also. However, this implies that T(y) = 0 and hence a contradiction since x = T(y) ≠ 0. ♦
I'd appreciate if someone could look at it. Thanks a lot!