Final answer:
To prove range T = null φ, we need to show that any vector x in the range of T is also in the null space of φ, and vice versa. An example of such a pair T=0 and φ=0 in V = R₂ and W = R₃ would be T: R₂ → R₃, where T(x) = 0 for all x in R₂, and φ: R₃ → R, where φ(y) = 0 for all y in R₃.
Step-by-step explanation:
To prove that range T = null φ, we need to show that any vector x in the range of T is also in the null space of φ, and vice versa.
Let y be a vector in the range of T. This means there exists a vector x in V such that T(x) = y. Since null T' = span(φ), we know that for any vector z in V, T'(z) = 0 if and only if z is in the null space of φ.
Now let's consider T(x). Since y is in the range of T, there exists a vector x in V such that T(x) = y. So, T'(y) = T'(T(x)) = (T'∘T)(x) = 0, since T'∘T is the zero linear transformation. This means that y is also in the null space of φ, which shows that range T is a subset of null φ.
For the other direction, let z be a vector in the null space of φ. We want to show that there exists a vector x in V such that T(x) = z. Since null T' = span(φ), z is in the range of T' and can be written as T'(w) for some vector w in V. Then T(z) = T(T'(w)) = (T∘T')(w) = 0, since T∘T' is the zero linear transformation. This shows that z is also in the range of T, which means null φ is a subset of range T. Therefore, range T = null φ.
An example of such a pair T=0 and φ=0 in V = R₂ and W = R₃ would be:
- T: R₂ → R₃, where T(x) = 0 for all x in R₂. This means the range of T is the zero subspace.
- φ: R₃ → R, where φ(y) = 0 for all y in R₃. This means the null space of φ is the entire space R₃.