Question

5.1.10. If \( T \) is a linear transformation from one vector space to another, then the set of vectors mapped to 0 under \( T \) is called the kernel of \( T \). Prove that the kernel of \( T \) is a

Answer 1

The kernel of a linear transformation T is proven to be a subspace by showing it contains the null vector, and is closed under addition and scalar multiplication, which are the three main properties of a subspace.

Step-by-step explanation:

To prove that the kernel of a linear transformation T is a subspace, we need to verify that it satisfies the three properties of a subspace in the context of vector algebra. Those properties are that the set must contain the null vector, be closed under addition, and be closed under scalar multiplication.

Firstly, by the definition of a linear transformation, T(0) = 0, where 0 is the null vector in the domain space, so the null vector is in the kernel of T.

Secondly, suppose v and w are in the kernel of T. This means T(v) = 0 and T(w) = 0. By the properties of a linear transformation, T(v + w) = T(v) + T(w) = 0 + 0 = 0, which shows that the kernel is closed under addition.

Lastly, for any scalar a and any vector v in the kernel of T, T(av) = aT(v) = a0 = 0, so the kernel is closed under scalar multiplication. Therefore, these properties confirm that the kernel of T is indeed a subspace of the domain vector space.