Final answer:
To prove that a vector in the null space of a normal linear transformation T in ℝ³ has components that sum to zero, consider that (1,1,1) is an eigenvector of T, and any vector orthogonal to it is in the null space of T. By taking the dot product of any vector in the null space with (1,1,1) and setting it to zero, it follows that the sum of the vector's components must be zero.
Step-by-step explanation:
The question is asking to demonstrate that for a normal linear transformation T from ℝ³ to itself, if a vector (z₁, z₂, z₃) is in the null space of T, then the sum of its components z₁ + z₂ + z₃ must be zero. This question is rooted in the properties of linear transformations and vector spaces, specifically the concept of a normal transformation, which is one that commutes with its adjoint.
Let's start with the given vector (1,1,1) that, when transformed by T, yields the vector (2,2,2). We can use this information to show that any vector in the null space of T, which means T(z₁, z₂, z₃) = (0,0,0), must have components that sum to zero.
Since T is normal, it can be diagonalized, and the eigenvectors form an orthonormal basis. That implies that (1,1,1) is an eigenvector corresponding to a nonzero eigenvalue, and any vector orthogonal to (1,1,1) is in the null space of T.
Let (z₁, z₂, z₃) be in the null space. The null space is orthogonal to any vector mapped to a nonzero vector by T. Thus, we have (z₁, z₂, z₃) ⋅ (1,1,1) = 0. Expanding this dot product, we get z₁ + z₂ + z₃ = 0, which proves the required assertion.