Discussion
Let T be a linear transformation from R3 to R given by
\[ T[x, y, z]^{T} = 2x - 5y + 9z, \]
where [x, y, z]T donote the transpose of the vector [x, y, z]. Then, the dimension of the null space of T is :
Answer: C
The null space has dimension 2.
T maps \(\mathbf{R}^{3}\) onto R: it is not the zero map, since \(T(1, 0, 0) = 2\), so its rank is 1. By the rank-nullity theorem
\[ \dim N(T) = 3 - \text{rank}(T) = 3 - 1 = 2 \]
Geometrically the null space is the plane \(2x - 5y + 9z = 0\) through the origin, which is two-dimensional. 1 would be a line and 3 would make T the zero map; 0 is impossible for a map from a 3-dimensional space to a 1-dimensional one.
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