Question

Why is this true?
If the linear transformation T :Rn→Rn is onto and
T (u − v) = 0, then u = v.

Answer 1

By the rank-nullity theorem, we have\[\dim \ker T + \dim \mathop{\mathrm{im}}T=\dim \mathbb{R}^n = n.\]Because the transformation is onto, we have\[\mathop{\mathrm{im}}T = \mathbb{R}^n\]so\[\dim \ker T + n = n \Rightarrow \dim \ker T = 0\]which means the transformation is injective. Therefore,\[T(u - v) = 0 \Rightarrow T(u) - T(v) = 0 \Rightarrow T(u) = T(v) \Rightarrow u = v.\]