The answer is “No”.

We show this by establishing that \(\dim(\ker(T)) > 0.\)

Applying the rank-nullity theorem to \(T\), we have \(\dim(\operatorname{range}(T)) + \dim(\ker(T)) = \dim(\mathbb{R}^4).\) Hence, \(\dim(\ker(T)) = \dim(\mathbb{R}^4) - \dim(\operatorname{range}(T)).\) Note that \(\dim(\mathbb{R}^4) = 4\). In addition, \(\operatorname{range}(T)\) is a subspace of \(\mathbb{R}^3\) and therefore has dimension at most \(3\). It follows that \(\dim(\ker(T)) = 4 - \dim(\operatorname{range}(T)) \geq 4 - 3 =1.\)