Question

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\) be a linear transformation.

a. What is the dimension of range of T if T is one-to-one mapping? Explain.

b. What is the dimension of the kernel of T (see section 4.2) if T maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^m}\)? Explain.

Short answer

a. The range of T is n.

b. The dimension of the kernel of T is \(n - m\).

Step-by-step solution

Find the dimension of Null A

As T is a one-to-one transformation, the columns of A are linearly independent. So,

\(\dim \,\;{\rm{Nul}}\,A = 0\).

Find the dimension of Column space A

Using the rank theorem,

\(\begin{aligned} \dim \left( {{\rm{col}}\,A} \right) + \dim \,{\rm{Nul}}\,A &= n\\\dim \left( {{\rm{col}}\,A} \right) + 0 &= n\\\dim \left( {{\rm{col}}\,A} \right) &= n\end{aligned}\).

The dimension of range of T is n.

Find the dimension of Null A for onto mapping

If T maps \({\mathbb{R}^n}\) to \({\mathbb{R}^m}\), then the columns of A span \({\mathbb{R}^m}\). By the rank theorem,

\(\begin{aligned} \dim \left( {{\rm{col}}\,A} \right) + \dim \,{\rm{Nul}}\,\,A &= n\\m + \dim \,{\rm{Nul}}\,\,A &= n\\\dim \,{\rm{Nul}}\,\,A &= n - m\end{aligned}\).

Find the dimension of Kernel of T

The kernel of T is a null space of A. So, the dimension of the kernel of T is \(n - m\) and \(n - m\) must be nonnegative, i.e., \(n \ge m\).

Since H is a subspace of V and B is also in V, B is a linearly independent set in V. So, B must also be a basis for V. Hence, Hand V are the same.