Subsection IVLT Invertible Linear Transformations

We have seen, both in theorems and in examples, that questions about linear transformations are often equivalent to questions about matrices. It is the matrix representation of a linear transformation that makes this idea precise. Here is our final theorem that solidifies this connection.

Theorem IMR Invertible Matrix Representations

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation, $B$ is a basis for $U$ and $C$ is a basis for $V$. Then $T$ is an invertible linear transformation if and only if the matrix representation of $T$ relative to $B$ and $C$, $\matrixrep{T}{B}{C}$ is an invertible matrix. When $T$ is invertible, \begin{equation*} \matrixrep{\ltinverse{T}}{C}{B}=\inverse{\left(\matrixrep{T}{B}{C}\right)} \end{equation*}

Proof

By now, the connections between matrices and linear transformations should be starting to become more transparent, and you may have already recognized the invertibility of a matrix as being tantamount to the invertibility of the associated matrix representation. The next example shows how to apply this theorem to the problem of actually building a formula for the inverse of an invertible linear transformation.

Example ILTVR Inverse of a linear transformation via a representation

You might look back at Example AIVLT, where we first witnessed the inverse of a linear transformation and recognize that the inverse ($S$) was built from using the method of Example ILTVR with a matrix representation of $T$.

Theorem IMILT Invertible Matrices, Invertible Linear Transformation

Suppose that $A$ is a square matrix of size $n$ and $\ltdefn{T}{\complex{n}}{\complex{n}}$ is the linear transformation defined by $\lt{T}{\vect{x}}=A\vect{x}$. Then $A$ is an invertible matrix if and only if $T$ is an invertible linear transformation.

Proof

This theorem may seem gratuitous. Why state such a special case of Theorem IMR? Because it adds another condition to our NMEx series of theorems, and in some ways it is the most fundamental expression of what it means for a matrix to be nonsingular — the associated linear transformation is invertible. This is our final update.

Theorem NME9 Nonsingular Matrix Equivalences, Round 9

Suppose that $A$ is a square matrix of size $n$. The following are equivalent.

  1. $A$ is nonsingular.
  2. $A$ row-reduces to the identity matrix.
  3. The null space of $A$ contains only the zero vector, $\nsp{A}=\set{\zerovector}$.
  4. The linear system $\linearsystem{A}{\vect{b}}$ has a unique solution for every possible choice of $\vect{b}$.
  5. The columns of $A$ are a linearly independent set.
  6. $A$ is invertible.
  7. The column space of $A$ is $\complex{n}$, $\csp{A}=\complex{n}$.
  8. The columns of $A$ are a basis for $\complex{n}$.
  9. The rank of $A$ is $n$, $\rank{A}=n$.
  10. The nullity of $A$ is zero, $\nullity{A}=0$.
  11. The determinant of $A$ is nonzero, $\detname{A}\neq 0$.
  12. $\lambda=0$ is not an eigenvalue of $A$.
  13. The linear transformation $\ltdefn{T}{\complex{n}}{\complex{n}}$ defined by $\lt{T}{\vect{x}}=A\vect{x}$ is invertible.

Proof
Sage NME9 Nonsingular Matrix Equivalences, Round 9
(in context)