Why is this theorem (on eigenvalues and invertibility) important? Theorem 5.1.5 in this book says

A square matrix $A$ is invertible if and only if $\lambda=0$ is not an eigenvalue of $A$.

I'd like to know if this theorem is important for practical purposes. Are there cases in which the above theorem is the easiest way to determine if a matrix is invertible?
Thanks.
 A: It tells you that invertibility is one of the properties that is determined by the eigenvalues of a square matrix. Other such properties (if you include complex eigenvalues) are the determinant $\det(A)$, the trace $\operatorname{Tr}(A)$ and its characteristic polynomial $\chi_A$.
Hence, the eigenvalues of a square matrix determine quite some important properties.
The proof of the theorem is pretty easy by the way. Let $f\colon V\to V$ be a linear map on a finite dimensional vector space $V$,
\begin{align*}
&\text{$f$ is invertible} \\
\Leftrightarrow\quad& \text{$f$ is injective} \\
\Leftrightarrow\quad& \ker f = \{0\} \\
\Leftrightarrow\quad& \text{there exists no $v\neq 0$ such that $f(v)=0$} \\
\Leftrightarrow\quad& \text{there exists no $v\neq 0$ such that $f(v)=0v$} \\
\Leftrightarrow\quad& \text{$0$ is not an eigenvalue of $f$}.
\end{align*}
A: It's not exactly a direct application, but for example if you find a non-zero vector in the kernel of a square matrix then this vector is an eigenvector that has eigenvalue zero so the matrix is not invertible. Also the determinant is the product of eigenvalues so if the determinant is zero then one of the eigenvalues is zero so the matrix is not invertible. Finally, if you don't know that the determinant of a product is equal to the product of determinants, you can note that if $v$ is an eigenvector with eigenvalue $0$ of $A$ then $v$ is also an eigenvector of eigenvalue 0 of the product $BA$ (assuming $A,B$ are square). So $BA$ is not invertible for any $B$.
A: In my opinion this is just a fancy way of saying that a square matrix is invertible if and only if its kernel consists of only the zero vector, and this basically follows directly from the rank-nullity theorem. So given that you know the theorem that a square matrix is invertible iff its kernel is trivial I can't think of a situation where this would really help since it's really just a rewording of the latter theorem. However if you didn't know it, then yes this is a useful theorem.
