Problem
Prove that a triangular matrix is invertible iff its eigenvalues are real and nonzero.
Attempt
Let's call this triangular matrix $A$.
From intuition (from the invertability of A), I quickly noted that:
$$ A\vec{x} = \lambda I\vec{x} $$
$$ A^{-1}A\vec{x} = \lambda I A^{-1}I\vec{x} $$
$$ \vec{x} = \lambda IA^{-1}\vec{x} $$
$$ \frac{1}{\lambda}I \vec{x} = A^{-1}\vec{x} $$
$$ A^{-1}\vec{x} = \frac{1}{\lambda}I \vec{x} $$
So, the eigenvalues of $ A^{-1} $ are the reciprocals of the eigenvalues of $ A $. However, not sure how to proceed from here.
Thanks in advance!
Best Answer
I believe the assumption here is that $A$ is a real triangular matrix.
$$\det(A-\lambda I)=0$$
$$\prod_{i=1}^n (A_{ii}-\lambda)=0$$
The diagonal entries are the eigenvalues.
If $A$ is invertible, its determinant is non-zero. Hence, all the diagonal entries are non-zero and hence all the eigenvalues are non-zero.
Also, if the eigenvalues are real non-zero, then all the diagonal entries ae non-zero and hence $A$ is invertible.