Proof: A Triangular Matrix is Invertible $ \Longleftrightarrow $ its Eigenvalues are Real and Nonzero 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!
 A: 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.
A: Notice that for a triangular matrix eigen values are nothing but diagonal entries and determinant of a tiangular matrix is product of diagonal enties. Note another thing a matrix is invertible iff it's determinant is non zero.
It is enough to prove the second statement as for a triangular matrix $A$ the matrix $A-\lambda I$ is also tiangular. Where $I$ is the identity matrix of order same as order of $A$ and $\lambda$ is an indeterminate. 
Notice the formula of determinant $$det(B)=\sum_{\sigma \in S_n} b_{1,\sigma(1)}.....b_{n,\sigma(n)}$$ ,where $B$ is a $n×n$ matrix and $S_n$ is the symmetric group of $\{1,2,....,n\}$. Notice that whenever $B$ is triangular then each term of summation is zero ,except one term ( this particular term is nothing but product of diagonal entries), as each term of summation contains exactly one element from each column and exactly one element from each row.
