Prove that a triangular matrix is invertible iff its eigenvalues are real and nonzero.


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!


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.

  • $\begingroup$ Woah, this is clear as he$\mathcal{L}$$\mathcal{L}$. Thanks for the help! $\endgroup$ – Anthony Krivonos Dec 2 '18 at 5:00

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.


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