Inverse of Triangular matrix [duplicate]

I want to know if the inverse of a Triangular matrix (Cholesky decomposition of a symmetric, positive-definite matrix in my particular case) is still triangular. If so could someone provide a proof as well?

marked as duplicate by Jonas Meyer, user147263, dustin, Adam Hughes, Przemysław ScherwentkeMar 9 '15 at 0:36

Yes, if $A$ is triangular matrix and invertible, then the inverse $A^{-1}$ of $A$ is also triangular. To show this, note that (see here) $$adj(A)\cdot A=\det(A)\cdot I$$ where $adj(A)$ is the adjugate of $A$ and $I$ is the identity matrix with same size as $A$. Therefore, we have $$A^{-1}=\frac{1}{\det(A)}adj(A).$$ Since $A$ is triangular, we can show that $adj(A)$ is also triangular.
More precisely, if $A$ is upper triangular, then the cofactor $C_{ij}$ of $A$ is zero for $i>j$, i.e. $adj(A)$ is upper triangular. Similarly, if $A$ is lower triangular, then the cofactor $C_{ij}$ of $A$ is zero for $i<j$, i.e. $adj(A)$ is lower triangular.
Without determinants: If $L$ is $n\times n$ triangular (say, lower) nonsigular matrix it can be written in the partitioned form $$L = \begin{bmatrix}\lambda & 0 \\ l & \tilde{L}\end{bmatrix},$$ where $\tilde{L}$ is an $(n-1)\times (n-1)$ lower triangular matrix and, in addition, $\lambda\neq 0$ and $\tilde{L}$ is nonsingular (otherwise it is easy to show, that $L$ is singular). A simple calculation shows that $$L^{-1}=\begin{bmatrix}\lambda^{-1} & 0\\ -\lambda^{-1}\tilde{L}^{-1}l & \tilde{L}^{-1}\end{bmatrix}.$$ Hence $L^{-1}$ is lower triangular provided that the smaller lower triangular matrix $\tilde{L}$ has a lower triangular inverse (one smells an easy induction argument here).