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The inverse of an (invertible) upper triangular matrix is always upper triangular and the inverse of an (invertible) lower triangular matrix is always lower triangular. Why?

I don't have any work to show and this isn't homework.

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A flagged vector space of dimension $n$ is a vector space $V$ equipped with a filtration $$0\subseteq V_1 \subseteq \dots \subseteq V_n,$$ where the dimension increases by $1$ at each step. A morphism $V \to W$ of flagged vector spaces of the same dimension is a linear transformation which respects the flags.

A triangular matrix is nothing but an endomorphism of the flagged vector space

$$0\subseteq \mathbf F^1 \subseteq \dots \subseteq \mathbf F^n.$$

It is immediate that among these, the isomorphisms correspond to invertible triangular matrices, which therefore form a group under matrix multiplication.

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You can find the inverse of $A$ by row reduction in the extended matrix $(A|I)$. You should just trace what is happening to the identity matrix $I$ on the right while doing the row reduction process.

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Some hints and proofs can be found on the linked post in the comments above.

Another simple proof can be made by induction. It is clear that the statement is true for $n=1$. Assume that it is true as well for matrices of dimension $n-1$ and consider an $n\times n$ upper triangular and nonsingular matrix $U$ in the block form $$ U = \begin{bmatrix}\nu & u^* \\ 0 & \tilde{U}\end{bmatrix}, $$ where $\tilde{U}$ is $(n-1)\times (n-1)$. Since $U$ is assumed to be nonsingular and so $\nu\neq 0$ and $\tilde{U}$ is nonsingular. Since the statement is true for $(n-1)\times(n-1)$ matrices, $\tilde{U}^{-1}$ is upper triangular. Now let $$ U^{-1}=\begin{bmatrix}\alpha&b^*\\c&D\end{bmatrix} $$ be the partitioning of $U^{-1}$ conforming to the partitioning of $U$. Then $UU^{-1}=I$ is equivalent to $$ \alpha\nu+u^*c=1, \quad \nu b+D^*u=0, \quad \tilde{U}c=0, \quad \tilde{U}D=I. $$ The last two equations imply that $D=\tilde{U}^{-1}$ and $c=0$, which already implies that $U^{-1}$ is upper triangular.

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Since we are assuming that $A$ is invertible we know that for any linear independent set of vectors $x_k$ $\>(1\leq k\leq r)$ the vectors $Ax_k$ are again linearly independent.

If the matrix of $A$ with respect to the basis $(e_k)_{1\leq k\leq n}$ is upper triangular then for each $r\in[n]$ one has $$A\bigl(\langle e_1,e_2,\ldots, e_r\rangle\bigr)\subset \langle e_1,e_2,\ldots, e_r\rangle\ ,$$ and on account of the preliminary remark we even can say that $$A\bigl(\langle e_1,e_2,\ldots, e_r\rangle\bigr) = \langle e_1,e_2,\ldots, e_r\rangle\ .$$ Applying $A^{-1}$ on both sides of the last equation gives $$\langle e_1,e_2,\ldots, e_r\rangle =A^{-1}\bigl( \langle e_1,e_2,\ldots, e_r\rangle\bigr)\qquad(1\leq r\leq n)\ ,$$ and this says that the matrix of $A^{-1}$ with respect to the basis $(e_k)_{1\leq k\leq n}$ is upper triangular.

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