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Let $B$ denote the $n \times n$ invertible upper triangular matrices. I am trying to duplicate the work done here where I asked a similar question for $GL_{n}(\mathbb{R})$.

My thought is: Let $C$ be the space of $n \times n$ upper triangular matrices, then $C \cong \mathbb{R}^{n(n + 1)/2}$. If $B$ is an open subset of $C$, then I am done, by the similar reasoning as in the $GL_{n}(\mathbb{R})$ case. However, I can't seem to think of a continuous map and a set that would give my such a result.

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    Open in where, $C$ or $M(n)$?2012-03-07

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If you are only interested in triangular matrices, there is a fully elementary solution. Namely, consider the natural mapping $\phi: C \to \mathbb{R}^{n(n+1)/2}$ that identifies them with the subset of the appropriate vector space.

Now, a triangular matrix is invertible iff all of its diagonal elements are non-zero (there are many arguments possible to see that, perhaps the simplest is that the diagonal elements are exactly the eigenvalues). So, if $x \in C$ is a triangular matrix, then ti is invertible iff $\phi(x)$ has non-zero elements on some specific $n$ coordinates. Another way of saying this is that $\phi(B) = \mathbb{R}^{n(n-1)/2} \times (\mathbb{R} \setminus \{0\})^n$ (perhaps up to rearrangement of coordinates). It is hopefully quite clear that this second set is open.

If you want to stick with determinant, I believe you can also do it, as indicated in comments.

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    +1 Nice complete answer. Just one comment about "perhaps the simplest is that the diagonal elements are exactly the eigenvalues". You certainly don't need to know about eigenvalues (or determinants) here: a triangular system of linear equations is solvable (independently of its right-hand side) by back-substitution if and only if its diagonal entries are all non-zero. I think this is one of the first things usually taught in linear algebra courses (although maybe not formulated exactly this way).2012-12-20