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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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How about the determinant map again, this time restricted to $C$? –  Neal Mar 7 '12 at 19:08
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Can you characterize the set of invertible upper triangular matrices inside the set of all upper triangular matrices? –  Mariano Suárez-Alvarez Mar 7 '12 at 19:08
    
The invertible upper triangular matrices are those which have nonzero determinant in $C$. Then apply the same argument as before and hence we have that $B$ is an open set? –  198203 Mar 7 '12 at 19:53
    
Open in where, $C$ or $M(n)$? –  Neal Mar 7 '12 at 21:11
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1 Answer

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). –  Marc van Leeuwen Dec 20 '12 at 7:24
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