Proving Triangular Matrix Inverse is Also Triangular

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SUMMARY

The inverse of a triangular matrix is indeed triangular, as established through induction and row operations. For an invertible upper triangular matrix A, the product AA^{-1} equals the identity matrix I, confirming that the entries of A^{-1} maintain the triangular structure. Specifically, the last row of A multiplied by the last column of A^{-1} yields the identity's last entry, while subsequent rows ensure that non-diagonal entries remain zero. This proof elegantly demonstrates the triangular nature of the inverse without relying solely on brute force methods.

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It is stated in almost every linear algebra text i could find that the inverse of a triangular matrix is also triangular, but no proofs accompanied such statements.

I am convinced that it is the truth, but I have not been able to write anything down that I am satisfied with that doesn't rely on the argument that row operations on the matrix (A|I) to obtain (I|A^{-1}).

Since this would only be the forward pass(if A is lower triangular) and the backwards pass(if A is upper triangular) and these operations ultimately do not introduce non zero terms above/below the diagonal entries(depending on what A was), thus A^{-1} would be a triangular matrix of the same flavor.

Has anyone come across anything a little more elegant than simply brute forcing it?
 
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Assume you have a invertible upper triangular matrix. Consider

AA^{-1} = I

You can use induction, starting from the last row of A times the last column of A^{-1}. gives you the entry. lower left entry 1.

Then again take the last row of A and n-1 column of A^{-1}. To be able to get a zero in the identity matrix, (n,n-1) entry of A^{-1} must be zero.
...
Carry on to the upper left corner and you are done.
 

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