Defactoring: Difference between revisions

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\end{array} \right] \end{array} </math>

And we've completed the first step! The original matrix is now in HNF. So the next step is to take the other matrix we've been working on — the unimodular one from the Hermite decomposition — and invert it. Again, since we're in a transposed state, we're going to do the by-hand inversion technique, but to the bottom using elementary column operations rather than to the right using elementary row operations.

And we've completed the first step!<ref>Note that while the HNF is unique, the unimodular matrix is not. Because the 3rd row of the left matrix — the one in HNF — is all 0's, any number of that row can be added to either of the other two rows without altering the HNF at all, but with affecting the unimodular matrix on the right. Please feel free to experiment yourself, but I expect you will find that the inverse of any matrix you come up with this way, transposed, trimmed, and HNF'd, will give you the same canonical form — no need to worry about the exact path you happen to take to the HNF in the first step.</ref> The original matrix is now in HNF. So the next step is to take the other matrix we've been working on — the unimodular one from the Hermite decomposition — and invert it. Again, since we're in a transposed state, we're going to do the by-hand inversion technique, but to the bottom using elementary column operations rather than to the right using elementary row operations.

For our first step, let's add the 1st column to the 2nd column. That will get us a 0 in the top-center entry. Remember, we're trying to get the top-right matrix to look like an identity matrix.

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 \end{array} \right] \end{array} </math>
-And we've completed the first step! The original matrix is now in HNF. So the next step is to take the other matrix we've been working on — the unimodular one from the Hermite decomposition — and invert it. Again, since we're in a transposed state, we're going to do the by-hand inversion technique, but to the bottom using elementary column operations rather than to the right using elementary row operations.
+And we've completed the first step!<ref>Note that while the HNF is unique, the unimodular matrix is not. Because the 3rd row of the left matrix — the one in HNF — is all 0's, any number of that row can be added to either of the other two rows without altering the HNF at all, but with affecting the unimodular matrix on the right. Please feel free to experiment yourself, but I expect you will find that the inverse of any matrix you come up with this way, transposed, trimmed, and HNF'd, will give you the same canonical form — no need to worry about the exact path you happen to take to the HNF in the first step.</ref> The original matrix is now in HNF. So the next step is to take the other matrix we've been working on — the unimodular one from the Hermite decomposition — and invert it. Again, since we're in a transposed state, we're going to do the by-hand inversion technique, but to the bottom using elementary column operations rather than to the right using elementary row operations.
 For our first step, let's add the 1st column to the 2nd column. That will get us a 0 in the top-center entry. Remember, we're trying to get the top-right matrix to look like an identity matrix.