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As part of a larger problem, I want to compute the Smith Normal Form of $xI-B$ over $\mathbb{Q}[x]$ where $$ B=\begin{pmatrix} 5 & 2 & -8 & -8 \\ -6 & -3 & 8 & 8 \\ -3 & -1 & 3 & 4 \\ 3 & 1 & -4 & -5\end{pmatrix}. $$

So I do some elementary row and column operations and get to

$$\begin{pmatrix} 1+x & -2 & 0 & 0 \\ -3(x+1) & x+3 & 0 & 0 \\ 0 & 1 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix}. $$ Then I work with the upper left 3x3 matrix, and ultimately get:

$$\begin{pmatrix} x-3 & 0 & 0 & 0 \\ 0 & x+1 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix}. $$

So now I have a diagonal matrix (and I'm pretty sure I didn't mess anything up in performing row and column operations), except according to http://mathworld.wolfram.com/SmithNormalForm.html, the diagonal entries are supposed to divide each other, but obviously x-3 does not divide x+1. This means that: either I did something wrong, or diagonal matrix is not unique. Any ideas for how to transform my final matrix into a matrix whose diagonal entries divide each other?

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    Add column 2 to column 1. Subtract row 2 from row 1. Now you have a scalar in the (1,1) position -- rescale to 1. Wipe out everything in its row and column. Now your diagonal is 1,x+1,x+1,x+1.2011-10-30
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    Also, Smith Normal Form is unique (if you rescale all polynomials to monic polynomials at the end).2011-10-30
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    Wait, do you meant $(x+1)(x+5)$?2011-10-30
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    I didn't actually work out the details, so if you came up with a factor "$(x+1)(x+5)$" that's probably right :)2011-10-30
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    @BillCook Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-11
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    Just in case someone is wondering: The 2 unimodular matrices, which you multiply with your original matrix from left and from right, in order to get the Smith normal form: those 2 matrices are NOT unique. Only the normal form itself is unique (up to unit factors from one diag elt to the next.)2014-03-11

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To expand my comment...Add column 2 to column 1. Subtract row 2 from row 1. Now you have a scalar in the (1,1) position -- rescale to 1.

$$\begin{pmatrix} x-3 & 0 & 0 & 0 \\ 0 & x+1 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim \begin{pmatrix} x-3 & 0 & 0 & 0 \\ x+1 & x+1 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim $$ $$\begin{pmatrix} -4 & -x-1 & 0 & 0 \\ x+1 & x+1 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim \begin{pmatrix} 1 & (1/4)(x+1) & 0 & 0 \\ x+1 & x+1 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim $$

Now add $(-1/4)(x+1)$ times column 1 to column 2 (to clear everything beside 1).

$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ x+1 & x+1-(1/4)(x+1)^2 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim $$

Add $-(x+1)$ times row 1 to row 2 (to clear everything below 1) & simplify the (2,2)-entry. Then rescale row 2 (so the polynomial is monic).

$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -(1/4)(x+1)(x-3) & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & (x+1)(x-3) & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim $$

Finally swap columns 2 and 4 and then rows 2 and 4 to switch the positions of $(x+1)(x-3)$ and $x+1$. We are left with the Smith normal form.

$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & x+1 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & (x+1)(x-3)\end{pmatrix} $$