Equivalences for Linearizations of Matrix Polynomials

Given a potential linearization $\boldsymbol{L}(z)=z\boldsymbol{C}_{1}-\boldsymbol{C}_{0}$, it is possible to discover the matrices $\boldsymbol{E}(z)$ and $\boldsymbol{F}(z)$ by computing the generic Hermite Form⁵⁵5The Smith form, which is related, is also useful here; but we found that the Maple implementation of the Hermite Form Storjohann (1994) gave a simpler answer, although we had to compute the matrix $\boldsymbol{F}(z)$ separately ourselves because the Hermite form is upper triangular, not diagonal., with respect to the variable $z$, for instance by using a modestly sized example of $\boldsymbol{L}(z)$ and a symbolic computation system such as Maple. The generic form that we obtain is of course not correct on specialization of the symbolic coefficients, and in particular is incorrect if the leading coefficient is zero; but we may adjust this by hand to find the following construction, which we will not detail in this section but will in the next.

By using the five-by-five scalar case, with symbolic coefficients which we write as $a_{k}$ instead of $\boldsymbol{A}_{k}$ because they are scalar, we find that if

$$\boldsymbol{E}(z)=\left[\begin{array}[]{ccccc}1&h_{4}&h_{3}&h_{2}&h_{1}\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&0&0&0&-1\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&0&0&-1&-z\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&0&-1&-z&-{z}^{2}\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&-1&-z&-{z}^{2}&-{z}^{3}\end{array}\right]$$

(3.2)

where $h_{5}=a_{5}$ and $h_{k}=a_{k}+zh_{k+1}$ for $k=4$, $3$, $2$, $1$ are the partial Horner evaluations of $p(z)=a_{5}z^{5}+\cdots+a_{0}$, and if

$$\boldsymbol{F}(z)=\left[\begin{array}[]{ccccc}{z}^{4}&0&0&0&1\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr{z}^{3}&0&0&1&0\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr{z}^{2}&0&1&0&0\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr z&1&0&0&0\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 1&0&0&0&0\end{array}\right]\>,$$

(3.3)

and if

$$\boldsymbol{L}(z)=\left[\begin{array}[]{ccccc}za_{{5}}+a_{{4}}&a_{{3}}&a_{{2}}&a_{{1}}&a_{{0}}\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr-1&z&0&0&0\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&-1&z&0&0\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&0&-1&z&0\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&0&0&-1&z\end{array}\right]\>,$$

(3.4)

then we have $\boldsymbol{E}(z)\boldsymbol{L}(z)\boldsymbol{F}(z)=\mathrm{diag}{(}p(z),1,1,1,1)$. Moreover, $\det\boldsymbol{E}(z)=\pm 1$ and $\det\boldsymbol{F}(z)=\pm 1$, depending only on dimension.

Once this form is known, it is easily verified for general grades and quickly generalizes to matrix polynomials, establishing (as is well-known) that this form (known as the second companion form) is a linearization.

Note that the polynomial coefficients $a_{k}$ appear linearly in $\boldsymbol{E}(z)$ and the unimodular matrix polynomials $\boldsymbol{E}$ and $\boldsymbol{F}$ are thus universally valid.

The monomial basis, the shifted monomial basis, the Taylor basis, the Newton interpolational bases, and many common orthogonal polynomial bases all have three-term recurrence relations that, for $k\geq 1$, can be written

$$z\phi_{k}(z)=\alpha_{k}\phi_{k+1}(z)+\beta_{k}\phi_{k}(z)+\gamma_{k}\phi_{k-1}(z)\>.$$

(3.5)

All such polynomial bases require $\alpha_{k}\neq 0$. For instance, the Chebyshev polynomial recurrence is usually written $T_{n+1}(z)=2zT_{n}(z)-T_{n-1}(z)$ but is easily rewritten in the above form by isolating $zT_{n}(z)$, and all Chebyshev $\alpha_{k}=1/2$ for $k>1$. We refer the reader to section 18.9 of the Digital Library of Mathematical Functions (dlmf.nist.gov) for more. See also Gautschi (2016).

For all such bases, we have the linearization⁶⁶6For exposition, we follow Peter Lancaster’s dictum, namely that the $5\times 5$ case almost always gives the idea. $\boldsymbol{L}(z)=zC_{1}-C_{0}$ where

$$C_{1}=\left[\begin{array}[]{c|cccc}\dfrac{a_{5}}{\alpha_{4}}&&&&\\ \hline\cr&\boldsymbol{I}_{4}\\ \end{array}\right]$$

(3.6)

and

$$C_{0}=\left[\begin{array}[]{c|cccc}-a_{4}+\dfrac{\beta_{4}}{\alpha_{4}}a_{5}&-a_{3}+\dfrac{\gamma_{4}}{\alpha_{4}}a_{5}&-a_{2}&-a_{1}&-a_{0}\\ \hline\cr\alpha_{3}&\beta_{3}&\gamma_{3}&&\\ &\alpha_{2}&\beta_{2}&\gamma_{2}&\\ &&\alpha_{1}&\beta_{1}&\gamma_{1}\\ &&&\alpha_{0}&\beta_{0}\end{array}\right]\>,$$

(3.7)

In Maple we compute the Hermite normal form of a grade $5$ example with symbolic coefficients by the following commands:

That procedure does not use the same convention we use here and so we apply the standard involutory permutation (SIP) matrix to it and transpose it to place the polynomial coefficients in the top row.

Now compute the generic Hermite normal form:

That returns matrices such that $\boldsymbol{U}\boldsymbol{L}=\boldsymbol{H}$, or $\boldsymbol{L}=\boldsymbol{U}^{-1}\boldsymbol{H}$. We now look at their structure.

This produces

$$\left[\begin{array}[]{ccccc}x&0&0&0&x\\ 0&x&0&0&x\\ 0&0&x&0&x\\ 0&0&0&x&x\\ 0&0&0&0&x\end{array}\right]$$

(3.8)

and a separate investigation shows the diagonal entries are (as is correct in the generic case) all just $1$, until the lower corner entry which is a monic version of the original polynomial. The matrix $\boldsymbol{U}$ has a simple enough shape in this case, but $\boldsymbol{U}^{-1}$ is more convenient:

produces

$$\left[\begin{array}[]{ccccc}x&x&x&x&x\\ x&x&x&0&0\\ 0&x&x&x&0\\ 0&0&x&x&0\\ 0&0&0&x&0\end{array}\right]\>.$$

(3.9)

Because the first $m-1$ columns of $\boldsymbol{H}$ is a subset of the identity matrix, the first $m-1$ columns of $\boldsymbol{U}^{-1}$ are the same as the first $m-1$ columns of $\boldsymbol{L}$. Since the final column of $\boldsymbol{U}^{-1}$ has only one nonzero entry, at the top, call that $u_{0}$. Call the entries in the final column of $\boldsymbol{H}$ in descending order $h_{m-1}$, $h_{m-2}$,$\ldots$, $h_{1}$; let us suppose that the final entry is a multiple $cp(z)$ of the scalar polynomial.

Multiplying out $\boldsymbol{U}^{-1}\boldsymbol{H}$ and equating its final column to the final column of $\boldsymbol{L}(z)$ gives us a set of (triangular, as it happens) equations in the unknowns. These allow us to deduce a general form for $\boldsymbol{U}^{-1}(z)$, and to prove it is unimodular. Once we have $\boldsymbol{U}$ and $\boldsymbol{H}$ with $\boldsymbol{U}\boldsymbol{L}(z)=\boldsymbol{H}$, construction of unimodular $\boldsymbol{E}(z)$ and $\boldsymbol{F}(z)$ so that $\boldsymbol{E}(z)\boldsymbol{L}(z)\boldsymbol{F}(z)=\mathrm{diag}{(}\boldsymbol{P}(z),\boldsymbol{I}_{n},\ldots,\boldsymbol{I}_{n})$ is straightforward.

The results are slightly simpler in this case to present as inverses: Here we show the Chebyshev scalar case.

$$\boldsymbol{E}^{-1}=\left[\begin{array}[]{ccccc}1&a_{1}&a_{2}&a_{3}-a_{5}&2za_{5}+a_{4}\\ 0&0&-\frac{1}{2}&z&-\frac{1}{2}\\ 0&-\frac{1}{2}&z&-\frac{1}{2}&0\\ 0&z&-\frac{1}{2}&0&0\\ 0&-1&0&0&0\end{array}\right]$$

(3.10)

and

$$\boldsymbol{F}^{-1}=\left[\begin{array}[]{ccccc}0&0&0&0&1\\ 0&0&0&1&-T_{1}\!\left(z\right)\\ 0&0&1&0&-T_{2}\!\left(z\right)\\ 0&1&0&0&-T_{3}\!\left(z\right)\\ 1&0&0&0&-T_{4}\!\left(z\right)\end{array}\right].$$

(3.11)

It is not immediately obvious that $\boldsymbol{E}(z)$ is unimodular, but it is. For $\boldsymbol{F}(z)$ it is obvious.

Notice again that the result is linear in the unknown polynomial coefficients, and that we therefore have a universal equivalence (once generalized to the matrix polynomial case, and of arbitrary dimension, which is straightforward).

We use the same approach as before, probing with a small example and computing its Hermite normal form explicitly to suggest the correct form.

The resulting suggested form for $\boldsymbol{U}^{-1}$ is

$$\left[\begin{array}[]{ccccc}x&x&x&x&x\\ x&x&0&0&x\\ 0&x&x&0&x\\ 0&0&x&x&x\\ 0&0&0&x&x\end{array}\right]$$

(4.14)

which is more complicated than before, because the final column is full (as before, the first $\ell-1$ columns merely copy the companion pencil $\boldsymbol{L}(z)$). Moreover, this time the entries in $\boldsymbol{U}^{-1}$ are polynomial functions of the $y_{k}$, $0\leq k<\ell$, not just linear; but involve negative powers of $y_{\ell}$ (this equivalence appears to be new: the treatment in Mackey and Perović (2016) is implicit, while the treatment in Amiraslani et al. (2008) only gives a local linearization). This means that in the case $\boldsymbol{Y}_{\ell}$ is singular, $z=1$ is an eigenvalue and something else must be done to linearize the matrix polynomial.

We find a recurrence relation for the unknown blocks in both the purported Hermite normal form and in the inverse of the cofactor. Put $\boldsymbol{U}^{-1}=$

$$\begin{bmatrix}z\boldsymbol{Y}_{\ell}/\ell-(z-1)\boldsymbol{Y}_{\ell-1}&-(z-1)\boldsymbol{Y}_{\ell-2}&\cdots&-(z-1)\boldsymbol{Y}_{1}&\boldsymbol{W}_{\ell}\\ -(z-1)\boldsymbol{I}_{n}&2z\boldsymbol{I}_{n}/(\ell-1)&&&\boldsymbol{W}_{\ell-1}\\ &-(z-i)\boldsymbol{I}_{n}&\ddots&&\vdots\\ &&&-(z-1)\boldsymbol{I}_{n}&\boldsymbol{W}_{0}\\ \end{bmatrix}$$

(4.15)

which, apart from the final column, is the same as the linearization $\boldsymbol{L}(z)$, and the Hermite normal form analogue as

$$\boldsymbol{H}=\begin{bmatrix}\boldsymbol{I}_{n}&&&&\boldsymbol{H}_{\ell-1}\\ &\boldsymbol{I}_{n}&&&\boldsymbol{H}_{\ell-2}\\ &&\ddots&&\vdots\\ &&&\boldsymbol{I}_{n}&\boldsymbol{H}_{1}\\ &&&&\boldsymbol{P}(z)\end{bmatrix}\>.$$

(4.16)

Multiplying out $\boldsymbol{U}^{-1}\boldsymbol{H}$ we find that for the lower right corner block

$$-(z-1)\boldsymbol{H}_{1}+\boldsymbol{W}_{0}\boldsymbol{P}(z)=\ell z\boldsymbol{I}_{n}\>.$$

(4.17)

A separate investigation shows that $\boldsymbol{W}_{0}$ must be a constant matrix. Evaluating that equation at $z=1$ shows that $\boldsymbol{W}_{0}=\ell\boldsymbol{P}^{-1}(1)$, so $\boldsymbol{P}(1)$ must be nonsingular for this form to hold. Once $\boldsymbol{W}_{0}$ is identified, equation (4.17) can be solved uniquely for the grade $\ell-1$ matrix polynomial $\boldsymbol{H}_{1}(z)$:

$$\boldsymbol{H}_{1}(z)=\frac{\ell}{z-1}\left(\boldsymbol{P}^{-1}(1)\boldsymbol{P}(z)-z\boldsymbol{I}_{n}\right)\>.$$

(4.18)

Division is exact there, as can be checked by expanding the numerator of the right-hand side in Taylor series about $z=1$.

The other block entries in the multiplied-out equation give a triangular set of recurrences for the $\boldsymbol{W}_{k}$ and the $\boldsymbol{H}_{k}(z)$ that are solvable in the same way and under the same condition, namely that $\boldsymbol{P}(1)$ must be nonsingular.

Once we have $\boldsymbol{U}$ and $\boldsymbol{H}$ with $\boldsymbol{U}\boldsymbol{L}(z)=\boldsymbol{H}$, construction of unimodular $\boldsymbol{E}(z)$ and $\boldsymbol{F}(z)$ so that $\boldsymbol{E}(z)\boldsymbol{L}(z)\boldsymbol{F}(z)=\mathrm{diag}{(}\boldsymbol{P}(z),\boldsymbol{I}_{n},\ldots,\boldsymbol{I}_{n})$ is straightforward.

For the pencil in equation (4.13) we exhibit the strict equivalence by the unimodular matrices $\boldsymbol{U}$ and $\boldsymbol{W}$ giving $\boldsymbol{L}_{m}(z)=\boldsymbol{U}\boldsymbol{L}_{B}(z)\boldsymbol{W}$, below; for ease of understanding, we actually show $\boldsymbol{U}^{-1}$ which shows its relationship to $\boldsymbol{W}$ more clearly:

$$\boldsymbol{W}=\left[\begin{array}[]{ccccc}5&0&0&0&0\\ -10&10&0&0&0\\ 10&-20&10&0&0\\ -5&15&-15&5&0\\ 1&-4&6&-4&1\end{array}\right]$$

(4.19)

and

$$\boldsymbol{U}^{-1}=\left[\begin{array}[]{ccccc}1&h_{3}&h_{2}&h_{1}&-y_{0}\\ 0&5&0&0&0\\ 0&-10&10&0&0\\ 0&10&-20&10&0\\ 0&-5&15&-15&5\end{array}\right]\>,$$

(4.20)

where $h_{3}=-10y_{3}+20y_{2}-15y_{1}+4y_{0}$, $h_{2}=-10y_{2}+15y_{1}-6y_{0}$, and $h_{1}=-5y_{1}+4y_{0}$. Generalizing these to matrix polynomials is straightforward, as is generalizing these to arbitrary grade.

The paper Mackey and Perović (2016) contains, in its equation (3.6), a five by five example including tensor products that shows how to find a strict equivalence of the pencil here to the strong linearizations they construct in their paper. Both the results of that paper and the construction given by example above demonstrate that the Bernstein linearization here is a strong linearization, independently of the singularity of $\boldsymbol{P}(1)$ and indeed independently of the regularity of the matrix polynomial.

We here present a slightly different reversal, namely rev $p(z)=(z+1)^{\ell}p(1/(z+1))$ of a polynomial of grade $\ell$ expressed in a Bernstein basis, instead of the standard reversal $z^{\ell}p(1/z)$. This new reversal has a slight numerical advantage if all the coefficients of $p(z)$ are the same sign. We also give a proof that the linearization of this reversal is the corresponding reversal of the linearization, thus giving a new independent proof that the linearization is a strong one. This provides the details of the entries in the unimodular matrix polynomials $\boldsymbol{E}(z)$ and $\boldsymbol{F}(z)$ with $\boldsymbol{E}(z)\boldsymbol{L}(z)\boldsymbol{F}(z)=\mathrm{diag}{\boldsymbol{P}(z),\boldsymbol{I}_{n},\ldots,\boldsymbol{I}_{n}}$ constructed above.

A short computation shows that if

$$p(z)=\sum_{k=0}^{\ell}y_{k}B_{k}^{\ell}(z)$$

(4.21)

then

$$\mathrm{rev}\,p(z)=(z+1)^{\ell}p\left(\frac{1}{z+1}\right)=\sum_{k=0}^{\ell}d_{k}B_{k}^{\ell}(z)$$

(4.22)

where

$$d_{k}=\sum_{j=0}^{k}\binom{k}{j}y_{\ell-j}\>,$$

(4.23)

whereas the coefficients of the standard reversal are, in contrast,

$$e_{k}=\sum_{m=0}^{\ell-k}(-1)^{m}\binom{\ell-k}{m}y_{\ell-m-k}$$

(4.24)

which has introduced sign changes, which may fail to preserve numerical stability if all the $y_{k}$ are of one sign. A further observation is that the coefficient $d_{0}$ only involves $y_{\ell}$, while $e_{0}$ involves all $y_{k}$; $d_{1}$ involves $y_{\ell}$ and $y_{\ell-1}$ while $e_{1}$ involves all but $y_{0}$, and so on; in that sense, this new reversal has a more analogous behaviour to the monomial basis reversal, which simply reverses the list of coefficients.

For interest, we note that if $(\boldsymbol{A},\boldsymbol{B})$ is a linearization for $p(z)$ so that $p(z)=\det(z\boldsymbol{B}-\boldsymbol{A})$, then reversing the linearization by this transformation is not a matter of simply interchanging $\boldsymbol{B}$ and $\boldsymbol{A}$:

	$\displaystyle(z+1)^{\ell}p\left(\frac{1}{z+1}\right)$	$\displaystyle=(z+1)^{\ell}\det\left(\frac{1}{z+1}\boldsymbol{B}-\boldsymbol{A}\right)$
		$\displaystyle=\det(\boldsymbol{B}-(z+1)\boldsymbol{A})$
		$\displaystyle=\det(\boldsymbol{B}-\boldsymbol{A}-z\boldsymbol{A})$		(4.25)

and so the corresponding reversed linearization is $(\boldsymbol{A},\boldsymbol{B}-\boldsymbol{A})$. The sign change is of no importance.

Suppose that the Bernstein linearization of $p(z)$ is $(\boldsymbol{A},\boldsymbol{B})$ and that the Bernstein linearization of rev $p(z)$ is $(\boldsymbol{A}_{R},\boldsymbol{B}_{R})$. That is, the matrices $\boldsymbol{A}_{R}$ and $\boldsymbol{B}_{R}$ have the same form as that of $\boldsymbol{A}$ and $\boldsymbol{B}$, but where $(\boldsymbol{A},\boldsymbol{B})$ contain $y_{k}$s the matrices $(\boldsymbol{A}_{R},\boldsymbol{B}_{R})$ contain $d_{k}$s. To give a new proof that the Bernstein linearization is actually a strong linearization, then, we must find a pair of unimodular matrices $(\boldsymbol{U},\boldsymbol{W})$ which have $\boldsymbol{U}\boldsymbol{A}_{R}\boldsymbol{W}=\boldsymbol{B}-\boldsymbol{A}$ and $\boldsymbol{U}\boldsymbol{B}_{R}\boldsymbol{W}=\boldsymbol{A}$, valid for all choices of coefficients $y_{k}$ (which determine the corresponding reversed coefficients $d_{k}$ by the formula above).

First, it simplifies matters to deal not with $\boldsymbol{W}$ but rather with $\boldsymbol{W}^{-1}$. Then, our defining conditions become

	$\displaystyle\boldsymbol{U}\boldsymbol{A}_{R}$	$\displaystyle=\left(\boldsymbol{B}-\boldsymbol{A}\right)\boldsymbol{W}^{-1}$		(4.26)
	$\displaystyle\boldsymbol{U}\boldsymbol{B}_{R}$	$\displaystyle=\boldsymbol{A}\boldsymbol{W}^{-1}\>,$		(4.27)

which are linear in the unknowns (the entries of $\boldsymbol{U}$ and of $\boldsymbol{W}^{-1}$). By inspection of the first few dimensions, we find that $\boldsymbol{U}$ and $\boldsymbol{W}^{-1}$ have the following form (using the six-by-six case, for variation, to demonstrate). The anti-diagonal of the general $\boldsymbol{U}$ has entries $-(\ell-i+1)/i$ for $i=1$, $2$, $\ldots$, $\ell$.

$$\boldsymbol{U}=\left[\begin{array}[]{cccccc}0&0&0&0&0&-6\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&0&0&0&-{\frac{5}{2}}&u_{{2,6}}\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&0&0&-{\frac{4}{3}}&u_{{3,5}}&u_{{3,6}}\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&0&-{\frac{3}{4}}&u_{{4,4}}&u_{{4,5}}&u_{{4,6}}\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&-{\frac{2}{5}}&u_{{5,3}}&u_{{5,4}}&u_{{5,5}}&u_{{5,6}}\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr-{\frac{1}{6}}&u_{{6,2}}&u_{{6,3}}&u_{{6,4}}&u_{{6,5}}&u_{{6,6}}\end{array}\right]$$

(4.28)

and

$$\boldsymbol{W}^{-1}=\left[\begin{array}[]{cccccc}0&0&0&0&z_{{1,5}}&z_{{1,6}}\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&0&0&z_{{2,4}}&z_{{2,5}}&z_{{2,6}}\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&0&z_{{3,3}}&z_{{3,4}}&z_{{3,5}}&z_{{3,6}}\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&z_{{4,2}}&z_{{4,3}}&z_{{4,4}}&z_{{4,5}}&z_{{4,6}}\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr z_{{5,1}}&z_{{5,2}}&z_{{5,3}}&z_{{5,4}}&z_{{5,5}}&z_{{5,6}}\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&0&0&0&0&1\end{array}\right]$$

(4.29)

One can then guess the explicit general formulae

	$\displaystyle u_{i,j}$	$\displaystyle=-\frac{(\ell-i+1)}{i}\binom{i}{\ell+1-j}\qquad 1\leq i,j\leq\ell$		(4.30)
	$\displaystyle z_{i,j}$	$\displaystyle=-\frac{(\ell-i)}{j}\binom{i}{\ell-j}\qquad 1\leq i,j\leq\ell-1$		(4.31)
	$\displaystyle z_{i,\ell}$	$\displaystyle=d_{i}-\frac{(\ell-i)}{\ell}y_{\ell}\>,\qquad 1\leq i\leq\ell-1$		(4.32)

and then prove that these are not only necessary for the equations above, but also sufficient. The matrix $\boldsymbol{B}-\boldsymbol{A}$ is diagonal and gives a direct relationship between the triangular block in $\boldsymbol{W}^{-1}$ and a corresponding portion of $\boldsymbol{U}$; the other equation gives a recurrence relation for the entries of $\boldsymbol{U}$. Comparison of the final columns of the products gives an explicit formula for the final column of $\boldsymbol{W}^{-1}$ and an explicit formula for the entries of $\boldsymbol{U}$ by comparison of the coefficients of the symbols $y_{k}$; this formula can be seen to verify the recurrence relation found earlier, closing the circle and establishing sufficiency. Both matrices $\boldsymbol{U}$ and $\boldsymbol{W}$ have determinant $\pm 1$: $\lfloor\ell/2\rfloor$ row-permutations brings $\boldsymbol{U}$ to upper triangular form and the determinant $(-1)^{\ell}$ (times $(-1)^{\lfloor\ell/2\rfloor}$) can be read off as the product of the formerly anti-diagonal elements, and similarly for the $[1:\ell-1,1:\ell-1]$ block of $\boldsymbol{W}^{-1}$ which gives a sign $(-1)^{\ell-1+\lfloor(\ell-1)/2\rfloor}$.

As we did for the monomial basis, we start with a scalar version. We compute the Hermite normal form $\boldsymbol{H}$, and the transformation matrix $\boldsymbol{U}$ so that $\boldsymbol{U}\boldsymbol{L}(z)=\boldsymbol{H}$, with Maple to give clues to find the general form. When we do this for the grade $3$ example above we find that the form of $\boldsymbol{U}$ is not helpful, but that the form of $\boldsymbol{U}^{-1}$ and $\boldsymbol{H}$ are:

$$L(z)=\left[\begin{array}[]{ccccc}0&x&x&x&0\\ x&x&0&0&x\\ x&0&x&0&x\\ x&0&0&x&x\\ x&0&0&0&0\end{array}\right]\left[\begin{array}[]{ccccc}1&0&0&0&x\\ 0&1&0&0&x\\ 0&0&1&0&x\\ 0&0&0&1&x\\ 0&0&0&0&x\end{array}\right]\>.$$

(5.36)

As is usual, the generic Hermite normal form contains $p(z)$ in the lower right corner; on specialization of the polynomial coefficients this can change, of course. We return to this point later.

This gives us enough information to conjecture the general form in the theorem below, and a proof follows quickly.