hermiteForm

Find the Hermite form of an inverse Hilbert matrix.

A = sym(invhilb(5))

A = 
$$\left(\begin{array}{ccccc}25& -300& 1050& -1400& 630\\ -300& 4800& -18900& 26880& -12600\\ 1050& -18900& 79380& -117600& 56700\\ -1400& 26880& -117600& 179200& -88200\\ 630& -12600& 56700& -88200& 44100\end{array}\right)$$

H = hermiteForm(A)

H = 
$$\left(\begin{array}{ccccc}5& 0& -210& -280& 630\\ 0& 60& 0& 0& 0\\ 0& 0& 420& 0& 0\\ 0& 0& 0& 840& 0\\ 0& 0& 0& 0& 2520\end{array}\right)$$

Create a 2-by-2 matrix, the elements of which are polynomials in the variable x.

syms x
A = [x^2 + 3, (2*x - 1)^2; (x + 2)^2, 3*x^2 + 5]

A = 
$$\left(\begin{array}{cc}{x}^2+3& {\left(2\hspace{0.17em}x-1\right)}^2\\ {\left(x+2\right)}^2& 3\hspace{0.17em}{x}^2+5\end{array}\right)$$

Find the Hermite form of this matrix.

H = hermiteForm(A)

H = 
$$\left(\begin{array}{cc}1& \frac{4\hspace{0.17em}{x}^3}{49}+\frac{47\hspace{0.17em}{x}^2}{49}-\frac{76\hspace{0.17em}x}{49}+\frac{20}{49}\\ 0& x^4+12\hspace{0.17em}{x}^3-13\hspace{0.17em}{x}^2-12\hspace{0.17em}x-11\end{array}\right)$$

Create a 2-by-2 matrix that contains two variables: x and y.

syms x y
A = [2/x + y, x^2 - y^2; 3*sin(x) + y, x]

A = 
$$\left(\begin{array}{cc}y+\frac{2}{x}& x^2-y^2\\ y+3\hspace{0.17em}\sin \left(x\right)& x\end{array}\right)$$

Find the Hermite form of this matrix. If you do not specify the polynomial variable, hermiteForm uses symvar(A,1) and thus determines that the polynomial variable is x. Because 3*sin(x) + y is not a polynomial in x, hermiteForm(A) throws an error.

Find the Hermite form of A specifying that all elements of A are polynomials in the variable y.

H = hermiteForm(A,y)

H = 
$$\left(\begin{array}{cc}1& \frac{x\hspace{0.17em}{y}^2}{3\hspace{0.17em}x\hspace{0.17em}\sin \left(x\right)-2}+\frac{x\hspace{0.17em}\left(x-x^2\right)}{3\hspace{0.17em}x\hspace{0.17em}\sin \left(x\right)-2}\\ 0& 3\hspace{0.17em}{y}^2\hspace{0.17em}\sin \left(x\right)-3\hspace{0.17em}{x}^2\hspace{0.17em}\sin \left(x\right)+y^3+y\hspace{0.17em}\left(x-x^2\right)+2\end{array}\right)$$

Find the Hermite form and the corresponding transformation matrix for an inverse Hilbert matrix.

A = sym(invhilb(3));
[U,H] = hermiteForm(A)

U = 
$$\left(\begin{array}{ccc}13& 9& 7\\ 6& 4& 3\\ 20& 15& 12\end{array}\right)$$

H = 
$$\left(\begin{array}{ccc}3& 0& 30\\ 0& 12& 0\\ 0& 0& 60\end{array}\right)$$

Verify that H = U*A.

isAlways(H == U*A)

ans = 3×3 logical array

   1   1   1
   1   1   1
   1   1   1

Find the Hermite form and the corresponding transformation matrix for a matrix of polynomials.

syms x y
A = [2*(x - y), 3*(x^2 - y^2);
    4*(x^3 - y^3), 5*(x^4 - y^4)];
[U,H] = hermiteForm(A,x)

U = 
$$\left(\begin{array}{cc}\frac{1}{2}& 0\\ 2\hspace{0.17em}{x}^2+2\hspace{0.17em}x\hspace{0.17em}y+2\hspace{0.17em}{y}^2& -1\end{array}\right)$$

H = 
$$\left(\begin{array}{cc}x-y& \frac{3\hspace{0.17em}{x}^2}{2}-\frac{3\hspace{0.17em}{y}^2}{2}\\ 0& x^4+6\hspace{0.17em}{x}^3\hspace{0.17em}y-6\hspace{0.17em}x\hspace{0.17em}{y}^3-y^4\end{array}\right)$$

Verify that H = U*A.

isAlways(H == U*A)

ans = 2×2 logical array

   1   1
   1   1

If a matrix does not contain a particular variable, and you call hermiteForm specifying that variable as the second argument, then the result differs from what you get without specifying that variable. For example, create a matrix that does not contain any variables.

A = [9 -36 30; -36 192 -180; 30 -180 180]

A = 3×3

     9   -36    30
   -36   192  -180
    30  -180   180

Call hermiteForm specifying variable x as the second argument. In this case, hermiteForm assumes that the elements of A are univariate polynomials in x.

syms x
hermiteForm(A,x)

ans = 3×3

     1     0     0
     0     1     0
     0     0     1

Call hermiteForm without specifying variables. In this case, hermiteForm treats A as a matrix of integers.

hermiteForm(A)

ans = 3×3

     3     0    30
     0    12     0
     0     0    60