This is machine translation

Translated by Microsoft
Mouseover text to see original. Click the button below to return to the English verison of the page.

Note: This page has been translated by MathWorks. Please click here
To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.

Symbolic Matrix Computation

This example shows how to perform simple matrix computations using Symbolic Math Toolbox™.

Generate a possibly familiar test matrix, the 5-by-5 Hilbert matrix.

H = sym(hilb(5)) 
H =

The determinant is very small.

d = det(H) 
d =

The elements of the inverse are integers.

X = inv(H) 
X =

Verify that the inverse is correct.

I = X*H
I =

Find the characteristic polynomial.

syms x; p = charpoly(H,x) 
p =

Try to factor the characteristic polynomial.

ans =

The result indicates that the characteristic polynomial cannot be factored over the rational numbers.

Compute the 50 digit numerical approximations to the eigenvalues.

e = eig(vpa(H)) 
e =

Create a generalized Hilbert matrix involving a free variable, .

t = sym('t'); 
[I,J] = meshgrid(1:5); 
H = 1./(I+J-t)
H =

Substituting retrieves the original Hilbert matrix.

ans =

The reciprocal of the determinant is a polynomial in .

d = 1/det(H) 
d =

d = expand(d)
d =

The elements of the inverse are also polynomials in .

X = inv(H) 
X =

Substituting generates the Hilbert inverse.

X = subs(X,t,'1') 
X =

X = double(X) 
X = 

          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

Investigate a different example.

A = sym(gallery(5)) 
A =

This matrix is "nilpotent". It's fifth power is the zero matrix.

ans =

Because this matrix is nilpotent, its characteristic polynomial is very simple.

p = charpoly(A,'lambda') 
p =

You should now be able to compute the matrix eigenvalues in your head. They are the zeros of the equation lambda^5 = 0.

Symbolic computation can find the eigenvalues exactly.

lambda = eig(A) 
lambda =

Numeric computation involves roundoff error and finds the zeros of an equation that is something like lambda^5 = eps*norm(A) So the computed eigenvalues are roughly lambda = (eps*norm(A))^(1/5) Here are the eigenvalues, computed by the Symbolic Toolbox using 16 digit floating point arithmetic. It is not obvious that they should all be zero.

lambda = eig(vpa(A)) 
lambda =

This matrix is also "defective". It is not similar to a diagonal matrix. Its Jordan Canonical Form is not diagonal.

J = jordan(A) 
J =

The matrix exponential, expm(t*A), is usually expressed in terms of scalar exponentials involving the eigenvalues, exp(lambda(i)*t). But for this matrix, the elements of expm(t*A) are all polynomials in t.

t = sym('t'); 
E = simplify(expm(t*A)) 
E =

By the way, the function "exp" computes element-by-element exponentials.

X = exp(t*A) 
X =

Was this topic helpful?