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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 = 

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