Symbolic Math Toolbox

Linear Algebraic Operations

The following examples show to do several basic linear algebraic operations using the Symbolic Math Toolbox.

The command

• H = hilb(3)
  

generates the 3-by-3 Hilbert matrix. With format short, MATLAB prints

• H =
  1.0000    0.5000    0.3333
  0.5000    0.3333    0.2500
  0.3333    0.2500    0.2000
  

The computed elements of H are floating-point numbers that are the ratios of small integers. Indeed, H is a MATLAB array of class double. Converting H to a symbolic matrix

• H = sym(H)
  

gives

• [  1, 1/2, 1/3]
  [1/2, 1/3, 1/4]
  [1/3, 1/4, 1/5]
  

This allows subsequent symbolic operations on H to produce results that correspond to the infinitely precise Hilbert matrix, sym(hilb(3)), not its floating-point approximation, hilb(3). Therefore,

• inv(H)
  

produces

• [  9,  -36,   30]
  [-36,  192, -180]
  [ 30, -180,  180]
  

and

• det(H)
  

yields

• 1/2160
  

You can use the backslash operator to solve a system of simultaneous linear equations. For example, the commands

• b = [1 1 1]'
  x = H\b    % Solve Hx = b
  

produce the solution

• [  3]
  [-24]
  [ 30]
  

All three of these results, the inverse, the determinant, and the solution to the linear system, are the exact results corresponding to the infinitely precise, rational, Hilbert matrix. On the other hand, using digits(16), the command

• V = vpa(hilb(3))
  

returns

• [               1., .5000000000000000, .3333333333333333]
  [.5000000000000000, .3333333333333333, .2500000000000000]
  [.3333333333333333, .2500000000000000, .2000000000000000]
  

The decimal points in the representation of the individual elements are the signal to use variable-precision arithmetic. The result of each arithmetic operation is rounded to 16 significant decimal digits. When inverting the matrix, these errors are magnified by the matrix condition number, which for hilb(3) is about 500. Consequently,

• inv(V)
  

which returns

• ans =
  [  9.000000000000179, -36.00000000000080,  30.00000000000067]
  [ -36.00000000000080,  192.0000000000042, -180.0000000000040]
  [  30.00000000000067, -180.0000000000040,  180.0000000000038]
  

shows the loss of two digits. So does

• det(V)
  

which gives

• .462962962962953e-3
  

and

• V\b
  

which is

• [ 3.000000000000041]
  [-24.00000000000021]
  [ 30.00000000000019]
  

Since H is nonsingular, calculating the null space of H with the command

• null(H)
  

returns an empty matrix, and calculating the column space of H with

• colspace(H)
  

returns a permutation of the identity matrix. A more interesting example, which the following code shows, is to find a value s for H(1,1) that makes H singular. The commands

• syms s
  H(1,1) = s
  Z = det(H)
  sol = solve(Z)
  

produce

• H =
  [  s, 1/2, 1/3]
  [1/2, 1/3, 1/4]
  [1/3, 1/4, 1/5]
  
  Z =
  1/240*s-1/270
  sol = 
  8/9
  

Then

• H = subs(H,s,sol)
  

substitutes the computed value of sol for s in H to give

• H =
  [8/9, 1/2, 1/3]
  [1/2, 1/3, 1/4]
  [1/3, 1/4, 1/5]
  

Now, the command

• det(H)
  

returns

• ans =
  0
  

and

• inv(H)
  

produces an error message

• ??? error using ==> inv
  Error, (in inverse) singular matrix
  

because H is singular. For this matrix, Z = null(H) and C = colspace(H) are nontrivial:

• Z =
  [   1]
  [  -4]
  [10/3]
  
  C =
  [     1,     0]
  [     0,     1]
  [ -3/10,   6/5]
  

It should be pointed out that even though H is singular, vpa(H) is not. For any integer value d, setting

• digits(d)
  

and then computing

• det(vpa(H))
  inv(vpa(H))
  

results in a determinant of size 10^(-d) and an inverse with elements on the order of 10^d.

Linear Algebra

Eigenvalues