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PINV Moore-Penrose Pseudoinverse

Section: Array Generation and Manipulations

# Usage

Calculates the Moore-Penrose pseudoinverse of a matrix. The general syntax for its use is

       y = pinv(A,tol)

or for a default specification of the tolerance `tol`,

       y = pinv(A)

For any `m x n` matrix `A`, the Moore-Penrose pseudoinverse is the unique `n x m` matrix `B` that satisfies the following four conditions

- `A B A = A`
- `B A B = B`
- `(A B)' = A B`
- `(B A)' = B A`

Also, it is true that `B y` is the minimum norm, least squares solution to `A x = y`. The Moore-Penrose pseudoinverse is computed from the singular value decomposition of `A`, with singular values smaller than `tol` being treated as zeros. If `tol` is not specified then it is chosen as

      tol = max(size(A)) * norm(A) * teps(A).

# Internals

The calculation of the MP pseudo-inverse is almost trivial once the svd of the matrix is available. First, for a real, diagonal matrix with positive entries, the pseudo-inverse is simply

One can quickly verify that this choice of matrix satisfies the four properties of the pseudoinverse. Then, the pseudoinverse of a general matrix `A = U S V'` is defined as

and again, using the facts that `U' U = I` and `V V' = I`, one can quickly verify that this choice of pseudoinverse satisfies the four defining properties of the MP pseudoinverse. Note that in practice, the diagonal pseudoinverse `S^{+}` is computed with a threshold (the `tol` argument to `pinv`) so that singular values smaller than `tol` are treated like zeros.

# Examples

Consider a simple `1 x 2` matrix example, and note the various Moore-Penrose conditions:

    --> A = float(rand(1,2))

    A =
        0.4840    0.0187

    --> B = pinv(A)

    B =
        2.0630
        0.0796

    --> A*B*A

    ans =
        0.4840    0.0187

    --> B*A*B

    ans =
        2.0630
        0.0796

    --> A*B

    ans =
        1.0000

    --> B*A

    ans =
        0.9985    0.0385
        0.0385    0.0015

To demonstrate that `pinv` returns the least squares solution, consider the following very simple case

    --> A = float([1;1;1;1])

    A =
     1
     1
     1
     1

The least squares solution to `A x = b` is just `x = mean(b)`, and computing the `pinv` of `A` demonstrates this

    --> pinv(A)

    ans =
        0.2500    0.2500    0.2500    0.2500

Similarly, we can demonstrate the minimum norm solution with the following simple case

    --> A = float([1,1])

    A =
     1 1

The solutions of `A x = 5` are those `x_1` and `x_2` such that `x_1 + x_2 = 5`. The norm of `x` is `x_1^ + x_2^2`, which is `x_1^2 + (5-x_1)^2`, which is minimized for `x_1 = x_2 = 2.5`:

    --> pinv(A) * 5.0

    ans =
        2.5000
        2.5000