Find the pseudoinverse of a machine-precision matrix:

Pseudoinverse of a complex matrix:

Pseudoinverse of an exact matrix:

Pseudoinverse of an arbitrary-precision matrix:

Compute a symbolic pseudoinverse:

The inversion of large machine-precision matrices is efficient:

Output formatting:

The pseudoinverse of a sparse matrix is returned as a normal matrix:

Format the result:

When possible, the pseudoinverse of a structured matrix is returned as another structured matrix:

This is not always possible:

IdentityMatrix [n] is its own pseudoinverse:

The pseudoinverse of IdentityMatrix [{m,n}] is a transposition:

Compute the pseudoinverse for HilbertMatrix :

m is a 16×16 Hilbert matrix:

Some singular values are below the default tolerance for machine precision:

Compute the pseudoinverse with the default tolerance:

It is not a true inverse since some singular values were considered to be effectively zero:

Compute the pseudoinverse with no tolerance:

Even though no singular values were considered zero, it is worse due to numerical error:

Solve the following system of equations using PseudoInverse :

Rewrite the system in matrix form:

The general solution is given by for an arbitrary vector :

Since the TemplateBox[{i}, CTraditional] dropped out, the solution is unique, as can be verified using SolveValues :

Find all solutions of the following system of equations:

First, write the coefficient matrix , vector variable and constant vector :

Verify the rewrite:

The general solution is given by for an arbitrary vector :

Verify the solution:

Although there are three parameters, this solution represents a line:

This is because the null space of is one-dimensional:

Hence it is possible to reparameterize to eliminate two of the parameters:

This parameterization gives the answer in the same form as SolveValues :

Find the minimum Frobenius-norm solution to , with and as follows:

The minimum norm solution is :

Compute the Frobenius norm of :

As in the vector case, the general solution is given by , with now an arbitrary matrix:

The minimum occurs when all the TemplateBox[{TemplateBox[{i, j}, RowDefault]}, CTraditional] are zero, confirming that is the minimum-norm solution:

In this case there is no solution to :

An approximate solution that minimizes the norm of is given by :

Compare to general minimization:

A more general solution is given by :

All these vectors minimize TemplateBox[{{{m, ., x}, -, b}}, Norm]:

Although there are three parameters in , it represents a line:

This is because the null space of is one-dimensional:

For the matrix and vector that follow, find a vector that minimizes TemplateBox[{{{m, ., x}, -, b}}, Norm]:

One solution, in this case unique, is given by :

This result could also have been obtained using LeastSquares [m,b]:

Confirm the answer using Minimize :

For the matrices and that follow, find a matrix that minimizes TemplateBox[{{{m, ., x}, -, b}}, Norm]_F:

One solution, in this case unique, is given by :

This result could also have been obtained using LeastSquares [m,b]:

Confirm the answer using Minimize :

PseudoInverse can be used to find a best-fit curve to data. Consider the following data:

Extract the and coordinates from the data:

Construct a design matrix, whose columns are and , for fitting to a line :

Get the coefficients and for a linear least‐squares fit:

Verify the coefficients using Fit :

Plot the best-fit curve along with the data:

Find the best-fit parabola to the following data:

Extract the and coordinates from the data:

Construct a design matrix, whose columns are , and , for fitting to a line :

Get the coefficients , and for a least‐squares fit:

Verify the coefficients using Fit :

Plot the best-fit curve along with the data: