Linear Algebra

Pseudo Inverse Matrix

If the columns of a matrix A are linearly independent, so  AT· A  is invertible and we obtain with the following formula the pseudo inverse:

A+ = (AT · A)-1 · AT

Here  A+  is a left inverse of  A , what means:  A+· A = E .

However, if the rows of the matrix are linearly independent, we obtain the pseudo inverse with the formula:

A+ = AT· (A · A T) -1

This is a right inverse of  A , what means:  A · A+ = E .

If both the columns and the rows of the matrix are linearly independent, then the matrix is invertible and the pseudo inverse is equal to the inverse of the matrix.

Example:

Matrix A
¯¯¯¯¯¯¯¯
  ⎧ 1  1  1  1 ⎫
  ⎩ 5  7  7  9 ⎭

AT· A
¯¯¯¯¯
  ⎧ 26  36  36  46 ⎫
  ⎪ 36  50  50  64 ⎪
  ⎪ 36  50  50  64 ⎪
  ⎩ 46  64  64  82 ⎭

AT· A is not invertible

A · AT
¯¯¯¯¯¯
  ⎧  4   28 ⎫
  ⎩ 28  204 ⎭

( A · AT )-1
¯¯¯¯¯¯¯¯¯¯¯¯
  ⎧ 6,375 -0,875 ⎫
  ⎩-0,875  0,125 ⎭

Right Inverse:  AT·( A·AT )-1
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
  ⎧    2 -0,25 ⎫
  ⎪ 0,25     0 ⎪
  ⎪ 0,25     0 ⎪
  ⎩ -1,5  0,25 ⎭

Proof by multiplication:

1. Matrix  ( A )
¯¯¯¯¯¯¯¯¯
  ⎧ 1  1  1  1 ⎫
  ⎩ 5  7  7  9 ⎭

2. Matrix  ( A+ )
¯¯¯¯¯¯¯¯¯
  ⎧    2 -0,25 ⎫
  ⎪ 0,25     0 ⎪
  ⎪ 0,25     0 ⎪
  ⎩ -1,5  0,25 ⎭

Produktmatrix ( A·A+)
¯¯¯¯¯¯¯¯¯¯¯¯¯
  ⎧ 1  0 ⎫
  ⎩ 0  1 ⎭

Pop-up Menu:

Right click to open a local menu, which offers you the following functions to manage the matrix.

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