Editing Inverse element (section)

==== Generalized inverses of matrices <span class="anchor" id="matrices"></span><!-- [[Generalized inverse]] links here.  Please do not change. --> ====
A [[square matrix]] <math>M</math> with entries in a [[field (mathematics)|field]] <math>K</math> is invertible (in the set of all square matrices of the same size, under [[matrix multiplication]]) if and only if its [[determinant]] is different from zero. If the determinant of <math>M</math> is zero, it is impossible for it to have a one-sided inverse; therefore a left inverse or right inverse implies the existence of the other one. See [[invertible matrix]] for more.

More generally, a square matrix over a [[commutative ring]] <math>R</math> is invertible [[if and only if]] its determinant is invertible in <math>R</math>.

Non-square matrices of [[full rank]] have several one-sided inverses:<ref>{{cite web| url = http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/detail/lecture33.htm| title = MIT Professor Gilbert Strang Linear Algebra Lecture #33 – Left and Right Inverses; Pseudoinverse.}}</ref>
* For <math>A:m\times n \mid m>n</math> we have left inverses; for example, <math>\underbrace{ \left(A^\text{T}A\right)^{-1} A^\text{T} }_{ A^{-1}_\text{left} } A = I_n</math>
* For <math>A:m\times n \mid m<n</math> we have right inverses; for example, <math>A \underbrace{ A^\text{T}\left(AA^\text{T}\right)^{-1} }_{ A^{-1}_\text{right} } = I_m</math>

The left inverse can be used to determine the least norm solution of <math>Ax = b</math>, which is also the [[least squares]] formula for [[regression analysis|regression]] and is given by <math>x = \left(A^\text{T}A\right)^{-1}A^\text{T}b.</math>

No [[rank deficient]] matrix has any (even one-sided) inverse.  However, the Moore–Penrose inverse exists for all matrices, and coincides with the left or right (or true) inverse when it exists.

As an example of matrix inverses, consider:

: <math>A:2 \times 3 =
  \begin{bmatrix}
    1 & 2 & 3 \\
    4 & 5 & 6
  \end{bmatrix}
</math>

So, as ''m'' < ''n'', we have a right inverse, <math>A^{-1}_\text{right} = A^\text{T} \left(AA^\text{T}\right)^{-1}.</math> By components it is computed as

: <math>\begin{align}
  AA^\text{T} &= \begin{bmatrix}
              1 & 2 & 3 \\
              4 & 5 & 6
            \end{bmatrix}
            \begin{bmatrix}
              1 & 4\\
              2 & 5\\
              3 & 6
            \end{bmatrix} =  
            \begin{bmatrix}
              14 & 32\\
              32 & 77
            \end{bmatrix} \\[3pt]
  \left(AA^\text{T}\right)^{-1} &= \begin{bmatrix}
       14 & 32\\
       32 & 77
\end{bmatrix}^{-1} = \frac{1}{54}
                   \begin{bmatrix}
                      77 & -32\\
                     -32 & 14
                   \end{bmatrix} \\[3pt]
  A^\text{T}\left(AA^\text{T}\right)^{-1} &= \frac{1}{54}
                   \begin{bmatrix}
                     1 & 4\\
                     2 & 5\\
                     3 & 6
                   \end{bmatrix}
                   \begin{bmatrix}
                      77 & -32\\
                     -32 & 14
                   \end{bmatrix} = \frac{1}{18}
                   \begin{bmatrix}
                     -17 & 8\\
                      -2 & 2\\
                      13 & -4
                   \end{bmatrix} = A^{-1}_\text{right}
\end{align}</math>

The left inverse doesn't exist, because

: <math>
  A^\text{T}A = \begin{bmatrix}
             1 & 4\\
             2 & 5\\
             3 & 6
           \end{bmatrix}
           \begin{bmatrix}
             1 & 2 & 3 \\
             4 & 5 & 6
           \end{bmatrix} =
           \begin{bmatrix}
             17 & 22 & 27 \\
             22 & 29 & 36\\
             27 & 36 & 45
           \end{bmatrix}
</math>

which is a [[singular matrix]], and cannot be inverted.