Generalized Inverse

Oct 21, 2020
4 min read
Mar 11, 2022 15:59 UTC
The generalized matrix inverse that applies to any $m \times n$ matrix.
Pragser Wildsee, Italy
Pragser Wildsee, Italy

In the previous chapter we discussed the matrix inverse, which was only meaningful for non-singular matrices. What about matrices that are not square, or don’t have full rank?

Let $\boldsymbol{A}$ be an $m \times n$ matrix with rank $r$. $\boldsymbol{G}: n \times m$ is the generalized inverse of $\boldsymbol{A}$, denoted by $\boldsymbol{A}^{-}$, that satisfies

$$ \boldsymbol{AGA} = \boldsymbol{A} $$

For example,

$$ \boldsymbol{A} = \begin{pmatrix} 1 & 3 & 2 \\ 2 & 6 & 4 \end{pmatrix}, \quad \boldsymbol{G}_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}, \quad \boldsymbol{G}_2 = \begin{pmatrix} -42 & -1 \\ 5 & 3 \\ 2 & 2 \end{pmatrix} $$

Obviously the generalized inverse is not unique. When we talk about generalized inverses, we refer to all the matrices that satisfy $\boldsymbol{AGA} = \boldsymbol{A}$, instead of some specific $\boldsymbol{G}$.

Finding the generalized inverse

We know that for matrix $\boldsymbol{A}$ with rank $r$, it can be decomposed into $\boldsymbol{B}: m \times r$ and $\boldsymbol{T}: r \times n$, both of rank $r$, such that $\boldsymbol{A} = \boldsymbol{BT}$.

Since $\boldsymbol{B}$ has full column rank, it has a left inverse, i.e. there exists $\boldsymbol{L}: r \times m$ such that $\boldsymbol{LB} = \boldsymbol{I}_r$. Since $\boldsymbol{T}$ has full row rank, there exists $\boldsymbol{R}: n \times r$ such that $\boldsymbol{TR} = \boldsymbol{I}_r$.

Let $\boldsymbol{G} = \boldsymbol{RL}$. Now we may check $\boldsymbol{AGA}$:

$$ \boldsymbol{AGA} = (\boldsymbol{BT})\boldsymbol{RL}(\boldsymbol{BT}) = \boldsymbol{BI}_r\boldsymbol{I}_r\boldsymbol{T} = \boldsymbol{BT} = \boldsymbol{A} $$


  • The first property we should know about is its relation to linear systems.

    If $\boldsymbol{Ax} = \boldsymbol{b}$ is consistent, then $\boldsymbol{x} = \boldsymbol{A}^{-}\boldsymbol{b}$ is a solution. Any solution can be written in this form because $\boldsymbol{b} \in \mathcal{C}(\boldsymbol{A})$, and it can be expressed as $\boldsymbol{b} = \boldsymbol{Av}$ for some $\boldsymbol{v}$. Thus,

    $$ \boldsymbol{AA}^{-}\boldsymbol{b} = \boldsymbol{AA}^{-}\boldsymbol{Av} = \boldsymbol{Av} = \boldsymbol{b} $$

  • Left and right inverses are generalized inverses.

    Let $\boldsymbol{A}: m \times n$ have rank $m$. If there exists $\boldsymbol{R}$ such that $\boldsymbol{AR} = \boldsymbol{I}_m$, then $\boldsymbol{ARA} = \boldsymbol{I}_m\boldsymbol{A} = \boldsymbol{A}$. Similarly, we can show the argument for the full column rank case.

  • Similar to the matrix inverse, we have

    • $(-\boldsymbol{A})^{-} = -\boldsymbol{A}^{-}$.
    • $(k\boldsymbol{A})^- = \frac{1}{k}\boldsymbol{A}^-$ where $k \neq 0$.
    • $(\boldsymbol{A}’)^- = (\boldsymbol{A}^-)’$. This can be easily shown by transposing both sides of $\boldsymbol{AGA} = \boldsymbol{A}$.
  • For matrices $\boldsymbol{A}$ and $\boldsymbol{B}$,

    $$ \mathcal{C}(\boldsymbol{B}) \subset \mathcal{C}(\boldsymbol{A}) \Longleftrightarrow \boldsymbol{B} = \boldsymbol{AA}^-\boldsymbol{B} $$

    Showing ($\Leftarrow$) is trivial. For the other direction, we know that $\boldsymbol{B}$ can be written as $\boldsymbol{B} = \boldsymbol{AF}$ for some $\boldsymbol{F}$. Then,

    $$ \boldsymbol{B} = \boldsymbol{AA}^-\boldsymbol{AF} = \boldsymbol{AA}^-\boldsymbol{B} $$

    In terms of vectors, this implicates

    $$ \boldsymbol{x} \in \mathcal{C}(\boldsymbol{A}) \Longleftrightarrow \boldsymbol{x} = \boldsymbol{Ay} \text{ for some } \boldsymbol{y} \Longleftrightarrow \boldsymbol{x} = \boldsymbol{AA}^-\boldsymbol{x} $$

  • A very important property: $\mathcal{C}(\boldsymbol{AA}^-) = \mathcal{C}(\boldsymbol{A})$.

    Showing ($\subset$) is trivial because $\mathcal{C}(\boldsymbol{AB}) \subset \mathcal{C}(\boldsymbol{A})$. For the other direction, see that

    $$ \mathcal{C}(\boldsymbol{A}) = \mathcal{C}(\boldsymbol{AA}^-\boldsymbol{A}) \subset \mathcal{C}(\boldsymbol{AA}^-) $$

  • $r(\boldsymbol{A}^-) \geq r(\boldsymbol{A})$. We know that the rank cannot be gained by matrix multiplication. Multiplying a full-rank matrix is the only way of not losing rank.

    $$ r(\boldsymbol{A}^-) \geq r(\boldsymbol{AA}^-) = r(\boldsymbol{A}) $$

  • If $\mathcal{R}(\boldsymbol{B}) \subset \mathcal{R}(\boldsymbol{A})$, and there exists $\boldsymbol{C}$ such that $\mathcal{C}(\boldsymbol{C}) \subset \mathcal{C}(\boldsymbol{A})$, then $\boldsymbol{BA}^-\boldsymbol{C}$ is invariant to the choice of the generalized inverse.

    $$ \begin{gathered} \mathcal{R}(\boldsymbol{B}) \subset \mathcal{R}(\boldsymbol{A}) \Longleftrightarrow \boldsymbol{B} = \boldsymbol{MA} \text{ for some } M \\ \mathcal{C}(\boldsymbol{C}) \subset \mathcal{C}(\boldsymbol{A}) \Longleftrightarrow \boldsymbol{C} = \boldsymbol{AF} \text{ for some } F \end{gathered} $$

    Then we have

    $$ \boldsymbol{BA}^- \boldsymbol{C} = \boldsymbol{MAA}^- \boldsymbol{AF} = \boldsymbol{MAF}, $$

    which is not dependent on $\boldsymbol{A}^-$. This property is used in the regression setting of $\boldsymbol{y} = \boldsymbol{X\beta} + \boldsymbol{\epsilon}$, where

    $$ \boldsymbol{A} = \boldsymbol{X}’\boldsymbol{X}, \quad \boldsymbol{B} = \boldsymbol{X}, \quad \boldsymbol{C} = \boldsymbol{X}' $$

    See that $\mathcal{R}(\boldsymbol{B}) \subset \mathcal{R}(\boldsymbol{A})$ and $\mathcal{C}(\boldsymbol{C}) \subset \mathcal{C}(\boldsymbol{A})$ (they are actually equal). The predicted values for the response variable are

    $$ \hat{\boldsymbol{y}} =\boldsymbol{X}(\boldsymbol{X}’\boldsymbol{X})^- \boldsymbol{X}’\boldsymbol{y} = \boldsymbol{BA}^- \boldsymbol{Cy} $$

    This means that when $\boldsymbol{X}’\boldsymbol{X}$ is rank-deficient, although there’s infinitely number of choices for the generalized inverse, the result of $\boldsymbol{X}(\boldsymbol{X}’\boldsymbol{X})^- \boldsymbol{X}’$ is invariant.

  • If $\boldsymbol{A}$ has full row rank $m$, then $\boldsymbol{A}’(\boldsymbol{AA}’)^{-1}$ is a right inverse.

    If $r(\boldsymbol{A}) = n$, then $(\boldsymbol{A}’\boldsymbol{A})^{-1}$ is a left inverse.

  • $\boldsymbol{A}’(\boldsymbol{AA}’)^-$ is a generalized inverse of $\boldsymbol{A}$, and so is $(\boldsymbol{A}’\boldsymbol{A})^- \boldsymbol{A}’$.

    To show that a matrix $\boldsymbol{A}$ is the generalized inverse of some other matrix $\boldsymbol{B}$, the only way is to show $\boldsymbol{ABA} = \boldsymbol{A}$. In this case, we need to show $\boldsymbol{A} \boldsymbol{A}’(\boldsymbol{AA}’)^- \boldsymbol{A} = \boldsymbol{A}$.

    Note the fact that $(\boldsymbol{AA}’)^-$ is the generalized inverse of $\boldsymbol{AA}’$, so

    $$ (\boldsymbol{AA}’) (\boldsymbol{AA}’)^- (\boldsymbol{AA}’) = \boldsymbol{AA}' $$

    Recall that $\boldsymbol{BAA}’ = \boldsymbol{CAA}’ \Longleftrightarrow \boldsymbol{BA} = \boldsymbol{CA}$, so if we set $\boldsymbol{B} = \boldsymbol{AA}’(\boldsymbol{AA}’)^-$ and $\boldsymbol{C} = \boldsymbol{I}$, we can use this fact to get the equation that we want to prove.

    This is another method to find the generalized inverse. Let’s say we have

    $$ \boldsymbol{A} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 4 & 2 \\ 0 & -2 & -1 \\ 0 & 3 & 3 \\ 0 & 8 & 4 \end{pmatrix} $$

    and we want to find its generalized inverse. We may instead find $\boldsymbol{A}’\boldsymbol{A}$, and $(\boldsymbol{AA}’)^-\boldsymbol{A}’$ would be a generalized inverse of $\boldsymbol{A}$.

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