---- > [!definition] Definition. ([[Moore-Penrose pseudoinverse]]) > Let $A \in \mathbb{F}^{M \times N}$ be arbitrary, where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$. The **Moore-Penrose Pseudoinverse** of $A$, denoted $A^{+}$ is the (unique) matrix following the four properties below: > 1. $(A A^{+} \text{ preserves } A )$ $AA ^{+}A=A$; ([[generalized matrix inverse|see]]) > 2. $(A^{ +} \text{ acts like a weak inverse})$ $A^{+}A A^{+}=A ^{+}$; > 3. $A A^{+}$ is [[conjugate symmetric|Hermitian]]; > 4. $A^{+}A$ is also [[conjugate symmetric|Hermitian]]. > \ > In general, $A^{+}=V_{r}\Sigma_{r}^{-1}U_{r}'=V \Sigma^{+}U'$ > where $V_{r}$, $\Sigma_{r}^{-1}$, and $U_{r}$ are obtained from the [[compact svd]] of $A$ and $V, \Sigma, U$ are obtained from the [[Singular Value Decomposition of a Matrix|full svd]] of $A$. And $\Sigma^{+}$ is characterized [[diagonal matrix pseudoinverse|here]]. > [!basicproperties] > - $A^{+}$ is unique; > - If $A$ is [[inverse matrix|invertible]], then $A^{+}=A^{^{-1}}$ (specialization) > - $A^{+}=A'(AA ')^{+}$ > - hence if $A$ has [[linearly independent]] columns, $A^{+}=(A'A)^{-1}A'$ (compare to [[normal equations]] when [[least squares]] has unique solution) > - In this case, $A^{+}$ is a *left inverse* of $A$ because $A^{+}A=I_{N}$ > - $A^{+}=A'(AA ')^{+}$ > - hence if $A$ has [[linearly independent]] rows, $A^{+}=A'(AA')^{-1}$ (compare to [[normal equations]] when [[least squares]] has unique solution) > - In this case, $A^{+}$ is a *right inverse* because $AA^{+}=I_{N}$ > - $(A')^{+}=(A^{+})'$ > - [[Moore-Penrose pseudoinverse and matrix products]] > [!justification] Closed Formula Derivation. > **Compact SVD Formulation.** Using the corollary in [[Moore-Penrose pseudoinverse and matrix products]] and the [[compact svd]], we have $\begin{align} A=U_{r}\Sigma_{r}V_{r}' \implies A^{+} = & (U_{r}\Sigma_{r}V_{r}')^{+} \\ = & \big(U_{r} (\Sigma_{r} V_{r}) \big)^{+} \\ = & (\Sigma_{r}V_{r})^{+} U_{r}^{+} \\ = & V_{r}^{+} \Sigma_{r}^{+} U_{r}^{+} \\ = & V_{r}' \Sigma_{r}^{-1} U_{r}^{'}. \end{align}$ \ **Full SVD Formulation.** Write $\begin{align} A=U\Sigma V' \implies & A^{+}=(U\Sigma V')^{+}=V \Sigma^{+} U'. \end{align}$ ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```