For the $m * n$ matrix $A$ the matrix $A^{+}$ is called its Moore-Penrose pseudo inverse, if for $A^{+}$ the Moore-Penrose conditions hold. The Moore-Penrose conditions are the following:
- $AA^{+}A=A$
- $A^{+}AA^{+}=A^{+}$
- $(AA^{+})^{T}=AA^{+}$
- $(A^{+}A)^{T}=A^{+}A$
A nice property of Moore-Penrose pseudo inverses is that they are unique. I was able to prove uniqueness by only making use of the first and third condition as follows.
Assume that $A_{1}^{+}$ and $A_{2}^{+}$ are both Moore-Penrose pseudo inverses for the matrix A. Then we have: $(AA_{1}^{+})^{T}\frac{cnd.3}{=}AA_{1}^{+} \frac{cnd.1}{=}AA_{2}^{+}AA_{1}^{+} \frac{cnd.3}{=} (AA_{2}^{+})^{T}(AA_{1}^{+})^{T} = (AA_{1}^{+}AA_{2}^{+})^{T} \rightarrow AA_{1}^{+} = AA_{1}^{+}AA_{2}^{+} \frac{cnd.1}{=}AA_{2}^{+} \rightarrow AA_{1}^{+}=AA_{2}^{+} \rightarrow A_{1}^{+}=A_{2}^{+}$
Because in this proof only the first and third condition are used, it made me think that the Moore-Penrose conditions could be reduced to just the first and third one, while no loss of properties of a Moore-Penrose pseudo inverse will occur. But that would have to mean that condition number 2 and 4 somehow follow from number 1 and 3. For the second one it is easy to see that it follows from the first condition ($AA^{+}A\frac{cnd.1}{=}A \rightarrow A^{+}AA^{+}A\frac{cnd.1}{=}A^{+}A \rightarrow A^{+}AA^{+}=A^{+}$). For the fourth one I hadn't figured it out yet.
What do you guys think? Could these conditions be compressed, or am I making errors in my thought process?