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I have understood the intuition of projection with an arbitrary basis given by the formula: $P=A(A'A)^{-1}A'$ However when my literature works with projection for example it doesn't use said formula but uses the pseudo inverse i.e. $A^+=(A'A)^{-1}A'$ and in most cases, it is interpreted as the projection matrix. What is the relation?

Hamzalihi
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  • You can create a projection matrix using the pseudoinverse, i.e. $,P=AA^+;$ but the pseudoinverse itself is not a projection matrix. – greg Jun 04 '22 at 13:21
  • If $A\in\operatorname{mat}(\mathbb{C},m, n)$, then $b\mapsto AA^+b$ is the orthogonal projection (in the canonical dot product in $\mathbb{C}^m$) from $\mathbb{C}^m$ onto $A(\mathbb{C}^n)$. See for example this posting. One can sustitute $\mathbb{C}$ for $\mathbb{R}$ of one only works with real numbers. – Mittens Jun 04 '22 at 14:41

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Say the matrix $A:R^n \to R^m$, for each $ y \in R^m$, the pseudo-inverse gives a $x \in R^n$ satisfies:
(i) $A$ mapping $x$ to the $y'$:the least square of $y$, i.e., the projection of $y$ onto the range of $A$ (the column space of $A$),
(ii) $x$ is the minimal norm solution of the equation: $Ax=y'$.

Note that in (i), the originoal $y$ may not be in the column space of $A$, i.e., $A$ may have no inverse, this is why $A^+$ has the name of pseudo-inverse.
And in (ii), $A$ may not be 1-1, so we choose the minimal solution $x $, i.e., the unique solution on row space of $A$.

Finally, for a subspace $W \subseteq R^m$ , with dim($W$)=n, and the matrix $A$ whose colummn vectors are a fixed basis of $W$, then for each $y \in R^m$, the pseudo-inverse of $A$ gives the $x$ so that $Ax=y'$, where $y'$ is the projection of $y$ onto the range of $A$, i.e., the subspace $W$, hence the projection matrix is the pseudo-inverse product $A$ from left.

claudy
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