Let $A$ be a real $n\times n $ matrix, and $P_S$ the orthogonal projector into a subspace $S\subset R^n$.

I'm looking for a geometric interpretation of $$P_SA=0$$ in terms of the a relation between $S$ and subspaces corresponding to $A$.

If $A$ is diagonalizable, $\operatorname{col}(A)\perp S$ is necessary and sufficient for $P_SA=0$. Is it possible to formulate a similar necessary and sufficient condition for $P_SA=0$ without assuming diagonalizability? (I can see that $\operatorname{col}(A)\perp S$ is sufficient but not necessary without assuming diagonalizability).

  • 1
    $\begingroup$ It means (equivalent to) that $\mbox{im} A$ is contained in $\ker P_S=S^\perp$. You can indeed state it as $\mbox{col}(A)\perp S$. It does not matter that $A$ be diagonalizable. $\endgroup$ – Julien Aug 3 '13 at 16:00
  • $\begingroup$ how can I see that? $\endgroup$ – mark Aug 3 '13 at 16:05
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    $\begingroup$ The matrix equation $P_SA=0$ means $P_S(Ax)=0$, i.e. $Ax$ belongs to $\ker P_S$ for every vector $x$. That is the range of $A$ is contained in $\ker P_S$. Now by definition of $P_S$, $\ker P_S=S^\perp$. $\endgroup$ – Julien Aug 3 '13 at 16:10

Write $A=[a_1 a_2 .. a_n]$, where $a_i$ are columns.

If $P_s A=0$, then $P_sa_i$=0, and so $col(A) \perp S$.

If $col(A) \perp S$, then $P_s a_i =0$ and then $P_s A =0$.


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