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- Homework Statement
- P1 and P2 are projections onto subspaces S and T. What is the requirement on those subspaces to have P1P2 = P2P1?

- Relevant Equations
- None

Textbook answer:

"If P1P2 = P2P1 then S is contained in T or T is contained in S."

My query:

If P1 = \begin{pmatrix}

1 & 0 & 0 \\

0 & 0 & 0 \\

0 & 0 & 0 \\

\end{pmatrix}

and P2 =\begin{pmatrix}

0 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & 0 \\

\end{pmatrix}

as far as I can tell this would make P1 a projection onto S, the subspace that is the X axis in R3, and P2 would be a projection onto T, the subspace that is the Y axis in R3.

So in this case P1P2 = P2P1 = the zero matrix

But S is not contained in T and T is not contained in S

I've seen other posts online about this question that don't refer to this special case so I assume I have misunderstood something. Could someone please tell me the source of the apparent inconsistency.

"If P1P2 = P2P1 then S is contained in T or T is contained in S."

My query:

If P1 = \begin{pmatrix}

1 & 0 & 0 \\

0 & 0 & 0 \\

0 & 0 & 0 \\

\end{pmatrix}

and P2 =\begin{pmatrix}

0 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & 0 \\

\end{pmatrix}

as far as I can tell this would make P1 a projection onto S, the subspace that is the X axis in R3, and P2 would be a projection onto T, the subspace that is the Y axis in R3.

So in this case P1P2 = P2P1 = the zero matrix

But S is not contained in T and T is not contained in S

I've seen other posts online about this question that don't refer to this special case so I assume I have misunderstood something. Could someone please tell me the source of the apparent inconsistency.