# Linear algebra projections commutativity

• Appleton2
S and T. What is the requirement on those subspaces to have P1P2 = P2P1?In summary, if P1P2 = P2P1 then S is contained in T or T is contained in S.

#### Appleton2

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
"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.

A projection ##P\, : \,V\longrightarrow V## is a surjective linear map such that ##P^2=P##. Hence you can write ##V\cong \operatorname{ker}P \oplus \operatorname{im}P## and ##\left.P\right|_{\operatorname{im}P}=\operatorname{id}_{\operatorname{im}P}.## If ##P=P_1## what ##P_2P_1=P_1P_2## mean for ##P_2##? How are ##\operatorname{ker}P_1## and ##\operatorname{im}P_1## mapped under ##P_2\,?##

@Appleton2 I agree that your textbook's answer is wrong as you stated it.

fresh_42 said:
A projection ##P\, : \,V\longrightarrow V## is a surjective linear map such that ##P^2=P##.
I don't think you mean to write "surjective" here (and I also suspect the OP means orthogonal projection).

Infrared said:
I don't think you mean to write "surjective" here (and I also suspect the OP means orthogonal projection).
I did, but I had the projection space, the image in mind. It is the standard example of a surjection. Of course you have to restrict the mapping onto its image, otherwise the terms injective and surjective don't make a lot of sense in vector spaces of equal finite dimension.

Last edited:
I search on the internet and didn't find such requirement as your textbook answer said. Then I think about this requirment, or we called it formally, this necessary condition, maybe you can view the linear projections on a basis which make the P1 could be described as Jordan normal form, we assume that we are working in the field of complex number C, and then you can use some matrix in block form to decide something about each block in P2, and I think each block in P2 should be like
1 0 0 0 0
4 1 0 0 0
2 4 1 0 0
8 2 4 1 0
5 8 2 4 1
7 5 8 2 4
that the column of matrix, which stand for the image of ei(element of basis), should "moving down from left to right(also I can describe it as increasing i)"
I need to add that I choose the Jordan normal form as "downside", like
1 0 0
1 1 0
0 1 1
not like
1 1 0
0 1 1
0 0 1
but then I didn't find any describing word to avoid involving the basis we choose, I mean I don't think you can state the requirement of the comutation of matrix multiply just using the concept of linear projection (like ker phi, I am phi...)

Thanks for your responses. It took me a while to decipher your notation. (@graphking, I think it will take me another week to decipher your comment, thank you)

I pasted the question from the textbook so I think it is accurately reproduced here. The book is "An introduction to linear algebra" by Gilbert Strang (4th edition). Disappointingly it doesn't seem to explain @fresh_42's notation so I've been on a bit of a Wikipedia odyssey.

If the question was changed to:
"P1 and P2 are orthogonal projections onto subspaces S and T. What is the requirement on those subspaces to have P1P2 = P2P1?"
as I think @Infrared has suggested, would this change the validity of the textbook answer? I suspect not. My comments above would still seem to apply, right?

If I've understood fresh_42 correctly it seems that if S is contained in T, P1 behaves like an identity matrix, and if T is in S, P2 behaves like an identity matrix and so P1P2 = P2P1. Also if S is orthogonal to T then P2 maps onto the nullspace of S and P1P2 = P2P1 = 0. How can I be sure these are the only conditions on S and T?

sorry, I didn't speak english

@graphking your English is fine, it's my limited understanding of maths that's the problem

Appleton2 said:
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