Question

If A is an idempotent matrix, A^2 = A. Let P and Q be nxn idempotent matrices.

a) Show that P transpose is idempotent.
b) Is P+Q idempotent? Justify your answer.
c) Show that PQ is idempotent if PQ=QP


For part (a) I start with P^2 = PP = P, and (PP)^t = P^t, but that doesn't seem like sufficient proof.

Part B I said that if P+Q is idempotent, (P+Q)^2 = P + Q, but (P+Q)^

Part (c) is similar, I have (PQ)^2 = PQPQ then by associative property we have P(QP)Q. I was thinking to say that if PQ/=/QP, couldn't have commuted PPQQ which is P^2*Q^2=PQ which doesn't seem right either.

Answer 1

Finishing part B, but (P+Q)^2 = P^2 + 2PQ + Q^2, which is only idempotent for the trivial case in which PQ is the 0 matrix, so P+Q can only trivially be idempotent. I am not confident in this argument, especially in the polynomial expansion

Answer 2

For a) \(P= PP\\P^T= (PP)^T=P^TP^T=((P^T))^2\)

Answer 3

For b) YOu can't do that since matrices are not commutative in multiplication.
Suppose \(P+ Q\) is idempotent, then \((P+Q)=(P+Q)^2=(P+Q)(P+Q)=P^2+PQ+QP+Q^2\)
and no way to get \(P^2+2PQ+Q^2\)

Answer 4

And \(P+Q\neq P^2+PQ+QP+Q^2\), why? because
\(P=P^2\\Q=Q^2\\P+Q= P^2+Q^2 \)
only

Answer 5

for c) if PQ = QP , \((PQ)^2=(PQ)(PQ)=PQQP=PQ=PPQQ=P^2Q^2\)
and you can see \(P=P^2\\Q=Q^2\\PQ=P^2Q^2\)
that shows PQ is idempotent if PQ=QP

Answer 6

The course is called "theoretical linear algebra" , right? hehehe... good luck

Answer 7

Yeah! This is the first time we've seen idempotence too. Thanks a lot! Really helpful.

Answer 8

Sidenote though - for (c), \[PQQP = PQ \]

Why is this the case? Because PQ is idempotent?