Transpose rules and algebra

Question

A, B and P are all squared matrices of the same order ($nXn$).

It is given that:

$PP^{T}=I$

$B^{T}B=I$

$A=P^{T}BP$

Which of the following is correct:

$(1) A=B$

$(2) AA^{T}=I$

$(3) (PB)^{-1}=(PB)^{T}$

I have started with (2). This is what I've done:

$(P^{T}BP)(P^{T}BP)$

$P^{T}BB^{T}P$

Something is not right. Can I say that

$B^{T}B=BB^{T}$ ?

According to the book, (2) and (3) are correct while (1) is not. I can't figure out why.

Answer 1

Hint: If $U$ and $V$ are square matrices (of the same size) , then

$UV=I \iff VU=I.$

Answer 2

To prove that (1) is false I will write a counterexample in terms of linear transformations instead of matrices. Let $P(x_1,x_2,\cdots,x_n)=(x_1,0,\cdots,0)$, $A=0$ and $B=I-P$. Then $A \neq B$ but $A=P^{T}BP$.

(2) follows from $AA^{T}=(P^{T}BP)(P^{T}BP)^{T}=I$ because $(MN)^{T}=N^{T}M^{T}$. [Observe that $P^{T}P=I$ and $BB^{T}=I$]. To prove (3) it is enough to show that $(PB)(PB)^{T}=I$ which is true since LHS $=B^{T}P^{T}PB=B^{T}B=I$.