# Matrices Proof

Need some help with the following problem:

Obviously question 11.

Thanks for all help!

#### chiro

MHF Helper

You really need to show some working out. Try getting an expression for QR_invQ_inv by getting R on one side, taking the inverse and then pre and post multiplying by the relevant matrices.

After that you use the hint.

#### Deveno

MHF Hall of Honor

If $PQ = QR$, then:

$P = PI = PQQ^{-1} = QRQ^{-1}$.

#### romsek

MHF Helper

If $PQ = QR$, then:

$P = PI = PQQ^{-1} = QRQ^{-1}$.
this isn't quite what was desired but the problem is still trivial

$(PQ)^{-1}=Q^{-1}P^{-1}=(QR)^{-1}=R^{-1}Q^{-1}$

Multiply both sides on the left by $Q$ to get

$P^{-1}=Q R^{-1}Q^{-1}$

for the rest of it just multiply again by $P^{-1}$ and do some more matrix manipulations, it's not hard.

1 person

Thanks, guys, all helpful. It was trivial, feel a bit embarrassed now!

#### Deveno

MHF Hall of Honor
this isn't quite what was desired but the problem is still trivial

$(PQ)^{-1}=Q^{-1}P^{-1}=(QR)^{-1}=R^{-1}Q^{-1}$

Multiply both sides on the left by $Q$ to get

$P^{-1}=Q R^{-1}Q^{-1}$

for the rest of it just multiply again by $P^{-1}$ and do some more matrix manipulations, it's not hard.
Well call $QR$, by $A$. Using the hint:

$(AQ)^{-1} = (Q^{-1})^{-1}A^{-1} = QA^{-1}$.

Using the hint again:

$A^{-1} = (QR)^{-1} = R^{-1}Q^{-1}$, so: $P^{-1} = QR^{-1}Q^{-1}$.

If one knows that $GL_n(V)$ is a group, one can simply verify that:

$P(QR^{-1}Q^{-1}) = (QRQ^{-1})(QR^{-1}Q^{-1}) = (QR)(Q^{-1}Q)(R^{-1}Q^{-1}) = (QR)(I)(R^{-1}Q^{-1}) = (QR)(R^{-1}Q^{-1}) = Q(RR^{-1})Q^{-1} = QIQ^{-1} = QQ^{-1} = I$.

and conclude $P^{-1} = QRQ^{-1}$ from the uniqueness of inverses.

In fact, it can be shown that $QAQ^{-1}QBQ^{-1} = Q(AB)Q^{-1}$, for any square matrices $A,B$ of the proper size and any invertible $Q$ of matching size. This, in turn, leads to $(QAQ^{-1})^k = QA^kQ^{-1}$.