Matrices Proof

Jul 2015
2
0
Perth
Need some help with the following problem:Screenshot_2015-07-28-16-34-04.jpg

Obviously question 11.

Thanks for all help!
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Hey elladan86.

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
Mar 2011
3,546
1,566
Tejas
To start with:

If $PQ = QR$, then:

$P = PI = PQQ^{-1} = QRQ^{-1}$.
 

romsek

MHF Helper
Nov 2013
6,837
3,079
California
To start with:

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.
 
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Jul 2015
2
0
Perth
Thanks, guys, all helpful. It was trivial, feel a bit embarrassed now!
 

Deveno

MHF Hall of Honor
Mar 2011
3,546
1,566
Tejas
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 $(QAQ^{-1})^k = QA^kQ^{-1}$.