Geometry

[Pearce_09]

Hello again,
This question confuses me for a reason. I read the questions and they sound to simple and to easy to answer. So maybe its something im reading wrong and not answering. Help would be greatly apreciated.

first off
Let O(n) = { A | A is an n x n matrix with A^t A = I } be the set of n by n orthogonal matrices. Show that

a) I "is in" O(n)
b) If A, B "is in" O(n), then AB "is in" O(n) and that
c) If A "is in" O(n), then A^-1 "is in" O(n)

now a) just seems so simple i just dont know how to answer somthing like that

and for b) i have
-- if A,B "is in" O(n)

AA^t = I
BB^t = I

if AA^t = I , and BB^t = I then,
AA^t = BB^t

-- show AB "is in" O(n)

AB(AB)^t = I
ABB^tA^t = I
since BB^t = AA^t
AA^tAA^t = I
therefore since AA^t = I then AA^tAA^t = I and
therefore AB = I

now does this last statement change the process of the question

(In other words, this problem asks you to show that using the operation of matrix multiplication, O(n) is a group.)
does this statement change the way i should aproach a)b)c)
thanks for you time
regards,
adam

[AKG]

Your proof for part ii) does too much:

(AB)t(AB) = BtAtAB = Bt(AtA)B = BtIB = BtB = I

You should know that in general, if X and Y are square matrices, and XY = I, then YX = I. To prove that A-1 is in O(n), use the fact that A-1 = At and the stuff in the previous sentence. In general, to prove that an n x n matrix X is in O(n) you need to prove that XtX = I.

[Pearce_09]

hello AKG,
thanks for the help, well for everything.
There is just one thing, part a). Isnt it completely obvious that I "is in" the set of orthogonal matrices. I just cant wrap my mind around proving somthing so simple.
thanks again
adam

[AKG]

Check the last sentence of my previous post for how to prove I is in O(n).

[Pearce_09]

oh yes, thats directed to a).. i see now. thanks again