Proving O(n) orthogonal matrices form a group

[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 I am 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 don't 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 approach a)b)c)
thanks for you time
regards,
adam

 

[AKG]

Your proof for part ii) does too much:

(AB)^t(AB) = B^tA^tAB = B^t(A^tA)B = B^tIB = B^tB = 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 = A^t 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 X^tX = 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 can't 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, that's directed to a).. i see now. thanks again