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Question 26683: Verify that thesetof orthogonal nxn matrices form a subgroup of the general linear group GLn(R).
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! ORTHOGONAL MATRICES ARE SQUARE MATRICES SUCH THAT PRODUCT OF AN ORTHOGONAL MATRIX WITH ITS TRANSPOSE IS AN IDENTITY MATRIX.THAT IS IF
A*A'=A'*A=I,THEN A IS AN ORTHOGONAL MATRIX.
TO PROVE THAT THEY FORM A SUB GROUP UNDER THE GENERAL LINEAR GROUP OF MATRICES,WE NEED TO SHOW THAT
I.ORTHOGONAL MATRICES ARE A SUB SET OF THE GENERAL GROUP OF MATRICES.
II.UNDER THE SAME COMPOSITION AS IN THE GENERAL GROUP,THE SUB GROUP ITSELF IS A GROUP.HERE WE CONSIDER THE COMPOSITION OF MULTIPLICATION.THIS MEANS THAT WE HAVE TO SHOW WITHIN THE SUBGROUP FOR MULTIPLICATION......
1.CLOSOURE
2.ASSOCIATIVITY
3.EXISTENCE OF IDENTITY
4.EXISTENCE OF INVERSE
NOW I CRITERIA IS OBVIOUS AS THE SET OF ORTHOGONAL MATRICES (CALLED S) AS DEFINED ARE INDEED A SUB SET OF GENERAL LINEAR GROUP OF MATRICES(CALLED G).
II....MULTIPLICATION IS INDEED WELL DEFINED AND EXISTS FOR ORTHOGONAL MATRICES AS IT IS FOR THE GENERAL GROUP OF MATRICES.
1 AND 2.ASSOCIATIVITY HOLDS FOR MATRIX MULTIPLCATION IN GENERAL SO WE SHALL PROVE CLOSOURE HERE.
LET A AND B BE ORTHOGONAL MATRICES IN THE SUB SET OF ORTHOGONAL MATRICES S .SINCE A' AND B' ARE ALSO ORTHOGONAL MATRICES,THEY ARE ALSO ELEMENTS OF S.
SO A*A'=A'*A=I...AND...B*B'=B'*B=I...
NOW..
IF WE SHOW THAT A*B IS ALSO AN ORTHOGONAL MATIX THEN IT WILL BE AN ELEMENT OF S WHICH PROVES CLOSOURE......
TST (A*B)*(A*B)'=I
WE KNOW FOR MATRIX MULTIPLICATION THAT (A*B)'=(B'*A')...HENCE
(A*B)*(A*B)'=(A*B)*(B'*A')=A*(B*B')*A'=A*I*A'=A*A'=AA'=I
HENCE A*B IS ALSO AN ORTHOGONAL MATRIX.HENCE IT IS AN ELEMENT OF S.
3. IDENTITY ELEMENT IN S IS THE UNIT MATRIX WHICH IS SAME AS IN G.SINCE,WE HAVE
A*I=I*A=A...WHICH FOLLOWS OBVIOUSLY FROM MULTIPLICATIVE PROPERTY.
4.WE HAVE A*A'=A'*A=I.....IF A IS AN ORTHOGONAL MATRIX ITS INVERSE A' IS ALSO AN ORTHOGONAL MATRIX. HENCE IT IS AN ELEMENT OF S.
THUS S IS A SUB GROUP OF G.
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