Linear Algebra -- Orthogonality?

They are two questions in one





a)Find an orthogonal basis for R3 one of whose vectors is


c = {2,1,-2}





b)Find an orthogonal matrix having a multiple of the vector c above as one of its columns.





Thank you for the help in advance!!

Linear Algebra -- Orthogonality?
Finding the second vector is easy: We can take any two integers we want as the first two components of our vector, and then the third can be found as the solution of a simple algebraic equation. Let's consider something of the form





(2,2,c)





to make the algebra easy, giving us





(2,2,c)* (2,1,-2) = 0





so: 2*2 + 2*1 + c*(-2) =0





4 + 2 - 2c = 0


6 - 2c = 0


6 = 2c


c = 3





(2,2,3) is one of the infinite number of vectors which are orthogonal to (2,1,-2). Finding the next such vector is almost as easy: we just take the cross product of the first two vectors, which in three space can be calculated by cranking out a simple determinant:





http://www.math.oregonstate.edu/home/pro...





We get (7,-10,2)





To get an orthogonal matrix including a scalar multiple of our first vector, we normalise these vectors, dividing each by its magnitude:





|| (2,1,-2) ||





= sqrt ( 2^2 + 1^2 + (-2)^2)


=sqrt (4 + 1 + 4) = sqrt (9) = 3





so the normalised form of our first vector is





(2/3,1/3,-2/3)








For our second vector, the semi-freebie:





||(2,2,3)||





= sqrt (2^2 + 2^2 + 3^2)





= sqrt (4 + 4 + 9) = sqrt (17)





producing the not quite as elegant normalisation





(2/sqrt(17),2/sqrt(17),3/sqrt(17))





or, in standard form, which is probably what a grader would want to see





(2sqrt(17)/17, 2sqrt(17)/17, 3sqrt(17)/17)





Never forget, a happy grader is an easy grader. To normalise the third vector, which we can easily see is orthogonal to the first two as we check our work, we compute





||(7,-10,2)||





= sqrt(7^2+ (-10)^2 + 2^2)


= sqrt(49+100+4)


=sqrt(153)


=sqrt(3*51)


=sqrt(3*3*17)


=3sqrt(17)





yeilding the normalised form





(7sqrt(17)/3*17, -10sqrt(17)/3*17,2sqrt(17)/3*17)





or (7sqrt(17)/51, -10 sqrt(17)/51, 2sqrt(17)/51)








Our orthogonal matrix, then, will be the one whose rows will read as follows:





First row, entries (in order, left to right):





2/3, 1/3, -2/3





Second row, entries





2sqrt(17)/17, 2sqrt(17)/17, 3sqrt(17)/17





Third row,





7sqrt(17)/51, -10 sqrt(17)/51, 2sqrt(17)/51





I hope this helps.














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Reply:I will help you with the first.





You have c is one basis vector. You know you need 3, so 2 more.





Can you find another vector that is orthogonal to c?





Well, you can do one by observation, say (1,0,1), but suppose you do not see that. Pick a vector, say (1,0,0), call it x.





Now, you wished x was orthogonal to c, so use Graham Schmidt to find the part orthogonal to c. Namely,





x - (c,x) c/(c,c) were (,) is dot product. so (c,c) = 9, and (c,x) = 2, Thus


(1,0,0) -(2/9) (2,1,-2) should be orthogonal to c. And checking, it is. For convenience, multiply by 9 and a second vector is


(9,0,0) - 2 (2,1, -2) = (5, -2, 4)








Now, how do you find a third? Use Graham Schmidt on a new vector, say (0,1,0). Subtract off the part orthogonal to c, and the part orthogonal to c and the part orthogonal to (5, -2,4) and you are done.
Reply:First, find a vector b, orthogonal to c:


How about b = (0, 2, 1)


That's clearly orthogonal since their dot product is 0.





Take their cross product to find the third vector:


a = b x c = det(i j k || 0 2 1 || 2 1 -2)


(-5, 2, -4)





The dot product of a with both b and c is 0 so a, b, c span R3.





To get your orthogonal matrix, have the three vectors above as columns (and I would first divide each vector by its magnitude, so that the resultant vector's magnitude is 1. That way your matrix will be orthonormal), and I'll flip the sign on a





5/√45 0 2/3





-2/√45 2/√5 1/3





4/√45 1/√5 -2/3