normalizing constant of a column

Good Morning All,
Probably one of the most simplest questions I have ever asked but when normalizing the first column of a matrix its simple just the square root of the sum of the elements of the column squared correct?
I have a torque roll axis direction defined as a normalizing constant, a, for the first column of matrix: a*J=M^-1
I have the mass matrix and need to find J and a. I am assuming that I am dividing inv(M) by a. and that a is the sqrt of the sum of the squares of inv(M).
Is this correct?
Thanks a bunch.

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Matt J
Matt J 2013 年 7 月 5 日
編集済み: Matt J 2013 年 7 月 5 日

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when normalizing the first column of a matrix its simple just the square root of the sum of the elements of the column squared correct?
It's correct assuming you want the Euclidean norm of the column set to 1. But there are other norms, so it really depends on what you're doing,e.g.
>> v=[1 2 3];
>> v/norm(v)
ans =
0.2673 0.5345 0.8018
>> v/norm(v,1)
ans =
0.1667 0.3333 0.5000
>> v/norm(v,inf)
ans =
0.3333 0.6667 1.0000

4 件のコメント

Melissa
Melissa 2013 年 7 月 5 日
Thanks Matt J. Another quick question if you can spare a moment (hopefully not inconveniencing you). According to a paper [Ux Uy Uz]=[Wx Wy Wz]/|[Wx Wy Wz]| where Wy and Wz are arbitrary values. I am assuming that the denominator is saying the norm of each element of that vector. Therefore I should be getting a matrix of 1's?
Matt J
Matt J 2013 年 7 月 5 日
Hi Melissa. The denominator denotes the norm of the 1x3 vector [Wx,Wy,Wz].
Melissa
Melissa 2013 年 7 月 5 日
Oh!!!! That makes more sense now I won't have a matrix full of ones. Thanks for clearing up my poor assumption. So W_vec=[Wx Wy Wz]; and Ux=Wx/norm(W_vec); Uy=Wy/norm(W_vec); Uz=Wz/norm(W_vec).
Matt J
Matt J 2013 年 7 月 5 日
Yes.

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