|Feb28-09, 09:28 PM||#1|
Linear Transformation Norm Preserving
1. The problem statement, all variables and given/known data
From Calculus on Manifolds by Spivak: 1-7
A Linear Transformation T:Rn -> Rn is Norm Preserving if |T(x)|=|x| and Inner Product Preserving if <Tx,Ty>=<x,y>.
Prove that T is Norm Preserving iff T is Inner Product Preserving.
2. Relevant equations
T is a Linear Transformation
=> For All x,y [tex]\in[/tex] Rn and scalar c
x is an n tuple i.e. x=(x1,...,xn)
<x,y>=[tex]\sum[/tex]xiyi where i=1,...,n
3. The attempt at a solution
I cannot see what to do here at all. I am definately missing something. I don't see how the definitions relate to help me here. It's probably something simple. Any direction or hint would be greatly appreaciated. Thank you.
|Mar1-09, 03:55 AM||#2|
This is the so-called polarization identity.
|Oct12-11, 09:12 PM||#3|
Calculus on Manifolds, Michael Spivak.
Comment: one direction is very easy, namely if a Linear Transformation preserves the inner product it also preserves the norm. Well, after all, the norm is just the square root of the inner product of a vector and itself. If T preserves the inner product of any two vectors it will certainly preserve the inner product of any vector with itself.
If<Tx,Ty> = <x,y>, then |Tx|=|y|.
Pf: |Tx| = √<Tx, Tx> (by definition)
= √ <x,x> (by assumption of the problem)
= |x| (again by definition)
Comment: the other direction is much more interesting. It says that if the norm of every vector is preserved by T then T also preserves the inner product of any two vectors. Since the inner product used here is just the good old "dot product" and since we remember (from the past) that the dot product of two vectors x and y equals the norm of x times the norm of y times the cosine of the smallest angle between x and y, we are saying here that if the norm is preserved by a linear transformation T, then T automatically also preserves the angle between any two vectors. Spivak hasn't introduced the angle idea yet in this problem 7, but does in the next problem 8! Here goes proof of the rest of 7(a):
If |Tx|=|x|, then <Tx,Ty> = <x,y>.
Pf: |x+y|squared = |x|squared + |y| squared + 2<x,y> (from the first line of Spivak's proof of Th.1-1(3))
so also, |Tx+Ty|squared = |Tx|squared + |Ty| squared + 2<Tx,Ty>
but, |Tx+Ty| = |T(x+y)| (by definition of linear transformation) note the notation: Tx = T(x)
and |T(x+y)| = |x+y| (by assumption of the problem)
Thus, |Tx+Ty| = |x+y|
so, |x+y|squared = |Tx+Ty|squared
and, |x|squared +|y| squared + 2<x,y> = |Tx|squared + |Ty| squared + 2<Tx,Ty>
forcing 2<x,y> = 2<Tx,Ty> (since T preserves the norm of both x and y)
so finally <x,y>=<Tx,Ty> QED.
b.) Comment: this is very easy. I won't write it out unless someone asks me.
Best wishes, Edward111
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