Math 206 Graded Problems/Proofs #14 Due Thursday Mar 28

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In this problem, Rⁿ is understood to have the usual operations of addition and scalar multiplication. You will also need the definition of (usual) dot product on Rⁿ, and you will probably want to use the usual notation for dot product, which is awkward to use in an html file such as this.

Let A be a (fixed but unspecified) subset of Rⁿ. A^perp is to be a particular subset of Rⁿ, namely the subset whose membership criterion is defined as follows:

A vector v in Rⁿ belongs to A^perp iff for each a belonging to A, the (usual) dot product of v with a is zero.

Thus the vectors in A^perp are precisely those vectors in Rⁿ that are perpendiculr to all of the vectors in A.

Prove: A^perp is a subspace of Rⁿ.


To carry out this proof, you should use the following outline.

• Step 1. Show that the zero vector for Rⁿ belongs to A^perp.
  (To do this, you must show that the zero vector satisfies the membership criterion for A^perp . So assume that a is a vector belonging to A. Now what do you have to show?)
  
  
• Step 2. Show that A^perp is closed under addition.
  • Assume that u and v belong to A^perp. What does this mean?
    
    
  • You must show that u+v belongs to A^perp. Thus you must show that:
    
    for each a belonging to A, the (usual) dot product of u+v with a is zero.
    
    
  • So assume that a belongs to A. (In other words, assume a is a fixed but unspecified vector belonging to A.)
    
    
  • Now try to argue, from the initial assumption for Step 2, that
    
    the (usual) dot product of u+v with a is zero.
    
    
• Step 3. Show that A^perp is closed under scalar multiplication.
   

• Alexia Sontag, Mathematics
• Wellesley College
• Date Created: January 4, 2001
• Last Modified: March 13, 2002
• Expires: June 30, 2002