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.

• Step 1. Show that the zero vector for Rⁿ belongs to A^perp.
  Assume that a is a vector belonging to A. We know from Math 205 (or from reviewing the dot-product definition x*a=x₁a₁+x₂a₂+...+x_na_n) that the dot product of the zero vector with any vector in Rⁿ is zero. In particular, 0*a=0. This completes Step 1.
  
• Step 2. Show that A^perp is closed under addition.
  • Assume that u and v belong to A^perp. Thus, for every a in A, u*a=v*a=0.
    
  • To show that u+v belongs to A^perp, assume that a belongs to A. Using properties of dot products again, we have
    
    (u+v)*a=u*a+v*a=0+0=0.
    

    Since a was given arbitrarily, we have shown that u+v belongs to A^perp , which finishes Step 2.

• Step 3. Show that A^perp is closed under scalar multiplication.
  • Assume that u belongs to A^perp. Thus, for every a in A, u*a=0. Also assume that k is a scalar.
  • To show that ku belongs to A^perp, assume that a belongs to A. Properties of dot products and properties of real numbers give us
    
    (ku)*a=k(u*a)=0k=0
    
    

    as required. Thus ku does belong to A^perp, and Step 3, now finished, completes the proof.

   

• Alexia Sontag, Mathematics
• Wellesley College
• Date Created: January 4, 2001
• Last Modified: April 17, 2002
• Expires: June 30, 2002