Let H be the set of all vectors of the form .a 3b; b a; a; b/, where a and b are arbitrary scalars. That is, let H D f. a 3b; b a; a; b/ W a and b in Rg. Show that H is a subspace of R

H is the span of the vectors (1,-1,1,0), (-3,1,0,1).

Step-by-step explanation:

To begin with remember the following theorem.

Theorem : Given a vector space V  let       be a set of vectors of the , then        is a subspace of V.

Notice that any vector, in the described subspace looks like this.

(a-3b , b-a , a , b ) = (a , -a , a , 0 )  +  ( -3b , b , 0 , b  )

                             = a( 1 , -1 , 1 , 0  ) + b( -3 , 1 , 0 , 1  )

Therefore H is the span of the vectors (1,-1,1,0), (-3,1,0,1) and according to the theorem it would be a subspace.

c. prove the lengths are the same.

step-by-step explanation:

you need to prove the segments are congruent. congruent segments have the same length. find the lengths of the segments using the distance formula, and compare them.

answer: c. prove the lengths are the same.