Show that the given set V is closed under addition and multiplication by scalars and is therefore a subspace of Rº. V is the set of all X y Z such that z = 0. The vector sum a + b is an element of the set V because it addition. (Type an integer or a fraction.) Let c be a scalar. Recall that a = ca = (Simplify your answer.) X y 0 Find the scalar multiple ca. has its third component equal to 0. Therefore, V is closed under

The set P = 6 = c = a + is a vector space under usual addition and multiplication. Find the negative vector. -a -a -а — d -d —а a + d a + d -d a —а +d —а +d d a -a d d

Let W be the set of all vectors of the form shown on the right, where a, b, and c represent arbitrary real numbers. Find a set S of vectors that spans W or give an example or an explanation to show that W is not a vector space. 4a + 2b 5b + 2c 8c + 8a 7b Select the correct choice and fill in the answer box as needed to complete your choice. O A. A spanning set is S={} (Use a comma to separate answers as needed.) O B. There is no spanning set of W because W does not contain the zero vector. O C. There is no spanning set of W because W is not closed under vector addition. O D. There is no spanning set of W because W is not closed under scalar multiplication.

Only part d please.

Let 4 be a linear combination of {u₁, 1₂, 13). Select the best statement. A. span{u₁, U₂, U3} = span{u₁, U2₂, U3, 14} when u4 is a scalar multiple of one of {U₁, U₂, U3}. B. span{u₁, U₂, U3} = span{U₁, U₂, U3, U4}. c. We only know that span{u₁, U₂, U3} span{u₁, U₂, U3, U4} . D. There is no obvious relationship between span{u₁, U₂, U3} and span{u₁, U₂, U3, U4} . E. none of the above

Please solve this linear algebra problem with a good explanation.