linear algebra question 1

Question

Consider 𝑬 = {(𝒙, 𝒚, 𝒛, 𝒕) ∈ ℝ𝟒 : 𝒙 = 𝒕, 𝒚 = 𝒛}

a)Determine whether the given set is subvector space or not

b) Show that this set is a basis for ℝ4 .

c) What is the dimension of 𝐸=?

Answer 1

a) To prove that something is a subspace you must prove (1) closure under addition, that for any u, v ∈ E, x+y ∈ E and (2) closure under scalar multiplication, that for any u ∈ E and α ∈ ℝ, αu ∈ E.

(1) Closure under addition

Let u=(t, z, z, t) and v=(s, w, w, s)

u+v=(t+s, z+w, z+w, t+s) ∈ E

(2) Closure under multiplication

Let u=(t, z, z, t) and α ∈ ℝ

αu=(αt, αz, αz, αt) ∈ E

Since both closure properties are satisfied, E is a subspace.

b) What set are you referring to? E itself cannot be a basis because E has infinitely many elements and ℝ⁴ is finite dimensional.

c) We need the middle two entries to be equal and the outer two entries to be equal. The outer two entries can be different from the middle two entries. Any multiple of (0, 1, 1, 0) will have middle entries that are equal and any multiple of (1, 0, 0, 1) will have outer entries that are equal; adding multiples of these together will give us any element of E, so we can form a basis of E with (0, 1, 1, 0) and (1, 0, 0, 1). E is two-dimensional.