MATH 415 Final: Lecture28gartlandfilled

Justify your answers. {[17, 6, -4]^t, [2, 3, 3]^t, [19, 9, -1]^t} does not span R^3. {(1.1]^t, [1, 2]^t, [4, 7]^t) spans R^2. Let W be a two-dimensional subspace of R^3. Then two of the following three, vectors span W: X = [1, 0, 0]^t, Y = [0, 1, 0]^t, Z = [0, 0, 1]^t. The following set of vectors is linearly independent. {[-36, 13, -11]^t, [22, pi, squareroot 2]^t, [41, -37/17, -2]^t, [squareroot 3, squareroot 7, -3)^t) The nullspace of a nonzero 4 Times 4 matrix cannot contain a set of four linearly independent vectors. Suppose that W is a four-dimensional subspace of R^7 and X^1, X^2, X^3, and X^4 are vectors that belong to W. Then {X^1, X^2, X^3, X^4) spans W. Suppose that {X^1, X^2, X^3, X^4, X^5} spans a four-dimensional vector space W of R^7. Then {X^1, X^2, X^3, X^4} also spans W. Suppose that S = {X^1, X^2, X^3, X^4, X^5} spans a four-dimensional subspace W o R7. Then S contains a basis for W. Suppose that {X^1, X^2, X^3, X^4, X^5) spans a four-dimensional subspace W of R7 Then one of the X_i must be a linear combination of the others.

(1 pt) Let V-p? For (u,12), (v1, U2) E V and a E R define vector addition by (u 1 , u2沺(V1Mz) := (u1 + U1+ 2, u2 + ข2-1 and scalar multiplication by a (m ,u2) :-(aul t. 2a-2, au2-a + 1). It can be shown that (V,田,U丬is a vector space over the scalar field R. Find the following yY on the sum: (-6,8) H (3,0) 1 the scalar multiple: 1 (-6,8)--6 the zero vector 0 the additive inverse of (x, y) (z, y) ( -2-x -y