10 010 a let a 1 0 0 show that the null space of a can

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Unformatted text preview: (the dimension of P2 ). Noting that the only solution to a(x2 − x + 2) + b(3x + 4) = 0 is a = b = 0, we see that {p1 , p2 } is linearly independent and is a basis of Span(X ), and so dim(Span(X )) = 2. 10. 010 a) Let A = 1 0 0 . Show that the null space of A can be interpreted geometrically as 000 points on the z −axis in 3-dimensional Cartesian space, and that the column space can be interpreted as the points on the xy −plane. b) Find a 3 × 3 matrix whose null space corresponds to the points on the x−axis in 3dimensional Cartesian space, and whose column space corresponds to the yz −plane. Math 2061: Tutorial 4 (week 5) — Solutions Page 5 Linear Mathematics Tutorial 4 (week 5) — Solutions Page 6 c) Find the null space and column space of the 3 × 3 identity matrix. Solution a) The null space of A consists of all column vectors x1 x2 x3 such that x2 = 0, x1 = 0, and x3 can take any real value. Therefore, x is in the null space if and only if x = 0 0 x3 , 0 0 where x3 ∈ R. Associating x with the point (0, 0, x3 ), we see that the null space can 3 be interpreted as the set of all points on the z -axis in three dimensional space. A basis for the column space of A is 0 1 0 , 1 0 0 . So, a typical element of the column space has the form a b 0 , where a, b ∈ R. Again, a b we can associate 0 with the point (a, b, 0) and so the column space corresponds to all points on the plane z = 0; that is, the points on the xy -plane. 000 b) 0 0 1 is such a matrix. 010 c) Since I x = x for all x ∈ R3 , the null space of the identity matrix I is the zero vector 100 subspace {0} of R3 . As I = 0 1 0 , the standard basis of R3 is a basis for the column 001 space of I . Hence, the column space of I is the whole of R3 . Math 2061: Tutorial 4 (week 5) — Solutions Page 6...
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