Linear Algebra

186 Chapter Three. Maps Between Spaces2.15 Theorem A linear map’s rank plus its nullity equals the dimension of itsdomain.Proof Let h: V → W be linear and let B N = 〈⃗β 1 , . . . , ⃗β k 〉 be a basis forthe null space. Expand that to a basis B V = 〈⃗β 1 , . . . , ⃗β k , ⃗β k+1 , . . . , ⃗β n 〉 forthe entire domain, using Corollary Two.III.2.12. We shall show that B R =〈h(⃗β k+1 ), . . . , h(⃗β n )〉 is a basis for the range space. With that, counting thesize of these bases gives the result.To see that B R is linearly independent, consider ⃗0 W = c k+1 h(⃗β k+1 ) + · · · +c n h(⃗β n ). The function is linear so we have ⃗0 W = h(c k+1⃗β k+1 + · · · + c n⃗β n )and therefore c k+1⃗β k+1 + · · · + c n⃗β n is in the null space of h. As B N is a basisfor the null space there are scalars c 1 , . . . , c k satisfying this relationship.c 1⃗β 1 + · · · + c k⃗β k = c k+1⃗β k+1 + · · · + c n⃗β nBut this is an equation among the members of B V , which is a basis for V, soeach c i equals 0. Therefore B R is linearly independent.To show that B R spans the range space, consider h(⃗v) ∈ R(h) and write ⃗v asa linear combination ⃗v = c 1⃗β 1 +· · ·+c n⃗β n of members of B V . This gives h(⃗v) =h(c 1⃗β 1 + · · · + c n⃗β n ) = c 1 h(⃗β 1 ) + · · · + c k h(⃗β k ) + c k+1 h(⃗β k+1 ) + · · · + c n h(⃗β n )and since ⃗β 1 , . . . , ⃗β k are in the null space, we have that h(⃗v) = ⃗0 + · · · + ⃗0 +c k+1 h(⃗β k+1 ) + · · · + c n h(⃗β n ). Thus, h(⃗v) is a linear combination of membersof B R , and so B R spans the range space.QED2.16 Example Where h: R 3 → R 4 is⎛ ⎞⎛ ⎞ xx⎜ ⎟⎝y⎠↦−→h0⎜ ⎟⎝y⎠z0the range space and null space are⎛ ⎞a⎛ ⎞00R(h) = { ⎜ ⎟ ∣ ⎜ ⎟a, b ∈ R} and N (h) = { ⎝0⎠ ∣ z ∈ R}⎝b⎠z0and so the rank of h is 2 while the nullity is 1.2.17 Example If t: R → R is the linear transformation x ↦→ −4x, then the rangeis R(t) = R 1 . The rank is 1 and the nullity is 0.2.18 Corollary The rank of a linear map is less than or equal to the dimension ofthe domain. Equality holds if and only if the nullity of the map is 0.We know that an isomorphism exists between two spaces if and only if thedimension of the range equals the dimension of the domain. We have now seen