Linear Algebra

Section II. Homomorphisms 1852.21 Theorem In an n-dimensional vector space V , these:(1) h is nonsingular, that is, one-to-one(2) h has a linear inverse(3) N (h) = {⃗0 }, that is, nullity(h) = 0(4) rank(h) = n(5) if 〈 ⃗ β 1 , . . . , ⃗ β n 〉 is a basis for V then 〈h( ⃗ β 1 ), . . . , h( ⃗ β n )〉 is a basis for R(h)are equivalent statements about a linear map h: V → W .Proof. We will first show that (1) ⇐⇒ (2). We will then show that (1) =⇒(3) =⇒ (4) =⇒ (5) =⇒ (2).For (1) =⇒ (2), suppose that the linear map h is one-to-one, and so has aninverse. The domain of that inverse is the range of h and so a linear combinationof two members of that domain has the form c 1 h(⃗v 1 ) + c 2 h(⃗v 2 ). On thatcombination, the inverse h −1 gives this.h −1 (c 1 h(⃗v 1 ) + c 2 h(⃗v 2 )) = h −1 (h(c 1 ⃗v 1 + c 2 ⃗v 2 ))= h −1 ◦ h (c 1 ⃗v 1 + c 2 ⃗v 2 )= c 1 ⃗v 1 + c 2 ⃗v 2= c 1 h −1 ◦ h (⃗v 1 ) + c 2 h −1 ◦ h (⃗v 2 )= c 1 · h −1 (h(⃗v 1 )) + c 2 · h −1 (h(⃗v 2 ))Thus the inverse of a one-to-one linear map is automatically linear. But this alsogives the (1) =⇒ (2) implication, because the inverse itself must be one-to-one.Of the remaining implications, (1) =⇒ (3) holds because any homomorphismmaps ⃗0 V to ⃗0 W , but a one-to-one map sends at most one member of Vto ⃗0 W .Next, (3) =⇒ (4) is true since rank plus nullity equals the dimension of thedomain.For (4) =⇒ (5), to show that 〈h( β ⃗ 1 ), . . . , h( β ⃗ n )〉 is a basis for the rangespacewe need only show that it is a spanning set, because by assumption the rangehas dimension n. Consider h(⃗v) ∈ R(h). Expressing ⃗v as a linear combinationof basis elements produces h(⃗v) = h(c 1β1 ⃗ + c 2β2 ⃗ + · · · + c nβn ⃗ ), which gives thath(⃗v) = c 1 h( β ⃗ 1 ) + · · · + c n h( β ⃗ n ), as desired.Finally, for the (5) =⇒ (2) implication, assume that 〈 β ⃗ 1 , . . . , β ⃗ n 〉 is a basisfor V so that 〈h( β ⃗ 1 ), . . . , h( β ⃗ n )〉 is a basis for R(h). Then every ⃗w ∈ R(h) a theunique representation ⃗w = c 1 h( β ⃗ 1 ) + · · · + c n h( β ⃗ n ). Define a map from R(h) toV by⃗w ↦→ c 1β1 ⃗ + c 2β2 ⃗ + · · · + c nβn⃗(uniqueness of the representation makes this well-defined). Checking that it islinear and that it is the inverse of h are easy.QEDWe’ve now seen that a linear map shows how the structure of the domain islike that of the range. Such a map can be thought to organize the domain spaceinto inverse images of points in the range. In the special case that the map is