3B Proof

Suppose \(V\) is finite-dimensional and \(T \in L(V,W)\). Prove that \(T\) is injective if and only if there exists \(S \in L(W,V)\) such that \(ST\) is the identity map on \(V\). NNOO!

3.16 injectivity is equivalent to null space equals {0} 3.15 injective: T: V -> W is injective if \(Tu = Tv\) implied \(u=v\) 3.8 product of linear maps:

identity map on v: Iv = v

product of linear maps:

if \(T \in L(U,V)\) and \(S \in L(V,W)\), then the product \(ST \in L(U,W)\) is defined by \[(ST)(u)=S(Tu)\] for all \(u \in U\).

Suppose \(V\) is finite-dimensional and \(T \in \mathcal{L}(V,W).\) Prove that \(T\) is surjective if and only if there exists \(S \in \mathcal{L}(W,V)\) such that \(TS\) is the identity map on \(W\).

3.20 :: def of surjective 3.17 :: def of range 3.2 :: def of linear map

• define a S that takes range of T back to V such that TS(v) = v
  • prove that this is linear, and we are done with the forwards direction
    • homogeneity
    • additivity
• additivity
  • \(T(u+v) = Tu + Tv\) for all \(u, v \in V\)
  • \(S(u+w) = Su + Sw\) for all \(u, w \in W\)

Since \(T\) is additive, and \(TS(v) = v\), we can say that: \[ TS(v + u) = v + u = T(Sv) + T(Su) = T(Sv + Su) \] Therefore, \(S\) is additive.

• homogeneity
  • \(T(\lambda V) = \lambda (Tv)\) for all \(\lambda \in F\) and all \(v \in V\).
• homogeneity \(T(S \lambda v) = \lambda v = \lambda T(Sv) = T(\lambda Sv)\)

backwards direction:

assume TS is the identity map. prove that T is surjective.