The inverse of a Linear Transformation September 10, 2007Recall:Let S and T be sets.A mapping or function from S to T is a rule which assigns to every element s of S a (well-defined) element f(s) of T.The set S is called the source or domain of f. The set T is called the target or codomain of f. A function f: S → T is said to be invertible if for every t in T there is a unique s in S such that f(s) = t.A closer look at f(s) =tf is said to be injective (one-to-one) if for every t there is at most one s such that f(s) = t.f is said to be surjective (onto) if for every t there is at least one s such that f(s) = t.f is said to be bijective if for every t there is exactly one s such that f(s) = t.Inverse functions f is surjective and injective.f is bijective.There exists a function g: T → S such thatg(f(s)) = s for all s and f(g(t)) = t for all t.This g is called the inverse of the function f.Theorem: Let f: S → T be a function. Then the following conditions are equivalent.The inverse of a linear transformationTheorem: Let A be an n x m matrix. Then TA: Rm → Rn is invertible if and only if n = m = rank(A). If this is the case, its inverse TA-1 is also linear.If r < m, then there are m-r free variables. Hence the equation TA(x) = 0 has infinitely many solutions and TA could not be injective.Thus if TA is injective m ≤ r, hence m = r.Conversely: if m = r, TA is injective.Proof: Let r be the rank of A.If r < n, then there is some y such that the equation TA(x) = y is inconsistent. Hence TA will not be surjective. Hence if TA is surjective n ≤ r, hence n = r.Conversely, if n=r, TA is surjective.Thus if TA is bijective, n = r = m.Conclusion: If A is an n x m matrix of rank r,Corollary: If n = m, then the following are equivalent:1. TA is injective (i.e., r = m).2. TA is surjective (i.e., r = n).3. TA is bijective.1.TA is injective if and only if m = r.2.TA is surjective if and only if n = r.If TA is invertible, then, (TA)-1 = T(A-1), another linear transformation, where A-1 is computed as follows: Form the matrix (A|In).Put (A|In) in reduced row echelon form. Then rref(A|In) = (In|A-1).See syllabus web page for a proof of this fact. Computing
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