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Unformatted text preview: 2.3 Characterizations of Invertible Matrices Suppose A is invertible and T (x) = Ax for linear transformation T : Rn → Rn . Q: What can we say about T (x)? A: Can show that T is one-to-one and onto, because
Ax = b has a unique solution for all b ∈ Rn . More generally, we have Invertible Matrix Theorem: For square (n × n) A,
the following are all equivalent: ·A is invertible ·A is row equivalent to In ·A has n pivots ·Ax = 0 has only the trivial soln x = 0 · The columns of A are linearly independent · The linear transformation T (x) = Ax is one-to-one ·Ax = b has at least one soln for each b ∈ Rn · The columns of A span Rn · The linear transformation T (x) = Ax is onto · There is n × n C such that CA = I 1 · There is n × n D such that AD = I · AT is invertible · The rows of A are linearly independent NOTE: For square (and only square) matrices A, all of
these must be either TRUE or FALSE Corollary: Given two (n × n) matrices A and B , if
AB = I then A and B are both invertible, with A−1 = B and B −1 = A EX: Explain why the matrices below are not invertible. (i) 135 216 2 40 (ii) 3 6 0 2 −1 0 23 2 (iii) 4 6 −1 00 0 243 (iv) 0 0 1 002 2 Given a linear transformation T : Rn → Rn . It is invertible if there exists a linear transformation S : Rn → Rn such that S (T (x)) = x for all x ∈ Rn and T (S (x)) = x for all x ∈ Rn T x ∈ Rn T (x) ∈ Rn S Theorem: T : Rn → Rn with standard matrix A is invertible iﬀ A is invertible. Its inverse S has the standard matrix A−1 Q: Given T : Rn → Rn. Is it invertible
(i) if T is onto? (ii) if T is one-to-one? A: Yes for both. Why?
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