MATH136 Lecture Notes - Lecture 26: Invertible Matrix, Mathematical Induction
14 May 2018
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Wednesday, June 28
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Lecture 26 : Properties of invertible matrices. (Refers to 5.1)
Concepts:
Properties of invertible matrices.
26.1 Definition − The negative power of a matrix. If k is a positive integer, and A has an
inverse A−1, we define A−k as A−k = (A−1)k. It can be shown that, with this definition, the
rules of exponents hold as usual for matrices. (This definition does not appear in text, we
state here just for convenience.) We define A0 = I.
26.2 Proposition − Suppose A0, A1, ..., Ak are invertible matrices. Then:
the product A0A1 ... Ak is invertible where (A0A1....Ak)−1 = Ak−1Ak-1−1....A0−1.
Proof : By mathematical induction.
Base case: Suppose A0, A1 are invertible.
Then (A0A1)(A1−1A0−1) = A0(A1 A1−1)A0−1 = (A0 I )A0−1 = I.
So (A0A1) −1 = A1−1A0−1.
Induction hypothesis: Suppose that for a particular n, (A0A1 … An) −1 = An−1An −1−1 … A0−1.
Then
(A0A1 … An An +1)( A n +1−1An −1 … A1−1)
= (A0A1 … An) (An +1 A n +1−1) (An −1 … A1−1)
= (A0A1 … An) (I) (An −1 … A1−1)
= (A0A1 … An) (An −1 … A1−1)
= I (by the induction hypothesis)
So (A0A1 … An An +1) −1 = ( A n +1−1An −1 … A1−1)
By the principle of mathematical induction, (A0A1 … An) −1 = An−1An −1−1 … A1−1, for all n.
26.3 Proposition − If A is an invertible matrix then [AT ]−1 = [A−1]T.
Proof :
Given: A is invertible and B = [A−1]T.
Required to show: B = [AT ]−1. Suffices to show: ATB = I.
We have ATB = AT [A−1]T = [A−1A ]T = I T = I (By a property of matrix algebra)
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Wednesday, june 28 lecture 26 : properties of invertible matrices. (refers to 5. 1) 26. 1 definition the negative power of a matrix. 26. 2 proposition suppose a0, a1, , ak are invertible matrices. 1a0 the product a0a1 ak is invertible where (a0a1ak) 1 = ak. Induction hypothesis: suppose that for a particular n, (a0a1 an) 1 = an. = (a0a1 an) (an +1 a n +1. So (a0a1 an an +1) 1 = ( a n +1. By the principle of mathematical induction, (a0a1 an) 1 = an. 26. 3 proposition if a is an invertible matrix then [at ] 1 = [a 1]t. Given: a is invertible and b = [a 1]t. We have atb = at [a 1]t = [a 1a ]t = i t = i (by a property of matrix algebra) 26. 4 proposition if a is an invertible matrix and c is any non-zero scalar, then [ca] 1 = (1/c)a 1. (proof presented in class).
