Invertible Matrix Theorem - Proof PDF

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Invertible Matrix Theorem Let A be an n × n matrix. Then the following are equivalent: (a) A is invertible (b) A!x = 0! has just the trivial solution (c) The Reduced Row Echelon Form of A is In Proof: (a =⇒ b) If A is invertible, we can multiply both sides of A!x = 0! on the left by A−1 to obtain A−1 A!x = A−1!0 or I!x = !0 so !x = !0 and we have just the trivial solution. (b =⇒ c) If A!x = !0 has just the trivial solution, then the system of equations in reduced form is = 0 = 0

x1 x2

..

. xn = 0

or 1x1 + 0x2 + . . . + 0xn = 0 0x1 + 1x2 + . . . + 0xn = 0 .. . 0x1 + 0x2 + . . . + 1xn = 0 or 

  1 0 ··· 0 x1 0 1 . . . 0  x2     =  ..  .  .. .   ..  . 0 0 ... 1 xn

  0 0   ..  .  0

and the RREF of A is I . (c =⇒ a) If the RREF of A is I, then there is a sequence of say k EROs that reduce A to I . Each ERO is equivalent to multiplying A on the left by an elementary matrix. Thus, there are k elementary matrices E1 , E2 , . . . Ek such that Ek · · · E2 E1 A = I This tells us A−1 = Ek · · · E2 E1 and so A is invertible.

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