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lecture_17 - MA 35100 LECTURE NOTES: WEDNESDAY, FEBRUARY 24...

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Unformatted text preview: MA 35100 LECTURE NOTES: WEDNESDAY, FEBRUARY 24 Relationship with Invertibility (cont’d) We summarize everything we have seen about invertibility: Theorem. Let A be an n × n matrix. Then the following are equivalent: i. A is invertible. ii. The linear system Avectorx = vector b has a unique solution vectorx for all vector b . iii. rref( A ) = I n . iv. rank( A ) = n . v. im( A ) = R n . vi. ker( A ) = { vector } . We explain why this is true: • (i) ⇐⇒ (ii): This is by Definitions 2.4.1 (pg. 79) and 2.4.2 (page 80). • (i) ⇐⇒ (iii): This is Theorem 2.4.3 (page 80). • (iii) ⇐⇒ (iv): This is Definition 1.3.2 (page 26). • (i) ⇐⇒ (vi): This is the previous theorem above. • (i) ⇐⇒ (v): We need to prove this. Say A is invertible; we show im( A ) = R n . Given vector b ∈ R n , the equation Avectorx = vector b has the solution vectorx = A- 1 vector b . Hence vector b ∈ im( A ), so that im( A ) = R n . Conversely, say im( A ) = R n ; we construct A- 1 ....
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This note was uploaded on 10/18/2010 for the course MATH 351 taught by Professor Egoins during the Spring '10 term at Purdue.

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lecture_17 - MA 35100 LECTURE NOTES: WEDNESDAY, FEBRUARY 24...

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