3.2 Rank of a Matrix

3.2 Rank of a Matrix - 3.2 3.2 The Rank of a Matrix and Matrix Inverse 1(a F(Theorem 3.5(b F ∵ If A ∈ M m × n F B ∈ M n × n F and B is

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Unformatted text preview: 3.2. 3.2. The Rank of a Matrix and Matrix Inverse 1. (a) F (Theorem 3.5) (b) F ( ∵ ) If A ∈ M m × n ( F ) , B ∈ M n × n ( F ) and B is invertible, then the rank(AB)=rank(A) (Example) A = B = 0 1 0 0 ¶ and rank ( A ) = rank ( B ) = 1 AB = 0 0 0 0 ¶ and rank ( AB ) = 0 (c) T (d) T (p.153, Corollary to the Theorem 3.4) (e) F (p.153, Corollary to the Theorem 3.4) (f) T (p.153, Theorem 3.4 and Theorem 3.5) (g) T (p.161) (h) T ( ∵ ) ∀ A ∈ M m × n ( F ) , rank ( A ) = dimR ( L A ) ≤ n (i) T 2. (a) A = 1 1 0 0 1 1 0 0 0 , rank ( A )=2 162 PNU-MATH 3.2. (b) A = 1 1- 1 1 1 , rank ( A )= 3 (c) A = 1 0 2 0 1 2 ¶ , rank ( A )= 2 (d) A = 1 2 1 0 0 0 ¶ , rank ( A )= 1 (e) A = 1 2 0 1 1 0 0 1 1- 2 0 0 0 0 1 0 0 0 0 , rank ( A )= 3 (g) A = 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 , rank ( A )= 1 3. ∀ A ∈ M m × n ( F ) , rank ( A ) = 0 iff A is the zero matrix ( ⇐ ) clear ( ⇒ ) let A = ( e 1 , , ··· , 0) , e 1 6 = 0 Since rank ( A ) = 0 , e 1 is dependent ∴ ∃ a ∈ F s.t e 1 = a It’s contradict to e 1 6 = 0 4. (a) 1 1 1 2 2 0- 1 2 1 1 1 2 ⇒ 1 0- 1 2 1 0 1 3 2 1 0 0 ⇒ 1 0 0 0 0 1 0 0 0 0 0 0 , 2 163 PNU-MATH 3.2. (b) 2 1- 1 2 2 1 ⇒ 2 0 0 1 0 0 ⇒ 12 0 1 , 2 5. ( A | I n ) ˆ ( I n | B ) , B = A- 1 (a) 1 2 1 0 1 1 0 1 ¶ ⇒ 1 0-1 2 0 1 1-1 ¶ ∴ A- 1 =- 1 2 1- 1 ¶ , rank ( A ) = 2 (b) 1 2 1 0 2 4 0 1 ¶ ⇒ 1 2 1 0 0-2 1 ¶ rank ( A ) = 1, so @ A- 1 (c) 1 2 1 1 0 0 1 3 4 0 1 0 2 3 -1 0 0 1 ⇒ 1 0 -5 3-2 0 0 1 3-1 1 0 0-3 1 1 rank ( A ) = 2, so @ A- 1 (d) 0 -2 4 1 0 0 1 1-1 0 1 0 2 4-5 0 0 1...
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This note was uploaded on 02/14/2012 for the course MAS 4105 taught by Professor Rudyak during the Spring '09 term at University of Florida.

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3.2 Rank of a Matrix - 3.2 3.2 The Rank of a Matrix and Matrix Inverse 1(a F(Theorem 3.5(b F ∵ If A ∈ M m × n F B ∈ M n × n F and B is

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