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Unformatted text preview: Math 416  Abstract Linear Algebra Fall 2011, section E1 Homework 10 solutions Section 5.5 5.2. (1 pt check) Since A has rank 2, let us find the orthogonal projections onto the 1dimensional subspaces Null A and Null A T . A = 1 1 1 1 3 2 2 4 3 ∼ 1 1 1 0 2 1 0 2 1 ∼ 1 1 1 0 2 1 0 0 0 ∼ 2 2 2 0 2 1 0 0 0 ∼ 2 0 1 0 2 1 0 0 0 Null A has basis v = 1 1 2 . The projection matrix onto Null A is therefore Proj Null A = v ( v T v ) 1 v T = 1 1 2 6 1 1 1 2 = 1 6 1 1 2 1 1 2 2 2 4 The projection onto Row A = (Null A ) ⊥ is Proj Row A = I Proj Null A = 1 6 6 0 0 0 6 0 0 0 6  1 1 2 1 1 2 2 2 4 = 1 6 5 1 2 1 5 2 2 2 2 . Likewise, let us find a basis of Null A T : A T = 1 1 2 1 3 4 1 2 3 ∼ 1 1 2 0 2 2 0 1 1 ∼ 1 1 2 0 1 1 0 0 0 ∼ 1 0 1 0 1 1 0 0 0 . 1 Null A T has basis w = 1 1 1 . The projection matrix onto Null A T is therefore Proj Null A T = w ( w T w ) 1 w T = 1 1 1 3 1 1 1 1 = 1 3 1 1 1 1 1 1 1 1 1 . The projection onto Col A = (Null A T ) ⊥ is Proj Col A = I Proj Null A T = 1 3 3 0 0 0 3 0 0 0 3  1 1 1 1 1 1 1 1 1 = 1 3 2 1 1 1 2 1 1 1 2 . 5.4. (2 pts) a. The ranknullity theorem applied to A and A * A yields ( rank A + dim ker A = n rank( A * A ) + dim ker( A * A ) = n ⇒ rank A + dim ker A = rank( A * A ) + dim ker( A * A ) ⇒ rank A = rank( A * A ) since ker A = ker( A * A ) . b. If A has trivial kernel, then so does A * A : ker( A * A ) = ker A = { } . Since A * A is square (i.e. its domain and codomain have the same dimension) and injective, it is invertible. Using the inverse of A * A , we obtain ( A * A ) 1 A * A = I so that ( A * A ) 1 A * is a left inverse of A ....
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This note was uploaded on 12/16/2011 for the course MATH 416 taught by Professor Frankland during the Fall '11 term at University of Illinois at Urbana–Champaign.
 Fall '11
 FRANKLAND
 Math, Linear Algebra, Algebra

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