ass7 - MATH 223, Linear Algebra Fall, 2007 Solutions to...

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MATH 223, Linear Algebra Fall, 2007 Solutions to Assignment 7 1. A linear operator T on a vector space V is called a projection if T 2 = T . (We will be looking at orthogonal projections later.) (a) If T = T A is represented by the matrix A = 2 0 2 0 1 0 - 1 0 - 1 , show that T is a projection. Solution: Since T 2 will be represented by A 2 , it’s enough to show that A 2 = A . I skip this very easy calculation. (b) Show that, if T is a projection on V , then so is I - T ; here I is the identity operator on V . Solution: For any ~v V , ( I - T ) 2 ~v = ( I - T )( I - T ) ~v = ( I - T )( ~v - T~v ) = ~v - T~v - T ( ~v - T~v ) = ~v - T~v - T~v + T 2 ~v. Using T 2 = T , this reduces to ~v - T~v = ( I - T ) ~v . So ( I - T ) 2 = I - T and I - T is a projection, too (if T is). (c) Show that, if T is a projection on V , then V = ker ( T ) im ( T ). (Recall that is the direct sum, so that not only do we have V = ker ( T ) + im ( T ) but also, ker ( T ) im ( T ) = ~ 0 } .) Solution: For any ~v V , of course T~v im ( T ). Now T ( ~v - T~v ) = T~v - T 2 ~v = ~ 0, so ~v - T~v ker ( T ). As ~v = ( ~v - T~v ) + T~v , we have ~v ker ( T ) + im ( T ) and so V = ker ( T ) + im ( T ). To show the sum is direct, suppose that ~v ker ( T ) im ( T ). Then T~v = ~ 0 and also ~v = T ~w for some ~w V . Then ~ 0 = T~v = T ( T ~w ) = T 2 ~w = T ~w = ~v . So ker ( T ) im ( T ) = { ~ 0 } , and thus V = ker ( T ) im ( T ). (d) Show that, if T is a projection on V , its only possible eigenvalues are 0 and 1. Solution: One can do this directly, using the definition of eigenvalues. Supposing that
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This note was uploaded on 04/29/2008 for the course MATH 223 taught by Professor Loveys during the Fall '07 term at McGill.

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ass7 - MATH 223, Linear Algebra Fall, 2007 Solutions to...

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