tXcjKA-01

tXcjKA-01 - m ) and c ∈ R . Solution. By de±nition, [ T...

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ASSIGNMENT 1 · SOLUTIONS MAT 473 · SPRING 2012 Problem 1 (1.1) . Given that R is complete, prove that R n is complete (as a metric space). Proof. First note that for any y =( y 1 ,y 2 ,...,y n ) R n ,andforany1 i n ,wehave | y i |≤ ° y 2 1 + y 2 2 + ··· + y 2 n = ° y °≤ n max {| y j || 1 j n } . Now suppose ( x ( k ) ) k =1 is a Cauchy sequence in R n .T h e nf o r1 i n and any °, m applying the ±rst inequality above to y = x ( ° ) x ( m ) gives | x ( ° ) i x ( m ) i |≤° x ( ° ) x ( m ) ° . It follows that each sequence ( x ( k ) i ) k =1 is Cauchy in R ,andhenceconve rge stosome x i in R .I fw el e t x =( x 1 ,x 2 ,...,x n ), then it follows by applying the second inequality above to y = x x ( k ) that x ( k ) x .Thu s R n is complete. ° Problem 2 (1.3) . Verify that [ S + T ]=[ S ]+[ T ] and [ cT ]=
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Unformatted text preview: m ) and c ∈ R . Solution. By de±nition, [ T ] is the unique m × n matrix such that T ( x ) = [ T ] x for all x ∈ R n . Since matrix algebra gives ([ S ] + [ T ]) x = [ S ] x + [ T ] x = S ( x ) + T ( x ) = ( S + T )( x ) , it follows that [ S + T ] = [ S ] + [ T ]. Similarly, since ( c [ T ]) x = c ([ T ] x ) = c ( T ( x )) = ( cT )( x ) , we have [ cT ] = c [ T ]. (Direct calculations using the formula [ T ] ij = ± e i , T ( e j ) ² are also possible.) Date : January 12, 2012. S. Kaliszewski, Department of Mathematics, Arizona State University. 1...
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This document was uploaded on 03/11/2012.

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