Page 31 - Vector Analysis

§1.7 Representation of Linear Transformations 27

mm

Assume max ÿ = ÿ for some 1ď k ď n. Let
1ďiďn |aij | |akj |

j=1 j=1

x = (sgn(ak1), sgn(ak2), ¨ ¨ ¨ , sgn(akn)) .

m

Then }x}8 = 1, and }Ax}8 = ř |akj|.

j=1

On the other hand, if }x}8 = 1, then

m ! mm m )
|anj| ;
ÿ max ÿ ÿ ÿ

|ai1x1 + ai2x2 + ¨ ¨ ¨ aimxm| ď |aij | ď |a1j |, |a2j |, ¨ ¨ ¨

j=1 j=1 j=1 j=1

! mm m )
|.
thus max ÿ ÿ ÿ In other words, is the largest

}A}8 = |a1j |, |a2j |, ¨ ¨ ¨ |anj }A}8

j=1 j=1 j=1

sum of the absolute value of row entries.

nn n
= max ! ÿ |ai1|, ÿ |ai2|, ¨ ¨ ¨ )
2. p = 1: ÿ |aim| .

}A}1 ,

i=1 i=1 i=1

Reason: Let (¨, ¨) denote the inner product in Rm. Then for x P Rn with }x}1 = 1, by

Hölder’s inequality (1.1) and Theorem 1.25 we have

}Ax}1 = sup (Ax, y) = sup (x, ATy) ď sup }x}1}ATy}8

}y}8=1 }y}8=1 }y}8=1

= sup }ATy}8 = }AT}8 ;

}y}8=1

thus }A}1 = sup }Ax}1 ď }AT}8. Similarly, if y P Rm and }y}8 = 1, then

}x}1=1

}ATy}8 = sup (x, ATy) = sup (Ax, y) ď sup }Ax}1}y}8

}x}1=1 }x}1=1 }x}1=1

= sup }Ax}1 = }A}1

}x}1=1

which implies that }AT}8 = sup }ATy}8 ď }A}1. As a consequence,

}y}8=1

! nn n )
|aim|
}AT}8 max ÿ ÿ ÿ

}A}1 = = |ai1|, |ai2|, ¨ ¨ ¨ , .

i=1 i=1 i=1