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matrix norm
For all column vectors x of dimension 2 x 1, can we say ||x||1
Solution
(b) The correct inequality is ||x||_inf <= ||x||_2 <= ||x||_1
let's verify that in 2x1 dimension vectors u=(x y)^T
||u||_2=sqrt(x^2+y^2)=sqrt( max|x|,|y|^2+ min|x|,|y| ^2) >= max|x|,|y| = ||u||_inf
||u||_1^2=(|x|+|y|)^2=x^2+y^2 + 2|x||y| >= x^2+y^2 = ||u||_2^2
(c) The correct answer is c)
remember ||A||_1 can be compute as max value of sum of column and ||A||_inf rows.
Take
A=
1 2
3 4
Here ||A||_1= 6 and ||A||_inf = 7, so ||A||_1 <= ||A||_inf here.
If we take the transpose matrix, we have the inverse result !
matrix norm For all column vectors x of dimension 2 x 1- can we say --.docx

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matrix norm For all column vectors x of dimension 2 x 1- can we say --.docx

  • 1. matrix norm For all column vectors x of dimension 2 x 1, can we say ||x||1 Solution (b) The correct inequality is ||x||_inf <= ||x||_2 <= ||x||_1 let's verify that in 2x1 dimension vectors u=(x y)^T ||u||_2=sqrt(x^2+y^2)=sqrt( max|x|,|y|^2+ min|x|,|y| ^2) >= max|x|,|y| = ||u||_inf ||u||_1^2=(|x|+|y|)^2=x^2+y^2 + 2|x||y| >= x^2+y^2 = ||u||_2^2 (c) The correct answer is c) remember ||A||_1 can be compute as max value of sum of column and ||A||_inf rows. Take A= 1 2 3 4 Here ||A||_1= 6 and ||A||_inf = 7, so ||A||_1 <= ||A||_inf here. If we take the transpose matrix, we have the inverse result !