Consider K n , K m , both with the | | . | | 1 -norm, where K = R or C .Let | | T | | = i n f { M ≥ 0 : | | T ( x ) | | ≤ M | | x | | ∀ x ∈ K n } be the operator norm of a linear transformation T : K n → K m .Show that the operator norm of the linear transformation T is also given by: | | T | | = m a x { ∑ i = 1 m | a i j | , 1 ≤ j ≤ n } =: | | A | | 1 where A is the transformation matrix of T and a i j it's entry in the i-th row and j-the column.

Question

Consider             K        n  ,             K        m  , both with the       |        |    .      |              |        1  -norm, where       K    =      R   or       C  .Let       |        |    T      |        |    =  i  n  f  {  M  ≥  0  :      |        |    T  (  x  )      |        |    ≤  M      |        |    x      |        |       ∀  x  ∈            K        n    } be the operator norm of a linear transformation   T  :            K        n    →            K        m  .Show that the operator norm of the linear transformation   T is also given by:      |        |    T      |        |    =  m  a  x  {      ∑          i      =      1        m        |        a          i      j            |    ,  1  ≤  j  ≤  n  }  =:      |        |    A      |              |        1  where   A is the transformation matrix of   T and       a          i      j       it&#039;s entry in the   i-th row and   j-the column.

Penelope Powers · Accepted Answer

Note that for any   x  ∈            K        n    ‖  T  (  x  )      ‖    1    =      ∑          i      =      1        m        |    T  (  x      )    i        |    ≤      ∑          i      =      1        m        ∑          j      =      1        n        |        a          i      ,      j            |        |        x    j        |      (  ∗  )For   1  ≤  j  ≤  n, set      α    j    =      ∑          i      =      1        m        |        a          i      ,      j            |  Then rearrange (∗) to get  ‖  T  (  x  )      ‖    1    ≤      ∑          j      =      1        n        α    j        |        x    j        |    ≤  ‖  A      ‖    1    ‖  x      ‖    1  This proves that   ‖  T  ‖  ≤  ‖  A      ‖    1  .For the reverse inequality assume first that       a          i      ,      j        ≥  0 for all   i  ,  j , and suppose      α    1    =      ∑          i      =      1        m        a          i      ,      1        =  ‖  A      ‖    1  Then for   x  =  (  1  ,  0  ,  0  ,  …  ,  0  ) we have   ‖  x      ‖    1    =  1 and  T  (  x  )  =  (      a          1      ,      1        ,      a          2      ,      1        ,  …  ,      a          m      ,      1        )  ⇒  ‖  T  (  x  )      ‖    1    =  ‖  A      ‖    1  and this prove that   ‖  T  ‖  ≥  ‖  A      ‖    1  .

Consider K_n, K_m, both with the ||.||1-norm, where K=R or C. Let ||T||=inf{M>=0:||T(x)||<=M||x|| ∀x∈K_n} be the operator norm of a linear transformation T:K_n->K_m

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