Is the result of a matrix transformation equivalent to the that of that same matrix but orthonormalized?Say we have the transformation matrix A of full rank such that A − 1 ≠ A t ,, i.e., the matrix A consists of linearly independent vectors which aren't orthogonal to each other, a vector v, and the orthonormalized transformation matrix A ′ . Is it true that A v = A ′ v ? And if not, is this unimportant?

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Is the result of a matrix transformation equivalent to the that of that same matrix but orthonormalized?Say we have the transformation matrix   A of full rank such that       A          −      1        ≠      A    t    ,,  i.e., the matrix   A consists of linearly independent vectors which aren&#039;t orthogonal to each other, a vector v, and the orthonormalized transformation matrix       A    ′    . Is it true that   A  v  =      A    ′    v  ? And if not, is this unimportant?

Karla Hull · Accepted Answer

One of the key features of the Gram-Schmidt process is that the vectors generated by it span the same   k-vector subspace as the original vectors. Put into the language of matrices, this means that the column space of   A (or the range of   T) is the same as the column space of       A    ′   (or the range of       T    ′  ). (Here, I assume that the linearly independent vectors we started with were the columns of   A  ..) Consequently, we have that  {  w  ∈  V    |    T  (  v  )  =  w   for some   v  ∈  V  }  =  {  w  ∈  V    |        T    ′    (  v  )  =  w   for some   v  ∈  V  }  .But that does not imply that   A  v  =  T  (  v  )  =      T    ′    (  v  )  =      A    ′    v for all   v  ∈  V  ..For instance, consider the   R -vector space         R    2    ,  i.e., the set of column vectors   (  a  ,  b      )    t   for   a  ,  b  ∈  R  . Observe that the vectors   v  =  (  1  ,  1      )    t   and   w  =  (  1  ,  2      )    t   are linearly independent (because the matrix with columns   v and   w has determinant 1). By Gram-Schmidt, we obtain an orthogonal basis       v    1    =  (  1  ,  1      )    t   and       v    2    =  (  −      1    2    ,      1    2        )    t    , and we can convert this into an orthonormal basis by taking       v    ′    =            v      1                  |          |              v        1                |          |       and       w    ′    =            v      2                  |          |              v        2                |          |        .Using these as columns, we can form the   2  ×  2 matrix       A    ′    . We will denote by       e    1    =  (  1  ,  0  ) and       e    2    =  (  0  ,  1  ) the standard basis vectors of         R    2    .  Observe that we have  A      e    1    =  v  ≠            v      1                  |          |              v        1                |          |        =      v    ′    =      A    ′        e    1    ,and the analogous statement holds for       e    2   in place of       e    1    .

Is the result of a matrix transformation equivalent to the that of that same matrix but orthonormali

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