Let a = ( a 1 , a 2 , a 3 ) be a fixed vector in R 3 . Define the cross product a × v of a and another vector v = ( v 1 , v 2 , v 3 ) ∈ R 3 as a × v = det [ e 1 e 2 e 3 a 1 a 2 a 3 v 1 v 2 v 3 ] Define a function T : R 3 → R 3 by T ( v ) = a × v for v ∈ R 3 .a) Show that T is a matrix transformation and calculate its representing matrix Mb) Find ker ⁡ ( T ) and interpret its answer geometricallyc) find range ⁡ ( T ) and interpret its answer geometrically

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Let   a  =  (  a  1  ,  a  2  ,  a  3  ) be a fixed vector in         R    3  . Define the cross product   a  ×  v of   a and another vector   v  =  (      v    1    ,      v    2    ,      v    3    )  ∈        R    3   as  a  ×  v  =  det      [                                        e            1                                                e            2                                                e            3                                                            a            1                                                a            2                                                a            3                                                            v            1                                                v            2                                                v            3                                ]  Define a function   T  :        R    3    →        R    3   by   T  (  v  )  =  a  ×  v for   v  ∈        R    3  .a) Show that   T is a matrix transformation and calculate its representing matrix   Mb) Find   ker  ⁡  (  T  ) and interpret its answer geometricallyc) find   range  ⁡  (  T  ) and interpret its answer geometrically

g2joey15 · Accepted Answer

T  :        R    3    →        R    3    is given by   T  (  v  )  =  a  ×  v  =  det      [                                                  e            1                                                          e            2                                                          e            3                                                            a            1                                                a            2                                                a            3                                                            v            1                                                v            2                                                v            3                                ]   where   {        e    1    ,        e    2    ,        e    3    } are the standard basis vectors for         R    3  . For instance,         e    3    =  (  0  ,  0  ,  1  )Because   T is a linear transformation, if we can figure out how it transforms any basis then we can figure out how it transforms any vector -- and more importantly for this question, what its standard matrix is.So let&#039;s see how   T transforms the standard basis (being the easiest one to check, usually).                    T        (                          e          1                )        =              a        ×                          e          1                                    =        det                  [                                                                                      e                  1                                                                                              e                  2                                                                                              e                  3                                                                                                      a                  1                                                                              a                  2                                                                              a                  3                                                                                    1                                            0                                            0                                              ]                                                  =        det                  [                                                                      a                  2                                                                              a                  3                                                                                    0                                            0                                              ]                                  e          1                −        det                  [                                                                      a                  1                                                                              a                  3                                                                                    1                                            0                                              ]                                  e          2                +        det                  [                                                                      a                  1                                                                              a                  2                                                                                    1                                            0                                              ]                                  e          3                                                  =                  a          3                                  e          2                −                  a          2                                  e          3                    Likewise, you can confirm for yourself that  T  (        e    2    )  =  −      a    3          e    1    +      a    1          e    3      T  (        e    3    )  =      a    2          e    1    −      a    1          e    2  Then we just need to find the matrix   M such that   M      [                            1                                      0                                      0                      ]    =      [                            0                                                  a            3                                                −                      a            2                                ]   and the similar expressions for the other basis vectors. But   M      [                            1                                      0                                      0                      ]    is just the first column of   M (Can you see why?). Thus  M  =      [                            M                      [                                                            1                                                                              0                                                                              0                                                      ]                                    M                      [                                                            0                                                                              1                                                                              0                                                      ]                                    M                      [                                                            0                                                                              0                                                                              1                                                      ]                                ]    =      [                            0                          −                      a            3                                                a            2                                                            a            3                                    0                          −                      a            1                                                −                      a            2                                                a            1                                    0                      ]  Here&#039;s how I&#039;d find the kernel and range of   T.From the properties of the determinant we can see that   a  ×  v  =  0 iff   a  ‖  v. Thus   ker  ⁡  (  T  )  =        R    3   if   a  =  0 or   ker  ⁡  (  T  )  =  span  ⁡  (  a  ) if   a  ≠  0.Then we can use the rank-nullity theorem to get an idea of the range of   T. If   a  =  0 then we can immediately see that   range  ⁡  (  T  )  =  {  0  }. If   a  ≠  0, then the range is 2-dimensional.Let&#039;s assume   a  ≠  0 and consider the dot product of   a  ×  v with   a.  a  ⋅  (  a  ×  v  )  =  (      a    1          e    1    +      a    2          e    2    +      a    3          e    3    )  ⋅      (    det          [                                                  a              2                                                          a              3                                                                          v              2                                                          v              3                                          ]                  e      1        −    det          [                                                  a              1                                                          a              3                                                                          v              1                                                          v              3                                          ]                  e      2        +    det          [                                                  a              1                                                          a              2                                                                          v              1                                                          v              2                                          ]                  e      3        )      =  det      [                                        a            1                                                a            2                                                a            3                                                            a            1                                                a            2                                                a            3                                                            v            1                                                v            2                                                v            3                                ]    =  0Thus   T  (  v  )     ⊥     a. But we know that   span  ⁡  (  a      )    ⊥   is 2-dimensional, thus this must be the range of   T.Geometrically speaking this means that   ker  ⁡  (  T  ) is the line parallel to   a and   range  ⁡  (  T  ) is the plane orthogonal to that line.

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