Finding ker ⁡ [ A ] T Let A = ( 1 1 − 1 2 1 1 0 3 − 1 0 1 0 ) .I have to calculate ker ⁡ [ A ] and ker ⁡ [ A ] T I proved that dim ⁡ ( I m [ A ] ) = 3 with basis [ 1 1 − 1 ] , [ 1 1 0 ] , [ − 1 0 1 ] and dim ⁡ ( ker ⁡ [ A ] ) = 1 with basis [ − 1 − 2 − 1 1 ] ,but i have problems with ker ⁡ [ A ] T . Do i have to calculate A T and then, with that matrix, finding the ker?

Question

Finding   ker  ⁡  [  A      ]          T      Let  A  =      (                            1                          1                          −          1                          2                                      1                          1                          0                          3                                      −          1                          0                          1                          0                      )    .I have to calculate   ker  ⁡  [  A  ] and   ker  ⁡  [  A      ]          T      I proved that   dim  ⁡  (      I    m    [  A  ]  )  =  3 with basis      [                            1                                      1                                      −          1                      ]    ,      [                            1                                      1                                      0                      ]    ,      [                            −          1                                      0                                      1                      ]  and   dim  ⁡  (  ker  ⁡  [  A  ]  )  =  1 with basis      [                            −          1                                      −          2                                      −          1                                      1                      ]    ,but i have problems with   ker  ⁡  [  A      ]    T  . Do i have to calculate       A    T   and then, with that matrix, finding the ker?

riveExedeer · Accepted Answer

There are a couple of ways to think about this. The most obvious is to take the transpose, then try to find the kernel, but this is the most inefficient way. The comments lay out another path, so let me add one more. If you know about fundamental subspaces, then you might know that   Ker  (      A    T    )  =  (  Im  (  A  )      )    ⊥   (where this orthogonal complement is in             R        3  ). As you stated, the dimension of the image is 3, which tells you that   Ker  (      A    T    )  =  {  0  }  .As the comments state, we can also think about it in the following way: the transpose is a map       A    T    :            R        3    →            R        4    , and since   3  =  rank  (  A  )  =  rank  (      A    T    )  ,, we know by the rank-nullity theorem that   nullity  (      A    T    )  =  dim  (            R        3    )  −  rank  (      A    T    )  =  3  −  3  =  0.

Finding ker[A]^T Let A=((1,1,-1,-2),(1,1,0,3),(-1,0,1,0)) I have to calculate ker[A] and ker[A]^T.

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