2
$\begingroup$

enter image description here What will be the dimension .seems easy but answer doesn't match mine

  • 3
    Show us your solution. – 2017-01-20
  • 1
    And preferably, give some explanation to how you arrived at your solution. – 2017-01-20
  • 0
    two. your turn. – 2017-01-20
  • 1
    Dim R^4= Dim U +DimW-Dim ( U internation W) which give 4 = 3+3-x(say) which gives x= 2. – 2017-01-20
  • 0
    Remember that the dimension of the intersection of subspaces is not the cardinality of the intersection of their bases. – 2017-01-20
  • 0
    Can u specify where I went wrong above – 2017-01-20
  • 0
    http://math.stackexchange.com/questions/2105340/how-many-subspaces-are-invariant-under-this-linear-map seems that the same question is being asked. – 2017-01-20

1 Answers 1

1

hint

You have two basis on the problem statement. For each of them, replace the second vector with a copy of itself minus the first one; replace the third one with itself minus the second one. Does this make the answer clearer?

  • 1
    Not that comfortable – 2017-01-20
  • 2
    Doing so I get two common vectors (0,1,0,0) and (0,0,1,0) .so this two will give the dim of the intersection of U and W? – 2017-01-20
  • 0
    Rserrao do you mean after the manipulation the cardinality of common vectors will be the dimension of the intersection? – 2017-01-20
  • 0
    @Khamba exactly. Note however that you can't always do this trick. But in this exercise it is clear that the $(1,0,0,0) $ from $U $ isn't in $V $ and that $(0,0,0,1) $ from $V $ isn't in $U $. – 2017-01-20
  • 0
    @shadowkh yes I do, but that isn't always the best method. It works this time because we can easily manipulate the vectors from each basis to give the vectors that are common to each basis. Generally one has to perform gaussian elimination and/or make use of the fact that $dim U + dim V = dim U\cup V - dim U\cap V $ – 2017-01-20