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Basis of linear transformations !

  1. Apr 19, 2012 #1
    http://dl.dropbox.com/u/33103477/linear%20transformations.png [Broken]

    My attempt was to first find the transformed matrices L1 and L2.

    L1= ---[3 1 2 -1]
    -------[2 4 1 -1]

    L2= ---[1 -1]
    -------[1 -3]
    -------[2 -8]
    -------[3 -27]

    Now reducing L1, I have

    -------[1 0 7/10 -3/10]
    -------[0 1 -1/10 -1/10]

    How do I find the kernel for this and how then how do I deduce the basis ?
    Also how do I find the image and then the basis for L2

    If you could just give the overview of the calculations required it would be helpful.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Apr 19, 2012 #2
    Kernel of L1 is {x} such that L1*x=0, find all x's such that L1*x=0 by mechanically solving the equation. Believe me there aren't that many other than x=[0,0,0,0]'.

    Image of L2 is even simpler, {y} such that y=L2*x for some x, so that {y} is the span of columns of L2.
     
  4. Apr 19, 2012 #3

    HallsofIvy

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    [itex]L_1[/itex] is defined by [itex]L_1(x_1, x_2, x_3 x_4)= (3x_1+ x_2+ 2x_3- x_4, 2x_1+ 4x_2+ x_3- x_4)[/itex]. The "kernel" of [itex]L_1[/itex], also called its "nullspace" is defined as the set or all [itex](x_1, x_2, x_3, x_4)[/itex] that are mapped into (0, 0). In other words, we must have [itex]3x_1+ x_2+ 2x_3- x_4= 0[/itex] and [itex]2x_1+ 4x_2+ x_3- x_4= 0[/itex]. Those are two equations in 4 variables- we can solve for two of them in terms of the other two.

    Since both equations have "[itex]-x_4[/itex]", we can subtract one equation form the other to get [itex]x_1- 3x_2+ x_3= 0[/itex] or [itex]x_1= 3x_2- x_3[/itex]. Replacing [itex]x_1[/itex] by that in either of the two equations, we will have an equation involving only [itex]x_2[/itex], [itex]x_3/itex], and [itex]x_4[/itex]. We can solve that for [itex]x_4[/itex] in terms of [itex]x_2[/itex] and [itex]x_4[/itex] and so have a two dimensional space in terms of those two parameters.

    [itex]L_2[/itex] is defined by [itex]L(y_1,y_2)= (y_1- y_2, y_1- 3y_2, 2y_1- 8y_2, 3y_1- 27y_2)[/itex]

    If we call that "<x, y, z, u>" we have [itex]x= y_1- y_2[/itex], [itex]y= y_1- 3y_2[/itex], [itex]z= 2y_1- 8y_2[/itex], and [itex]u= 3y_1- 27y_2[/itex]. Now we want to get relations among x, y, z, and u only. Subtracting y from x, x- y= 2y_2. Subtracting 2u from 3z, 3z- 2u= 30y_2= 14(x- y). That gives 3z- 2u= 14x- 14y so that 3z= 14x- 14y+ 2u or z= (14/3)x- (14/3)y+ (2/3)u.
     
    Last edited: Apr 19, 2012
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