Basis of linear transformations

[sid9221]

http://dl.dropbox.com/u/33103477/linear%20transformations.png

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.

 

[sunjin09]

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.

 

[HallsofIvy]

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

Since both equations have "-x_4", we can subtract one equation form the other to get x_1- 3x_2+ x_3= 0 or x_1= 3x_2- x_3. Replacing x_1 by that in either of the two equations, we will have an equation involving only x_2, x_3/itex], and x_4. We can solve that for x_4 in terms of x_2 and x_4 and so have a two dimensional space in terms of those two parameters.<br /> <br /> L_2 is defined by L(y_1,y_2)= (y_1- y_2, y_1- 3y_2, 2y_1- 8y_2, 3y_1- 27y_2)<br /> <br /> If we call that &quot;&lt;x, y, z, u&gt;&quot; we have x= y_1- y_2, y= y_1- 3y_2, z= 2y_1- 8y_2, and u= 3y_1- 27y_2. 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.