Find a set of vectors that span the image of the linear transformation.
Find a set of vectors that span the kernel of the linear transformation
rref the following:
[1 2 1 ]
[3 -2 4 ]
[1 -6 2 ]
[1 2 1]
[0 -8 1]
[0 -8 1]
[1 2 1]
[0 -8 1]
[0 0 0]
[4 0 5]
[0 -8 1]
[0 0 0]
[1 0 5/4]
[0 1 -1/8]
[0 0 0]
Column 1 and Column 2 present leading one's, so they are considered as vectors that span the image of the linear transformation.
[1] (correspond to)
[0] vector{1,3,1}
[0]
[0] (correspond to)
[1] vector{2,-2,-6}
[0]
Whereas, column 3 demonstrate a linear relation with both vector 1 and vector 2, meaning it span the kernel of linear transformation.
[ 5/4 ]
[-1/8 ] ----> $x_{3} $ = (5/4) $x_{1} $ + (-1/8) $x_{2} $
[ 0 ] 0 = (5/4) $x_{1} $ + (-1/8) $x_{2} $ - $x_{3} $
using the equation derived from linear relation, we can rewrite it as a vector that span the kernel
vector( $/frac{5}/{4}$ , $/frac {-1}/{8}$ , 1}
Ruiqi Huang
Disc 3B |