Find the the nullity of T and give a geometric description of the kernel and Range of T fo rthe following linear transformation: T is the projectiononto the xy- coordinate plane: T(x,y,z) = (x,y,0)
T = [1 0 0 0 1 0 0 0 0] That sound right? If so, you need to find its null space
amistre what do u think?
thats a good kernel alright
well, a kernel creates Ax=0 tho right?
yup
1 0 0 0 1 0 0 0 0 doesnt augment any but it does have a free variable: x1 0 0 x = x2 = 0 + x3 0 x3 0 1 to which the kernel i believe is the vectors from this 0 0 nullA = span: 0 , 0 0 1
so the nullity is the kernel?
http://www.wolframalpha.com/input/?i=kernel+%7B%7B1%2C0%2C0%7D%2C%7B0%2C1%2C0%7D%2C%7B0%2C0%2C0%7D%7D nullity is a dimension of the null space; in this case i believe its 1
so in this case the nullity is 2
good old wolfram i forgot all abt that
no, since rankT is 2, and T has 3 columns; nullity is 1
ohhhh ya i see
i spose the 0,0,0 vector is pointless :) 0 nullA = 0 1
ok then it makes sense
and that makes a little sense since the null space is orthogonal to the column space. the colA is the span that creates the surface for xy; and the zaxis itself is the otrhogonal vector to it
okk got it thanks amistre
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