1.
A=⎝⎛121110−110141⎠⎞
for null space: Ax=0
so, we have:
x1=−x4
x2=x3
−2x1+2x2=0⟹x1=x2
solution:
x=⎝⎛x1x2x3x4⎠⎞=⎝⎛x1x1x1−x1⎠⎞=x1⎝⎛111−1⎠⎞
then basis for the null space:
⎝⎛111−1⎠⎞
2.
⎝⎛1111−10002⎠⎞⎝⎛xyz⎠⎞=⎝⎛x+yx−yx+2z⎠⎞
⎝⎛1111−10002⎠⎞→⎝⎛121100002⎠⎞
basis of range T:
⎝⎛121⎠⎞,⎝⎛100⎠⎞,⎝⎛002⎠⎞
3.
i)
Transforming matrix T ;
⎝⎛2xx+yx−z⎠⎞ =x⎝⎛211⎠⎞ +y⎝⎛010⎠⎞ +z⎝⎛00−1⎠⎞
T=⎝⎛21101000−1⎠⎞
T∗=TT=⎝⎛20011010−1⎠⎞
⎝⎛20011010−1⎠⎞⎝⎛uvw⎠⎞=⎝⎛2u+v+wv−w⎠⎞
Therefore the adjoint operator T∗(u,v,w)
=(2u+v+w,v,−w)
Comments