線性代數之basis of the image/kernel

(1) A;R^2 to R^2. If [x,y]^T is in kernel of A then x=0, and y=0 . i.e Kernel of A=[0,0]^T, the zero vector. detA not 0 implies dim (image of A)=2 so {[0,1]^T,[1,0]^T}, the usual R^2 basis, forms a basis for image A. The notation [ V ]^T means the transpose of [V].

(2) A:R^3 to R^1=R, If [x,y,z]T is in kernek of A then y+2z=0 so {[x,-2z,z]^T }are in kernel of A, which has a basis {[1,0,0]^T, [0,-2,1]^T}. The image of A is R, which has a basis {[1]}.

(3) A:R^3 to R^3. det A not zero so if [x,y,z]^T is in kernek of A then [x,y,z]=[0,0,0]. Similar to (1), A basis of image of A can be selected as {[1,0,0]^T, [0,1,0]^T, [0,0,1]^T}.