
Eigenvectors/values
Let A = and by T(X) = AX.
Find eigenpairs for A, find a basis for Ker(T) using the eigenvectors previously found, find a basis for Im(T) using eigenvectors previously found.
The eigenpairs I got were...
(0, ), (1, ), and (2,
I wasn't sure about the Kernel or Image parts but for the kernel I got
t
And for the image I got
Could someone please verify if this is correct or not? I assume the kernel and image are not so could someone also please help me with that?
Thanks

Yes, that is correct. Your matrix is 3 by 3 so the base space is 3 dimensional and you have three independent eigenvectors. Since there are 3 and they are independent (eigevectors corresponding to distinct eigenvalues are always independent), they form a basis for the space. That means that any vector v, can be written as .
Applying A to that, .
That is, Av= 0 if and only if so is a basis for the kernel of A and so is a basis for the image.