Let I: V->V be the identity operator on an n-dimensional vector space V defined by I(v)=v for every v in V. Show that the matrix of I with respect to a basis S for V is .
Well it is the identity. So in particular, it takes the first basis element to the first basis element ( ), the second basis element to the second basis element ( ), etc.
Thus the first column is [1,0,0...0]
The second Column is [0,1,0,...0]
The nth column is [0,0,...,1]
So and for