- #1
JG89
- 728
- 1
Let T be a linear operator on an n-dimensional vector space V. Am I correct in saying that if T is diagonalizable then T is invertible? My reasoning is that if T is diagonalizable then there is an ordered basis of eigenvectors of T such that T's matrix representation is a diagonal matrix. Obviously the rank of the n*n diagonal matrix is n and so the rank of T is n. Now, from a well known theorem we have nullity(T) + rank(T) = dim(V), and since rank(T) = dim(V), we see that nullity(T) = 0, implying that T is both one-to-one and onto, and so T is invertible.