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.(adsbygoogle = window.adsbygoogle || []).push({});

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# Diagonalizability and Invertibility

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