# For a square matrix $A$ presence of $n$ linearly independent eigenvectors implies full rank, thereby invertible?

For any matrix $A$, $AA^T$ is symmetric, and thereby diagonalizable (Spectral Theorem). This implies that $AA^T$ has n linearly independent eigenvectors and therefore full rank.

Does this imply that for any matrix $A$, $AA^T$ is always invertible ? I know there is something wrong here because from the above chain we can conclude that a symmetric matrix is always invertible, which is definitely not true.

Could someone please point out where I am going wrong?

• So does having $n$ LI eigenvectors tell us anything at all then? – information_interchange Mar 26 at 3:21
• @information_interchange Having $n$ LI eigenvectors is the same thing as being diagonalizable. – spaceisdarkgreen Mar 26 at 4:42