# 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?

## 1 Answer

No, just cause it has linearly independent eigenvectors doesn't mean it's invertible / full rank. This tells you nothing about if any of those eigenvectors has eigenvalue zero.

In the diagonalizable case here, the subspace spanned by all the eigenvectors that have eigenvalue zero is the kernel. If the kernel is non trivial then the matrix is not invertible and is less than full rank.

If none of them have eigenvalue zero, then yes, it's invertible.