Show the rank of A equals to the number of non-zero eigenvalues of $A^TA$

Question

A is an $n\times n$ matrix with real entries. How to show that the rank of A equals to the number of non-zero eigenvalues of $A^TA$

Answer 1

It depends on what information you have available.

• If you have the singular value decomposition $A=VDU$, with $V,U$ orthogonal and $D$ diagonal, you have $$ A^TA=U^TDV^TVDU=U^TD^2U,$$ and since conjugation with an orthogonal preserves eigenvalues, the eigenvalues of $A^TA$ are precisely the squares of the diagonal entries of $D$. As $D$ is diagonal, its rank is equal to the number of nonzero diagonal entries. And since $U,V$ are invertible, the rank of $A$ is equal to the rank of $D$.

• If you have the rank-nullity theorem, you can prove that $\ker A=\ker A^TA$, and so $A$ and $A^TA$ have the same rank. As $A^TA$ is symmetric, it is orthogonally diagonalizable (equivalently, it admits an orthonormal basis of eigenvectors, cfr. the spectral theorem) and so its rank agrees with the number of nonzero eigenvalues.