Exercise 2.2.2

For all matrices, we have $A=U\mathrm{\Sigma }{V}^T$, and $A^{\text{+}}=V{\mathrm{\Sigma }}^{\text{+}}{U}^T$, the ranks of $\mathrm{\Sigma }$ and ${\mathrm{\Sigma }}^{\text{+}}$ are the same, and the ranks of $A$ and $A^{\text{+}}$ are the same as $\mathrm{\Sigma }$ and ${\mathrm{\Sigma }}^{\text{+}}$ respectively, so $A$ and $A^{\text{+}}$ have the same rank as well.

If $A$ is square, the non-zero singular values of $A^{\text{+}}$ are reciprocals of the non-zero singular values of $A$.

The eigenvectors and eigenvalues of $A$ and $A^{\text{+}}$ are different.