Exercise 6.3.1

Proof. Suppose the dimension of $A$ is $m\times n$. It is known that Ker $A$ = Ker $(A^{\text{∗}}A)$. Then Rank $A$ = $n-$ dim Ker $A$ = n - dim Ker $(A^{\text{∗}}A)$ = Rank $(A^{\text{∗}}A)$. The SVD of $A$ is $A=W\mathrm{\Sigma }{V}^{\text{∗}}$, then $A^{\text{∗}}A=V{\mathrm{\Sigma }}^{\text{∗}}{W}^{\text{∗}}W\mathrm{\Sigma }{V}^{\text{∗}}=V{\mathrm{\Sigma }}^2{V}^{\text{∗}}$. Then Rank $A$ = Rank $(A^{\text{∗}}A)$ = Rank $(V{\mathrm{\Sigma }}^2{V}^{\text{∗}})$ = Rank ${\mathrm{\Sigma }}^2$ = number of non-zero singular values, because $V$ is an orthogonal matrix (full rank). □