Exercise 6.10.11

If $A=\mathit{kB}$, then we have $A^{\text{∗}}A=k^{2}I$. So all the eigenvalues of $A^{\text{∗}}A$ are $k^2$. Thus we have $\parallel A\parallel =k$ and $\parallel A^{-1}\parallel =k^{-1}$. Hence the condition number of $A$ is $k\cdot k^{-1}=1$.

Conversely, if $\mathrm{cond}<!--\; FUNCTION\; APPLICATION\; -->(A)=1$, we have ${\lambda }_1={\lambda }_n$ by Theorem 6.44. This means that all the eigenvalues of $A^{\text{∗}}A$ are the same. Denote the value of these eigenvalue by $k$. Since $A^{\text{∗}}A$ is self-adjoint, we could find an orthonormal basis $\beta =\left\{v_i\right\}$ consisting of eigenvectors. But this means that

for all $i$. Since $\beta$ is a basis, we get that actually $A^{\text{∗}}A=\mathit{kI}$. This means that $B=\frac{1}{\sqrt{k}}A$ is unitary of orthogonal since $B^{\text{∗}}B=I$. Thus $A$ is a scalar multiple of $B$.