SVD and $\min ||A - B||$

Question

I have matrix $A$ and its SVD: $$ \begin{pmatrix} 11 & 14 & -4 \\ -8 & -2 & 7 \end{pmatrix} = \begin{pmatrix} -\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{pmatrix} \begin{pmatrix} 405 & 0 \\ 0 & 45 \end{pmatrix} \begin{pmatrix} -\frac{2}{27\sqrt{5}} & -\frac{2}{27\sqrt{5}} & \frac{1}{27\sqrt{5}} \\ -\frac{1}{9\sqrt{5}} & \frac{2}{9\sqrt{5}} & \frac{2}{9\sqrt{5}} \end{pmatrix} $$ I am asked to find B of rank 1 for which $||A - B||$ is minimal, where $||\cdot||$ is Frobenius norm. Am i right, that $$ B = \begin{pmatrix} -\frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}} \end{pmatrix} \begin{pmatrix} 405 \end{pmatrix} \begin{pmatrix} -\frac{2}{27\sqrt{5}} & -\frac{2}{27\sqrt{5}} & \frac{1}{27\sqrt{5}} \\ \end{pmatrix}? $$

Answer 1

Yes -- this is a simple application of the Eckart-Young theorem.