Chapter 7.1, Problem 1PS

To determine

What are the ranks r for these matrices with entries i times j and i plus j? Write $A$ and $B$ as the sum of r pieces $u{v}^T$ of rank one.

Answer to Problem 1PS

Rank(A)=1

Rank(B)=2

Explanation of Solution

Given information:

   $A=\left(\begin{array}{ccc}1& 2& \begin{array}{cc}3& 4\end{array}\\ 2& 4& \begin{array}{cc}6& 8\end{array}\\ \begin{array}{c}3\\ 4\end{array}& \begin{array}{c}6\\ 8\end{array}& \begin{array}{c}\begin{array}{cc}9& 12\end{array}\\ \begin{array}{cc}12& 16\end{array}\end{array}\end{array}\right) B=\left(\begin{array}{ccc}2& 3& \begin{array}{cc}4& 5\end{array}\\ 3& 4& \begin{array}{cc}5& 6\end{array}\\ \begin{array}{c}4\\ 5\end{array}& \begin{array}{c}5\\ 6\end{array}& \begin{array}{c}\begin{array}{cc}6& 7\end{array}\\ \begin{array}{cc}7& 8\end{array}\end{array}\end{array}\right)$

Not requiring $u_1^T{u}_2=v_1^T{v}_2=0$.

Calculation:

   $A=\left(\begin{array}{ccc}1& 2& \begin{array}{cc}3& 4\end{array}\\ 2& 4& \begin{array}{cc}6& 8\end{array}\\ \begin{array}{c}3\\ 4\end{array}& \begin{array}{c}6\\ 8\end{array}& \begin{array}{c}\begin{array}{cc}9& 12\end{array}\\ \begin{array}{cc}12& 16\end{array}\end{array}\end{array}\right)=\left(\begin{array}{c}1\\ 2\\ \begin{array}{c}3\\ 4\end{array}\end{array}\right)\left(\begin{array}{ccc}1& 2& \begin{array}{cc}3& 4\end{array}\end{array}\right)$

   $A=u{v}^T$ has rank 1 with $u^T=v^T=\left(\begin{array}{ccc}1& 2& \begin{array}{cc}3& 4\end{array}\end{array}\right)$

   ${\left|\left|u\right|\right|}^2=1^2+2^2+3^2+4^2=1+4+9+16=30$

Similarly, ${\left|\left|v\right|\right|}^2=30$.

So, the singular value decomposition has a division by $\sqrt{30}$ to reach $u_1$ and $v_1$.

Multiply by ${\sigma }_1=30$ to recover A.

Therefore, $A={\sigma }_1{u}_1{v}_1^T=30\frac{u}{\sqrt{30}}\frac{v^T}{\sqrt{30}}=u\text{Σ}{v}^T$.

Now consider the matrix $B=\left(\begin{array}{ccc}2& 3& \begin{array}{cc}4& 5\end{array}\\ 3& 4& \begin{array}{cc}5& 6\end{array}\\ \begin{array}{c}4\\ 5\end{array}& \begin{array}{c}5\\ 6\end{array}& \begin{array}{c}\begin{array}{cc}6& 7\end{array}\\ \begin{array}{cc}7& 8\end{array}\end{array}\end{array}\right)$

First two columns of $B$ are independent (pivot columns)

   $\begin{array}{l}col\left(3\right)=2\left(col2\right)–col\left(1\right)\hfill \\ col\left(4\right)=3\left(col2\right)–2col\left(1\right)\hfill \end{array}$

Therefore, $rank\left(B\right)=2$.

   $B=\left(\begin{array}{ccc}2& 3& \begin{array}{cc}4& 5\end{array}\\ 3& 4& \begin{array}{cc}5& 6\end{array}\\ \begin{array}{c}4\\ 5\end{array}& \begin{array}{c}5\\ 6\end{array}& \begin{array}{c}\begin{array}{cc}6& 7\end{array}\\ \begin{array}{cc}7& 8\end{array}\end{array}\end{array}\right)=\left(\begin{array}{cc}2& 3\\ 3& 4\\ \begin{array}{c}4\\ 5\end{array}& \begin{array}{c}5\\ 6\end{array}\end{array}\right)\left(\begin{array}{ccc}1& 0& \begin{array}{cc}-1& -2\end{array}\\ 0& 1& \begin{array}{cc}2& 3\end{array}\end{array}\right)$

Those pivot columns don't give the u's and v's of the SVD. For that we need eigenvalues and eigenvectors of $B^{T}B$ and $B{B}^T.$