How to transform a symmetric $N\times N$ matrix in a rank-1 matrix?

Question

If $A$ is a symmetric matrix, exists a transformation like $U\cdot A\cdot V = B$ ?

Here $B$ is a rank-1 matrix, or, more precisely, $B = b\cdot b^T$, where $b$ is a $N\times 1$ matrix.

Answer 1

$B=bb^T$ has rank $1$ if $b \ne 0$. If $b=0$, then the rank of $B$ is $0$.

If you don't mind $B$ being $0$, then just choose $U=V=0$.

If you want $B$ to have rank $1$, first $A$ has to be non-zero.

Suppose $A=WDW^T$ where $W$ is orthogonal where $D_{11} > 0$. Let $W= [ w_1, \ldots, w_n]$.

We let $U=\begin{bmatrix} w_1^T\\0_{(n-1)\times n}\end{bmatrix}$ and $V=U^T$.

$$UAV=(UW)D(W^TU^T)$$

I will leave the following verification as an exercise:

• Try to evaluate $UW$ first and then you can evaluate $UAV$ and obtain your solution.

In the event that all the diagonal entries of $D$ are negative, rather then letting $V=U^T$, you can let $V=-U^T$.

Answer 2

Here's a method that will work for any symmetric $n \times n$ matrix $A$ over any field of characteristic $\ne 2.$ To avoid trivialities, we'll assume that $A$ is not the 0 matrix. Then the rank of $A$, which we'll say is $r$, is not 0. Let $P$ be an invertible matrix such that $P^TAP=\text {diag }(d_1,...,d_n.)$ Let $$d_i'=1/d_i \text { if }d_i\ne 0, 1 \text{ if }d_i=0.$$Let $$D'=\text {diag }(d_1',...,d_n').$$Let $$D''=D'P^TAP=\text {diag }(d_1'',...,d_n''.)$$ $D''$is a diagonal matrix having $r$ 1's on the diagonal and all other entries 0. Let $v$ be the row-vector $v=[d_1'',...,d_n''].$ Let $M$ be the $n \times n$ matrix having its $i$-th row as $v$ if $d_i''=1$ or the 0 vector otherwise. Then $MD''$ is a matrix in which the element in row $i$ and column $j$ is 1 if $d_i''=1$ or $d_j''=1$ and 0 otherwise. Thus $$MD''=MD'P^TAP=v^Tv.$$ So $$U=MD'P^T,V=P,b=v^T.$$