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How we decompose or break matrix $A$ in the following form.

$$A = \begin{pmatrix} {a_1}^2 & a_1a_2 & ... a_1a_n \\ a_2a_1 & {a_2}^2 & ... a_2a_n \\.... \\ a_na_1 & a_na_2 & ... {a_n}^2 \end{pmatrix} = \begin{pmatrix} a_1 & a_1 & ... a_1 \\ a_2 & a_2 & ... a_2 \\.... \\ a_n & a_n & ... a_n\end{pmatrix}×\begin{pmatrix} a_1 & a_1 & ... a_1 \\ a_2 & a_2 & ... a_2 \\.... \\ a_n & a_n & ... a_n\end{pmatrix}$$

My Attempt:

Taking RHS = $$\begin{pmatrix} a_1 & a_1 & ... a_1 \\ a_2 & a_2 & ... a_2 \\.... \\ a_n & a_n & ... a_n\end{pmatrix}×\begin{pmatrix} a_1 & a_1 & ... a_1 \\ a_2 & a_2 & ... a_2 \\.... \\ a_n & a_n & ... a_n\end{pmatrix} = \begin{pmatrix} {a_1}^2+a_1a_2+...+a_1a_n & {a_1}^2+a_1a_2+...+a_1a_n & ... {a_1}^2+a_1a_2+...+a_1a_n \\ a_2a_1+{a_2}^2+...+a_1a_n & a_2a_1+{a_2}^2+...+a_1a_n & ... a_2a_1+{a_2}^2+...+a_1a_n \\.... \\ a_na_1+a_na_2+...+{a_n}^2 & a_na_1+a_na_2+...+{a_n}^2 & ... a_na_1+a_na_2+...+{a_n}^2 \end{pmatrix} $$

I couldn't proceed it further. Please help me.

Largest Prime
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1 Answers1

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The decomposition you post is wrong. A simple counterexample: Let $n=2$, $a_1=1$ and $a_2=0$. We have $$\begin{pmatrix}a_1 &a_1\\ a_2 & a_2\end{pmatrix}\begin{pmatrix}a_1 &a_1\\ a_2 & a_2\end{pmatrix}=\begin{pmatrix}1 &1\\ 0 & 0\end{pmatrix}\begin{pmatrix}1 &1\\ 0 & 0\end{pmatrix}=\begin{pmatrix}1 &1\\ 0 & 0\end{pmatrix}\neq\begin{pmatrix}1 &0\\ 0 & 0\end{pmatrix}=\begin{pmatrix}a_1^2 &a_2\\ a_2a_1 & a_2^2\end{pmatrix}.$$

A correct decomposition is $$A=\begin{pmatrix} a_1^2 & a_1a_2 & \dots & a_1a_n\\ a_2a_1 & a_2^2 &\dots & a_2a_n\\ \vdots & \vdots & \ddots & \vdots\\ a_na_1 & a_na_2 &\dots & a_n^2\end{pmatrix}=\begin{pmatrix} a_1 & 0 & \dots &0\\ 0& a_2 &\dots &0\\ \vdots & \vdots & \ddots & \vdots\\ 0& 0 &\dots & a_n\end{pmatrix}\begin{pmatrix} a_1 & a_2 & \dots & a_n\\ a_1 & a_2 &\dots & a_n\\ \vdots & \vdots & \ddots & \vdots\\ a_1 & a_2 &\dots & a_n\end{pmatrix}.$$ To obtain the above decomposition, simply note that all elements in the $j$-th row have a common factor $a_j$, for all $j=1,\dots,n$.

Or noting that all elements on the $j$-th column of $A$ have a common factor $a_j$, which yields $$A=\begin{pmatrix} a_1^2 & a_1a_2 & \dots & a_1a_n\\ a_2a_1 & a_2^2 &\dots & a_2a_n\\ \vdots & \vdots & \ddots & \vdots\\ a_na_1 & a_na_2 &\dots & a_n^2\end{pmatrix}=\begin{pmatrix} a_1 & a_1 & \dots & a_1\\ a_2 & a_2 &\dots & a_2\\ \vdots & \vdots & \ddots & \vdots\\ a_n & a_n &\dots & a_n\end{pmatrix}\begin{pmatrix} a_1 & 0 & \dots &0\\ 0& a_2 &\dots &0\\ \vdots & \vdots & \ddots & \vdots\\ 0& 0 &\dots & a_n\end{pmatrix}.$$

In fact, the second decomposition also follows from the first one right away, as $A$ is symmetric, i.e., $A=A^T$.

Frank Lu
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