Problem for the Olympiad. Prove that a non-zero square matrix has rank 1 if and only if it is a product of a column and a row (from left to right).
Proof. Let the nonzero square matrix
![\[C=\left( \begin{matrix} {{a}_{1}} \\ ... \\ {{a}_{n}} \\ \end{matrix} \right)\left( \begin{matrix} {{b}_{1}} & ... & {{b}_{n}} \\ \end{matrix} \right).\]](https://www.maturin.in.ua/wp-content/ql-cache/quicklatex.com-ae48b59d723dd0aae67123e21fe7b336_l3.png "Rendered by QuickLaTeX.com")
Тоді
![\[C=\left( \begin{matrix} {{a}_{1}}{{b}_{1}} & ... & {{a}_{1}}{{b}_{n}} \\ ... & ... & ... \\ {{a}_{n}}{{b}_{1}} & ... & {{a}_{n}}{{b}_{n}} \\ \end{matrix} \right).\]](https://www.maturin.in.ua/wp-content/ql-cache/quicklatex.com-99bc6060e5a6a4ece9cbc49762eb71eb_l3.png "Rendered by QuickLaTeX.com")
Without limiting the generality, we will assume that
![\[{{a}_{1}}\ne 0.\]](https://www.maturin.in.ua/wp-content/ql-cache/quicklatex.com-3a27a4ebd87fcb69c9ee4acd966c0942_l3.png "Rendered by QuickLaTeX.com")
Then it is obvious that the first row of the matrix C is non-zero, and each line of it is the product of the first line and some number. Therefore, the rank C is equal to 1.
On the contrary. Let the rank C is equal to 1. Then each row of such a matrix is the product of a fixed row and a number. Without limiting the generality, we can assume that this is the first row. Then C has the form
![\[C=\left( \begin{matrix} {{b}_{1}} & {{b}_{2}} & ... & {{b}_{n}} \\ {{a}_{2}}{{b}_{1}} & {{a}_{2}}{{b}_{2}} & ... & {{a}_{2}}{{b}_{n}} \\ ... & ... & ... & ... \\ {{a}_{n}}{{b}_{1}} & {{a}_{n}}{{b}_{2}} & ... & {{a}_{n}}{{b}_{n}} \\ \end{matrix} \right)=\left( \begin{matrix} 1 \\ {{a}_{2}} \\ ... \\ {{a}_{n}} \\ \end{matrix} \right)\left( \begin{matrix} {{b}_{1}} & {{b}_{2}} & ... & {{b}_{n}} \\ \end{matrix} \right).\]](https://www.maturin.in.ua/wp-content/ql-cache/quicklatex.com-6d6398b86380d45698ad3c4218294fb9_l3.png "Rendered by QuickLaTeX.com")