Matrix as a Multiplication of Non-Zero Vectors

Question

Let $u$ and $v$ be nonzero vectors in $R^{n*1}$. Is it possible to conclude that if a matrix $A$ is an $n*n$ matrix of rank $1$, then there are always vectors $u$ and $v$ in $R^{n*1}$ such that $A = uv^T$?

Here is my attempt:

We know that the size of matrix $uv^T$ is $n*n$ by definition of matrix multiplication. Furthermore, we know it has rank $1$, because as a matrix that results from the multiplication of $n*1$ and $1*n$ vectors, it follows that its $n-1$ rows not including the first are simply a linear combination of the first row. In $RREF$, these rows are zero rows. So, we simply choose $v^T$ to be the first row, and $u$ to be the first column of the matrix $A$.

Is this correctly reasoned? Please let me know, I would appreciate the help. Thanks.

Answer 1

If the matrix has rank 1 it has a row echelon form consisting of one row vector $a \ne 0$ and otherwise $0$ rows.

Then the $i$-th row of $A$ is a multiple $\lambda_i$ of that row vector $a$, thinking of reversing the steps of getting to the row echelon form.

So we can have $$ A = \lambda \, a^T $$ where $\lambda = (\lambda_i)$ is the column vector of the multiples $\lambda_i$.

Answer 2

Your definitions of vectors $u$ and $v$ may be affected by the matrix having $0$ as entries.

Pick a non-zero entry in the array. This is possible since $A$ is of rank $1$ (if it was filled with $0$, then it would be of rank $0$).

Let $v^T$ be the row of that non-zero entry, and $u$ be the column of that non-zero entry divided by the non-zero entry.