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If a $m \times n$ matrix has rank $1$, does it imply that it can be written as a product of one $m\times1$ and one $1\times n$ matrix. How to prove it ? Is this decomposition unique ? What are the entries of these matrices ?

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In a matrix of rank 1, every row is a scalar multiple of every other non-zero row (and there must be at least one non-zero row). For the $1 \times n$ matrix, take one of the non-zero rows; in the $m \times 1$ matrix, put the scalars from the first sentence. The decomposition is not unique: you can pick every non-zero scalar multiple of a non-zero row to start with.

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Yes, it is always possible. Let

$$A = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_m \end{bmatrix}$$

be a matrix of rank $1$, and let $a_i$ be its rows. Since $A$ is of rank $1$, each two of these rows are linearly dependent. Let us assume that $a_k \ne 0$ for some $k$ (such row must exist, or $A$ would be a zero matrix). Then for each $i$,

$$a_i = \alpha_i a_k,$$

for some (real or complex) numbers $\alpha_i$. Define

$$v = \begin{bmatrix} \alpha_1 \\ \vdots \\ \alpha_m \end{bmatrix}, \quad w := a_k.$$

Then $A = v w$.

Obviously, the decomposition is not unique. Let $\xi \not \in \{0,1\}$. Then $A = (\xi v)(\xi^{-1} w)$ is another such decomposition.

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Let $A = [a_1\ \cdots\ a_n]$ be the $m\times n$ where the $a_i$ are $m$-dimensional column vectors. If $A$ has rank $1$, then $\dim\operatorname{Col} = 1$. Therefore $\dim\,\operatorname{span}\{a_1, \dots, a_n\} = 1$. Choose $a_j \neq 0$ (such a column exists otherwise the $A = O$ which has rank zero), then $\operatorname{span}\{a_1, \dots, a_n\} = \operatorname{span}\{a_j\}$. In particular, $a_i \in \operatorname{span}\{a_j\}$ for $i = 1, \dots, n$. Therefore $a_i = c_ia_j$ for some $c_i \in \mathbb{R}$. Then

$$A = [a_1\ \cdots\ a_n] = [c_1a_j\ \cdots\ c_na_j] = a_j[c_1\ \cdots c_n].$$

Keep in mind, there is more to the last equality than first appears; $a_j$ is column vector not a constant.

More generally, an $m\times n$ matrix $A$ has rank $k$ if and only if the smallest $r$ such that $A = u_1v_1^T + \dots + u_rv_r^T$, where $u_1, \dots, u_r \in \mathbb{R}^m$, $v_1, \dots, v_r \in \mathbb{R}^n$, is $r = k$.

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The other answers have addressed the issue of existence quite well. Let me show that the decomposition is "essentially unique", i.e. up to scalar multiple of the two vectors.

Suppose that $A=\mathbf{x}_1\mathbf{y}_1^\mathrm{T} = \mathbf{x}_2\mathbf{y}_2^\mathrm{T}$. Note that $A\mathbf{v}=c\mathbf{w}$ for some vector $\mathbf{w}$ and some scalar $c$ since the image is one dimensional. Thus we have $$A\mathbf{y}_1 = \mathbf{x}_1\mathbf{y}_1^\mathrm{T}\mathbf{y}_1=\|\mathbf{y}_1\|^2\mathbf{x}_1=c_1\mathbf{w}$$ $$A\mathbf{y}_2 = \mathbf{x}_2\mathbf{y}_2^\mathrm{T}\mathbf{y}_2=\|\mathbf{y}_2\|^2\mathbf{x}_2=c_2\mathbf{w}$$ Therefore $\mathbf{x}_1$ and $\mathbf{x}_2$ are multiples of the same vector $\mathbf{w}$. The argument can be repeated for $\mathbf{y}_1$ and $\mathbf{y}_2$ by taking the transpose and repeating the argument.

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