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Any rank one matrix can be written in the form $uv^{t}$, where $u,v$ are column vectors considered as matrices and $t$ denotes transposition.

Why? How?

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3 Answers 3

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The matrix has at least a nonzero column, call it $u$. Then, having rank $1$, it must be of the form $$ \begin{bmatrix} \alpha_1 u & \alpha_2 u & \dots & \alpha_n u \end{bmatrix} $$ for scalars $\alpha_1,\alpha_2,\dots,\alpha_n$. Can you guess what $v$ should be?

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One way to see this is with SVD. This is overkill, though.

Another approach: suppose that $A$ is a rank $1$ matrix. Then $A$ has at least one non-zero row. Call this row $v^T$. Every row of $A$ must be a multiple of this row. That is, there exist coefficients $u_i$ such that the rows of $A$ are exactly $u_1v^T,u_2v^T,\dots u_nv^T$.

This is exactly the same as saying that $A= uv^T$, where $u$ is the (column-)vector $(u_1,\dots,u_n)$.

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Hint: Here's a rank-1 matrix: $$ \begin{bmatrix} 1 & 2 & 5 \\ 2 & 4 & 10 \\ 4 & 8 & 20 \\ 5 & 10 & 25 \end{bmatrix}. $$

Look closely at the columns and see if you notice anything about them. Then try to write this matrix in the form $u v^t$. Sub-hint: use the first column as $u$. What size will $v$ be?

Then generalize.

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