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Suppose you have a $n \times 1$ column vector

$$a=\begin{bmatrix}a_1\\{a_2}\\ \vdots\\{a_n}\end{bmatrix}$$

and a $1 \times m$ row vector

$$\quad b=\begin{bmatrix}b_1 & b_2 & \ldots & b_m\end{bmatrix}$$

If you then multiply these

$$A=ab=\begin{bmatrix} a_{1}b_1 & a_{1}b_2 & \ldots & a_{1}b_m\\ a_{2}b_1 & a_{2}b_2 & \ldots & a_{2}b_m \\ \vdots&&&\vdots \\ a_{n}b_1 & a_{n}b_2 & \ldots & a_{n}b_m\end{bmatrix}$$

How difficult (if possible) is it to find the original vectors $a$ and $b$ if you are given $A$?

2 Answers 2

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You cannot determine the original vectors $a$ and $b$, since they are not unique ($2a$ and $0.5b$ yield the same result as Rodrigo explains). However, you can construct one $a$ and $b$ such that $A=ab^T$. Just find any nonzero element in $A$. Let's assume that the (1,1) element is nonzero, but the method works in other cases too. The (1,1) element is $a_1 b_1$. Now let $a_1=1$, then you can compute $b_1$. By going through the first row and first column, you can compute all elements of the vectors $a$ and $b$.

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    So you would just need the first row and column to find a and b, because out of the assumption a1=1, you can find all b's and from all b's, you can't then find all a's? – 2017-01-26
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    Yes you can find all a's. Knowing $b1$ you can use the first column to find all a's. – 2017-01-26
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    Thanks! This problem has been going on in my mind all day yesterday, now I finally have the solution! – 2017-01-26
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Suppose I have vectors $\mathrm u \in \mathbb R^m$ and $\mathrm v \in \mathbb R^n$. I form the $m \times n$ rank-$1$ matrix

$$\mathrm A := \mathrm u \mathrm v^{\top}$$

and tell you what $\mathrm A$ is. Can you reconstruct vectors $\mathrm u$ and $\mathrm v$ from matrix $\mathrm A$? You cannot, because

$$\left( \gamma \mathrm u \right) \left( \frac{1}{\gamma} \mathrm v^{\top} \right) = \mathrm A$$

for all $\gamma \neq 0$. What you can recover from $\mathrm A$ is two lines passing through the origin and on which $\mathrm u$ and $\mathrm v$ live. More information is needed to recover $\mathrm u$ and $\mathrm v$.

For instance, if I tell you what $\| \mathrm u \|_2$ is, then you can find where the line whose direction vector is $\mathrm u$ intersects the Euclidean sphere of radius $\| \mathrm u \|_2$ centered at the origin. You find two points. One of them is $\mathrm u$ and the other is $-\mathrm u$, but you cannot know which is which, unfortunately. Sign ambiguity cannot be eliminated. Note that $(-\mathrm u) (-\mathrm v^{\top}) = \mathrm A$.

To summarize, matrix-valued function $\mathrm F (\mathrm x, \mathrm y) := \mathrm x \mathrm y^{\top}$ is not injective and, thus, the inputs $\mathrm x$ and $\mathrm y$ cannot be recovered from the output.